Median Filter Implementation In Python
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I implemented median filter in Python in order to remove the salt & pepper noise from the images. It is working fine and all but I would love to hear your advice or opinions.
def median_filter(data, filter_size):
temp =
indexer = filter_size // 2
for i in range(len(data)):
for j in range(len(data[0])):
for z in range(filter_size):
if i + z - indexer < 0 or i + z - indexer > len(data) - 1:
for c in range(filter_size):
temp.append(0)
else:
if j + z - indexer < 0 or j + indexer > len(data[0]) - 1:
temp.append(0)
else:
for k in range(filter_size):
temp.append(data[i + z - indexer][j + k - indexer])
temp.sort()
data[i][j] = temp[len(temp) // 2]
temp =
return data
python image iteration signal-processing
add a comment |Â
up vote
4
down vote
favorite
I implemented median filter in Python in order to remove the salt & pepper noise from the images. It is working fine and all but I would love to hear your advice or opinions.
def median_filter(data, filter_size):
temp =
indexer = filter_size // 2
for i in range(len(data)):
for j in range(len(data[0])):
for z in range(filter_size):
if i + z - indexer < 0 or i + z - indexer > len(data) - 1:
for c in range(filter_size):
temp.append(0)
else:
if j + z - indexer < 0 or j + indexer > len(data[0]) - 1:
temp.append(0)
else:
for k in range(filter_size):
temp.append(data[i + z - indexer][j + k - indexer])
temp.sort()
data[i][j] = temp[len(temp) // 2]
temp =
return data
python image iteration signal-processing
1
Your code doesn't make the complete sense to me. Could you explain a little what it does? Why do you loop throughfilter_size
on two of the three options in thefor z in range(filter_size)
loop?
â Peilonrayz
Apr 13 at 14:19
Sure, Median filter is usually used to reduce noise in an image. My code basically takes the array of the image which is corrupted by salt and pepper noise and remove the noise. I loop through "filter_size" because there are different sized median filters, like 3x3, 5x5. So there is more pixels that need to be considered. You can take a look at the GitHub repo, github.com/MeteHanC/Python-Median-Filter
â MeteHan
Apr 13 at 14:37
could you answer my second question? I don't really understand how you handle those median filters going out of bounds.
â Peilonrayz
Apr 13 at 14:40
I check the rows and columns going out of bounds with different if statements. So if a columns value is negative, I append zeros to the temp array. Adding zeros in the situation of an out bound, is a common approach. And if row value is not negative , I check the column value. And append zeros if out of bounds again. And about "i+z-indexer", with this way I put the current pixel in the hotspot(basically center of the array) and I check its surrounding pixels. For 3x3 median filter, lets say we got the [0][0] pixel. So program will check [-1][-1], [-1][0], [-1][1], [0][-1], [0][0], [0][1] so on
â MeteHan
Apr 13 at 14:47
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I implemented median filter in Python in order to remove the salt & pepper noise from the images. It is working fine and all but I would love to hear your advice or opinions.
def median_filter(data, filter_size):
temp =
indexer = filter_size // 2
for i in range(len(data)):
for j in range(len(data[0])):
for z in range(filter_size):
if i + z - indexer < 0 or i + z - indexer > len(data) - 1:
for c in range(filter_size):
temp.append(0)
else:
if j + z - indexer < 0 or j + indexer > len(data[0]) - 1:
temp.append(0)
else:
for k in range(filter_size):
temp.append(data[i + z - indexer][j + k - indexer])
temp.sort()
data[i][j] = temp[len(temp) // 2]
temp =
return data
python image iteration signal-processing
I implemented median filter in Python in order to remove the salt & pepper noise from the images. It is working fine and all but I would love to hear your advice or opinions.
def median_filter(data, filter_size):
temp =
indexer = filter_size // 2
for i in range(len(data)):
for j in range(len(data[0])):
for z in range(filter_size):
if i + z - indexer < 0 or i + z - indexer > len(data) - 1:
for c in range(filter_size):
temp.append(0)
else:
if j + z - indexer < 0 or j + indexer > len(data[0]) - 1:
temp.append(0)
else:
for k in range(filter_size):
temp.append(data[i + z - indexer][j + k - indexer])
temp.sort()
data[i][j] = temp[len(temp) // 2]
temp =
return data
python image iteration signal-processing
edited Apr 14 at 6:58
200_success
123k14142399
123k14142399
asked Apr 13 at 14:04
MeteHan
615
615
1
Your code doesn't make the complete sense to me. Could you explain a little what it does? Why do you loop throughfilter_size
on two of the three options in thefor z in range(filter_size)
loop?
â Peilonrayz
Apr 13 at 14:19
Sure, Median filter is usually used to reduce noise in an image. My code basically takes the array of the image which is corrupted by salt and pepper noise and remove the noise. I loop through "filter_size" because there are different sized median filters, like 3x3, 5x5. So there is more pixels that need to be considered. You can take a look at the GitHub repo, github.com/MeteHanC/Python-Median-Filter
â MeteHan
Apr 13 at 14:37
could you answer my second question? I don't really understand how you handle those median filters going out of bounds.
â Peilonrayz
Apr 13 at 14:40
I check the rows and columns going out of bounds with different if statements. So if a columns value is negative, I append zeros to the temp array. Adding zeros in the situation of an out bound, is a common approach. And if row value is not negative , I check the column value. And append zeros if out of bounds again. And about "i+z-indexer", with this way I put the current pixel in the hotspot(basically center of the array) and I check its surrounding pixels. For 3x3 median filter, lets say we got the [0][0] pixel. So program will check [-1][-1], [-1][0], [-1][1], [0][-1], [0][0], [0][1] so on
â MeteHan
Apr 13 at 14:47
add a comment |Â
1
Your code doesn't make the complete sense to me. Could you explain a little what it does? Why do you loop throughfilter_size
on two of the three options in thefor z in range(filter_size)
loop?
â Peilonrayz
Apr 13 at 14:19
Sure, Median filter is usually used to reduce noise in an image. My code basically takes the array of the image which is corrupted by salt and pepper noise and remove the noise. I loop through "filter_size" because there are different sized median filters, like 3x3, 5x5. So there is more pixels that need to be considered. You can take a look at the GitHub repo, github.com/MeteHanC/Python-Median-Filter
â MeteHan
Apr 13 at 14:37
could you answer my second question? I don't really understand how you handle those median filters going out of bounds.
â Peilonrayz
Apr 13 at 14:40
I check the rows and columns going out of bounds with different if statements. So if a columns value is negative, I append zeros to the temp array. Adding zeros in the situation of an out bound, is a common approach. And if row value is not negative , I check the column value. And append zeros if out of bounds again. And about "i+z-indexer", with this way I put the current pixel in the hotspot(basically center of the array) and I check its surrounding pixels. For 3x3 median filter, lets say we got the [0][0] pixel. So program will check [-1][-1], [-1][0], [-1][1], [0][-1], [0][0], [0][1] so on
â MeteHan
Apr 13 at 14:47
1
1
Your code doesn't make the complete sense to me. Could you explain a little what it does? Why do you loop through
filter_size
on two of the three options in the for z in range(filter_size)
loop?â Peilonrayz
Apr 13 at 14:19
Your code doesn't make the complete sense to me. Could you explain a little what it does? Why do you loop through
filter_size
on two of the three options in the for z in range(filter_size)
loop?â Peilonrayz
Apr 13 at 14:19
Sure, Median filter is usually used to reduce noise in an image. My code basically takes the array of the image which is corrupted by salt and pepper noise and remove the noise. I loop through "filter_size" because there are different sized median filters, like 3x3, 5x5. So there is more pixels that need to be considered. You can take a look at the GitHub repo, github.com/MeteHanC/Python-Median-Filter
â MeteHan
Apr 13 at 14:37
Sure, Median filter is usually used to reduce noise in an image. My code basically takes the array of the image which is corrupted by salt and pepper noise and remove the noise. I loop through "filter_size" because there are different sized median filters, like 3x3, 5x5. So there is more pixels that need to be considered. You can take a look at the GitHub repo, github.com/MeteHanC/Python-Median-Filter
â MeteHan
Apr 13 at 14:37
could you answer my second question? I don't really understand how you handle those median filters going out of bounds.
â Peilonrayz
Apr 13 at 14:40
could you answer my second question? I don't really understand how you handle those median filters going out of bounds.
â Peilonrayz
Apr 13 at 14:40
I check the rows and columns going out of bounds with different if statements. So if a columns value is negative, I append zeros to the temp array. Adding zeros in the situation of an out bound, is a common approach. And if row value is not negative , I check the column value. And append zeros if out of bounds again. And about "i+z-indexer", with this way I put the current pixel in the hotspot(basically center of the array) and I check its surrounding pixels. For 3x3 median filter, lets say we got the [0][0] pixel. So program will check [-1][-1], [-1][0], [-1][1], [0][-1], [0][0], [0][1] so on
â MeteHan
Apr 13 at 14:47
I check the rows and columns going out of bounds with different if statements. So if a columns value is negative, I append zeros to the temp array. Adding zeros in the situation of an out bound, is a common approach. And if row value is not negative , I check the column value. And append zeros if out of bounds again. And about "i+z-indexer", with this way I put the current pixel in the hotspot(basically center of the array) and I check its surrounding pixels. For 3x3 median filter, lets say we got the [0][0] pixel. So program will check [-1][-1], [-1][0], [-1][1], [0][-1], [0][0], [0][1] so on
â MeteHan
Apr 13 at 14:47
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
Code review
Peilonrays points out a mixup with the out-of-bounds testing that is valid. The statement if j ...
must be within the loop for k ...
. One of the results is that you add a different number of elements to temp
depending on which boundary you're at. But there are better ways to avoid out-of-bounds indexing, see below.
Your biggest bug, however, is that you write the result of the filter into the image you are processing. Median filtering cannot be done in-place. When you update data[i][j]
, you'll be reading the updated value to compute data[i][j+1]
. You need to allocate a new image, and write the result there.
I would suggest not adding zeros for out-of-bounds pixels at all, because it introduces a bias to the output pixels near the boundary. The clearest example is for the pixels close to any of the corners. At the corner pixel, with a 3x3 kernel, you'll have 4 image pixels covered by the kernel. Adding 5 zeros for the out-of-bounds pixels guarantees that the output will be 0. For larger kernels this happens in more pixels of course. Instead, it is easy to simply remove the temp.append(0)
statements, leading to a non-biased result. Other options are to read values from elsewhere in the image, for example mirroring the image at the boundary or extending the image by extrapolation. For median filtering this has little effect, IMO.
You set temp =
at the very beginning of your function, then reset it when you're done using it, in preparation for the next loop. Instead, initialize it once inside the main double-loop over the pixels:
for i in range(len(data)):
for j in range(len(data[0])):
temp =
# ...
You're looping over i
and j
as image indices, then over z
and c
or k
for filter kernel indices. c
and k
have the same function in two different loops, I would suggest using the same variable for that. z
doesn't really fit in with either c
or k
. I would pick two names that are related in the way that i
and j
are, such as m
and n
. The choice of variable names is always very limited if it's just one letter. Using longer names would make this code clearer: for example img_column
, img_row
, kernel_column
, kernel_row
.
Out-of-bounds checking
This concludes my comments on your code. Now I'd like to offer some alternatives for out-of-bounds checking. These tests are rather expensive when performed for every pixel -- it's a test that is done $n k$ times (with $n$ pixels in the image and $k$ pixels in the kernel). Maybe in Python the added cost is relatively small, it's an interpreted language after all, but for a compiled language these tests can easily amount to doubling processing time. There are 3 common alternatives that I know of. I will use border = filter_size // 2
, and presume filter_size
is odd. It is possible to adjust all 3 methods to even-sized filters.
Separate loops for image border pixels
The idea here is that the loop over the first and last border
pixels along each dimension are handled separately from the loop over the core of the image. This avoids all tests. But it does require some code duplication (all in the name of speed!).
for i in range(border):
# here we loop over the kernel from -i to border+1
for i in range(border, len(data)-border):
# here we loop over the full kernel
for i in range(len(data)-border, len(data)):
# here we loop over the kernel from -border to len(data)-i
Of course, within each of those loops, a similar set of 3 loops is necessary to loop over j
. The filter logic is thus repeated 9 times. In a compiled language, where this is the most efficient method, code duplication can be avoided with inlined functions or macros. I don't know how a Python function call compares to a bunch of tests for out-of-bounds access, so can't comment on the usefulness of this method in Python.
A separate code path for border pixels
The idea here is to do out-of-bounds checking only for those pixels that are close to the image boundary. For pixels within the border, you use a version of the filtering logic with out-of-bounds checking. For the pixels in the core of the image (which is the big majority of pixels), you use a second version of the logic without out-of-bounds checking.
for i in range(len(data)):
i_border = i < border or i >= len(data)-border
for j in range(len(data[0])):
j_border = j < border or j >= len(data)-border
if i_border or j_border:
# filtering with bounds checking
else:
# filtering without bounds checking
Padding the image
The simplest solution, and also the most flexible one, is to create a temporary image that is larger than the input image by 2*border
along each dimension, and copy the input image into it. The "new" pixels can be filled with zeros (to replicate what OP intended to do), or with values taken from the input image (for example by mirroring the image at the boundary or extrapolating in some other way).
The filter now never needs to check for out-of-bounds reads. When the filtering kernel is placed over any of the input image pixels, all samples fall within the padded image.
Since for this type of filtering it is necessary to create a new output image anyway (it is not possible to compute it in-place, as I mentioned before), this is not a huge cost: the original input image can now be re-used as output image.
This solution leads to the simplest code, allows for all sorts of boundary extension methods without complicating the filtering code, and often results in the fastest code too.
add a comment |Â
up vote
3
down vote
You seem to have a few bugs.
if i + z - indexer < 0 or i + z - indexer > len(data) - 1:
If
i
andz
are0
, whereindexer
is 1, then you'll have0 + 0 - 1 < 0
. This would mean that you'd replace the data in(-1, j)
,(0, j)
and(1, j)
to 0. Since 0 and 1 probably do contain data this is just plain wrong.if j + z - indexer < 0 or j + indexer > len(data[0]) - 1:
temp.append(0)This removes some data, meaning that the median is shifted. Say you should have
(0, 0, 0, 1, 2, 3)
, however you removed the first three because of this you'd have(0, 1, 2, 3)
. Now the median is1
rather than0
.
Your code would be simpler if you:
- Made a window list, that contained all the indexes that you want to move to.
- Have an if to check if the data in that index is out of bounds.
- If it's out of bounds default to 0.
- If it's not out of bounds use the data.
This could become:
def median_filter(data, filter_size):
temp =
indexer = filter_size // 2
window = [
(i, j)
for i in range(-indexer, filter_size-indexer)
for j in range(-indexer, filter_size-indexer)
]
index = len(window) // 2
for i in range(len(data)):
for j in range(len(data[0])):
data[i][j] = sorted(
0 if (
min(i+a, j+b) < 0
or len(data) <= i+a
or len(data[0]) <= j+b
) else data[i+a][j+b]
for a, b in window
)[index]
return data
That is the whole point, replacing those values with 0. As I said adding zeros, it is very common approach. Probably I couldnt make myself clear to you in the comment section, sorry about that Thank you for your comments
â MeteHan
Apr 13 at 15:15
@M.Han Please read my answer again. You're changing too many values to black. If changing valid data to black was 'the whole point', then you'd want to just change the code todata[i][j] = 0
...
â Peilonrayz
Apr 13 at 15:17
@M.Han Is(0, 0)
meant to be replaced to black?
â Peilonrayz
Apr 13 at 15:23
The window list not only makes the code simpler, it also allows for arbitrary window shapes. A circular window has much nicer properties than a square one. It is easy to implement that with this code. Nice!
â Cris Luengo
Apr 14 at 6:36
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Code review
Peilonrays points out a mixup with the out-of-bounds testing that is valid. The statement if j ...
must be within the loop for k ...
. One of the results is that you add a different number of elements to temp
depending on which boundary you're at. But there are better ways to avoid out-of-bounds indexing, see below.
Your biggest bug, however, is that you write the result of the filter into the image you are processing. Median filtering cannot be done in-place. When you update data[i][j]
, you'll be reading the updated value to compute data[i][j+1]
. You need to allocate a new image, and write the result there.
I would suggest not adding zeros for out-of-bounds pixels at all, because it introduces a bias to the output pixels near the boundary. The clearest example is for the pixels close to any of the corners. At the corner pixel, with a 3x3 kernel, you'll have 4 image pixels covered by the kernel. Adding 5 zeros for the out-of-bounds pixels guarantees that the output will be 0. For larger kernels this happens in more pixels of course. Instead, it is easy to simply remove the temp.append(0)
statements, leading to a non-biased result. Other options are to read values from elsewhere in the image, for example mirroring the image at the boundary or extending the image by extrapolation. For median filtering this has little effect, IMO.
You set temp =
at the very beginning of your function, then reset it when you're done using it, in preparation for the next loop. Instead, initialize it once inside the main double-loop over the pixels:
for i in range(len(data)):
for j in range(len(data[0])):
temp =
# ...
You're looping over i
and j
as image indices, then over z
and c
or k
for filter kernel indices. c
and k
have the same function in two different loops, I would suggest using the same variable for that. z
doesn't really fit in with either c
or k
. I would pick two names that are related in the way that i
and j
are, such as m
and n
. The choice of variable names is always very limited if it's just one letter. Using longer names would make this code clearer: for example img_column
, img_row
, kernel_column
, kernel_row
.
Out-of-bounds checking
This concludes my comments on your code. Now I'd like to offer some alternatives for out-of-bounds checking. These tests are rather expensive when performed for every pixel -- it's a test that is done $n k$ times (with $n$ pixels in the image and $k$ pixels in the kernel). Maybe in Python the added cost is relatively small, it's an interpreted language after all, but for a compiled language these tests can easily amount to doubling processing time. There are 3 common alternatives that I know of. I will use border = filter_size // 2
, and presume filter_size
is odd. It is possible to adjust all 3 methods to even-sized filters.
Separate loops for image border pixels
The idea here is that the loop over the first and last border
pixels along each dimension are handled separately from the loop over the core of the image. This avoids all tests. But it does require some code duplication (all in the name of speed!).
for i in range(border):
# here we loop over the kernel from -i to border+1
for i in range(border, len(data)-border):
# here we loop over the full kernel
for i in range(len(data)-border, len(data)):
# here we loop over the kernel from -border to len(data)-i
Of course, within each of those loops, a similar set of 3 loops is necessary to loop over j
. The filter logic is thus repeated 9 times. In a compiled language, where this is the most efficient method, code duplication can be avoided with inlined functions or macros. I don't know how a Python function call compares to a bunch of tests for out-of-bounds access, so can't comment on the usefulness of this method in Python.
A separate code path for border pixels
The idea here is to do out-of-bounds checking only for those pixels that are close to the image boundary. For pixels within the border, you use a version of the filtering logic with out-of-bounds checking. For the pixels in the core of the image (which is the big majority of pixels), you use a second version of the logic without out-of-bounds checking.
for i in range(len(data)):
i_border = i < border or i >= len(data)-border
for j in range(len(data[0])):
j_border = j < border or j >= len(data)-border
if i_border or j_border:
# filtering with bounds checking
else:
# filtering without bounds checking
Padding the image
The simplest solution, and also the most flexible one, is to create a temporary image that is larger than the input image by 2*border
along each dimension, and copy the input image into it. The "new" pixels can be filled with zeros (to replicate what OP intended to do), or with values taken from the input image (for example by mirroring the image at the boundary or extrapolating in some other way).
The filter now never needs to check for out-of-bounds reads. When the filtering kernel is placed over any of the input image pixels, all samples fall within the padded image.
Since for this type of filtering it is necessary to create a new output image anyway (it is not possible to compute it in-place, as I mentioned before), this is not a huge cost: the original input image can now be re-used as output image.
This solution leads to the simplest code, allows for all sorts of boundary extension methods without complicating the filtering code, and often results in the fastest code too.
add a comment |Â
up vote
3
down vote
accepted
Code review
Peilonrays points out a mixup with the out-of-bounds testing that is valid. The statement if j ...
must be within the loop for k ...
. One of the results is that you add a different number of elements to temp
depending on which boundary you're at. But there are better ways to avoid out-of-bounds indexing, see below.
Your biggest bug, however, is that you write the result of the filter into the image you are processing. Median filtering cannot be done in-place. When you update data[i][j]
, you'll be reading the updated value to compute data[i][j+1]
. You need to allocate a new image, and write the result there.
I would suggest not adding zeros for out-of-bounds pixels at all, because it introduces a bias to the output pixels near the boundary. The clearest example is for the pixels close to any of the corners. At the corner pixel, with a 3x3 kernel, you'll have 4 image pixels covered by the kernel. Adding 5 zeros for the out-of-bounds pixels guarantees that the output will be 0. For larger kernels this happens in more pixels of course. Instead, it is easy to simply remove the temp.append(0)
statements, leading to a non-biased result. Other options are to read values from elsewhere in the image, for example mirroring the image at the boundary or extending the image by extrapolation. For median filtering this has little effect, IMO.
You set temp =
at the very beginning of your function, then reset it when you're done using it, in preparation for the next loop. Instead, initialize it once inside the main double-loop over the pixels:
for i in range(len(data)):
for j in range(len(data[0])):
temp =
# ...
You're looping over i
and j
as image indices, then over z
and c
or k
for filter kernel indices. c
and k
have the same function in two different loops, I would suggest using the same variable for that. z
doesn't really fit in with either c
or k
. I would pick two names that are related in the way that i
and j
are, such as m
and n
. The choice of variable names is always very limited if it's just one letter. Using longer names would make this code clearer: for example img_column
, img_row
, kernel_column
, kernel_row
.
Out-of-bounds checking
This concludes my comments on your code. Now I'd like to offer some alternatives for out-of-bounds checking. These tests are rather expensive when performed for every pixel -- it's a test that is done $n k$ times (with $n$ pixels in the image and $k$ pixels in the kernel). Maybe in Python the added cost is relatively small, it's an interpreted language after all, but for a compiled language these tests can easily amount to doubling processing time. There are 3 common alternatives that I know of. I will use border = filter_size // 2
, and presume filter_size
is odd. It is possible to adjust all 3 methods to even-sized filters.
Separate loops for image border pixels
The idea here is that the loop over the first and last border
pixels along each dimension are handled separately from the loop over the core of the image. This avoids all tests. But it does require some code duplication (all in the name of speed!).
for i in range(border):
# here we loop over the kernel from -i to border+1
for i in range(border, len(data)-border):
# here we loop over the full kernel
for i in range(len(data)-border, len(data)):
# here we loop over the kernel from -border to len(data)-i
Of course, within each of those loops, a similar set of 3 loops is necessary to loop over j
. The filter logic is thus repeated 9 times. In a compiled language, where this is the most efficient method, code duplication can be avoided with inlined functions or macros. I don't know how a Python function call compares to a bunch of tests for out-of-bounds access, so can't comment on the usefulness of this method in Python.
A separate code path for border pixels
The idea here is to do out-of-bounds checking only for those pixels that are close to the image boundary. For pixels within the border, you use a version of the filtering logic with out-of-bounds checking. For the pixels in the core of the image (which is the big majority of pixels), you use a second version of the logic without out-of-bounds checking.
for i in range(len(data)):
i_border = i < border or i >= len(data)-border
for j in range(len(data[0])):
j_border = j < border or j >= len(data)-border
if i_border or j_border:
# filtering with bounds checking
else:
# filtering without bounds checking
Padding the image
The simplest solution, and also the most flexible one, is to create a temporary image that is larger than the input image by 2*border
along each dimension, and copy the input image into it. The "new" pixels can be filled with zeros (to replicate what OP intended to do), or with values taken from the input image (for example by mirroring the image at the boundary or extrapolating in some other way).
The filter now never needs to check for out-of-bounds reads. When the filtering kernel is placed over any of the input image pixels, all samples fall within the padded image.
Since for this type of filtering it is necessary to create a new output image anyway (it is not possible to compute it in-place, as I mentioned before), this is not a huge cost: the original input image can now be re-used as output image.
This solution leads to the simplest code, allows for all sorts of boundary extension methods without complicating the filtering code, and often results in the fastest code too.
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Code review
Peilonrays points out a mixup with the out-of-bounds testing that is valid. The statement if j ...
must be within the loop for k ...
. One of the results is that you add a different number of elements to temp
depending on which boundary you're at. But there are better ways to avoid out-of-bounds indexing, see below.
Your biggest bug, however, is that you write the result of the filter into the image you are processing. Median filtering cannot be done in-place. When you update data[i][j]
, you'll be reading the updated value to compute data[i][j+1]
. You need to allocate a new image, and write the result there.
I would suggest not adding zeros for out-of-bounds pixels at all, because it introduces a bias to the output pixels near the boundary. The clearest example is for the pixels close to any of the corners. At the corner pixel, with a 3x3 kernel, you'll have 4 image pixels covered by the kernel. Adding 5 zeros for the out-of-bounds pixels guarantees that the output will be 0. For larger kernels this happens in more pixels of course. Instead, it is easy to simply remove the temp.append(0)
statements, leading to a non-biased result. Other options are to read values from elsewhere in the image, for example mirroring the image at the boundary or extending the image by extrapolation. For median filtering this has little effect, IMO.
You set temp =
at the very beginning of your function, then reset it when you're done using it, in preparation for the next loop. Instead, initialize it once inside the main double-loop over the pixels:
for i in range(len(data)):
for j in range(len(data[0])):
temp =
# ...
You're looping over i
and j
as image indices, then over z
and c
or k
for filter kernel indices. c
and k
have the same function in two different loops, I would suggest using the same variable for that. z
doesn't really fit in with either c
or k
. I would pick two names that are related in the way that i
and j
are, such as m
and n
. The choice of variable names is always very limited if it's just one letter. Using longer names would make this code clearer: for example img_column
, img_row
, kernel_column
, kernel_row
.
Out-of-bounds checking
This concludes my comments on your code. Now I'd like to offer some alternatives for out-of-bounds checking. These tests are rather expensive when performed for every pixel -- it's a test that is done $n k$ times (with $n$ pixels in the image and $k$ pixels in the kernel). Maybe in Python the added cost is relatively small, it's an interpreted language after all, but for a compiled language these tests can easily amount to doubling processing time. There are 3 common alternatives that I know of. I will use border = filter_size // 2
, and presume filter_size
is odd. It is possible to adjust all 3 methods to even-sized filters.
Separate loops for image border pixels
The idea here is that the loop over the first and last border
pixels along each dimension are handled separately from the loop over the core of the image. This avoids all tests. But it does require some code duplication (all in the name of speed!).
for i in range(border):
# here we loop over the kernel from -i to border+1
for i in range(border, len(data)-border):
# here we loop over the full kernel
for i in range(len(data)-border, len(data)):
# here we loop over the kernel from -border to len(data)-i
Of course, within each of those loops, a similar set of 3 loops is necessary to loop over j
. The filter logic is thus repeated 9 times. In a compiled language, where this is the most efficient method, code duplication can be avoided with inlined functions or macros. I don't know how a Python function call compares to a bunch of tests for out-of-bounds access, so can't comment on the usefulness of this method in Python.
A separate code path for border pixels
The idea here is to do out-of-bounds checking only for those pixels that are close to the image boundary. For pixels within the border, you use a version of the filtering logic with out-of-bounds checking. For the pixels in the core of the image (which is the big majority of pixels), you use a second version of the logic without out-of-bounds checking.
for i in range(len(data)):
i_border = i < border or i >= len(data)-border
for j in range(len(data[0])):
j_border = j < border or j >= len(data)-border
if i_border or j_border:
# filtering with bounds checking
else:
# filtering without bounds checking
Padding the image
The simplest solution, and also the most flexible one, is to create a temporary image that is larger than the input image by 2*border
along each dimension, and copy the input image into it. The "new" pixels can be filled with zeros (to replicate what OP intended to do), or with values taken from the input image (for example by mirroring the image at the boundary or extrapolating in some other way).
The filter now never needs to check for out-of-bounds reads. When the filtering kernel is placed over any of the input image pixels, all samples fall within the padded image.
Since for this type of filtering it is necessary to create a new output image anyway (it is not possible to compute it in-place, as I mentioned before), this is not a huge cost: the original input image can now be re-used as output image.
This solution leads to the simplest code, allows for all sorts of boundary extension methods without complicating the filtering code, and often results in the fastest code too.
Code review
Peilonrays points out a mixup with the out-of-bounds testing that is valid. The statement if j ...
must be within the loop for k ...
. One of the results is that you add a different number of elements to temp
depending on which boundary you're at. But there are better ways to avoid out-of-bounds indexing, see below.
Your biggest bug, however, is that you write the result of the filter into the image you are processing. Median filtering cannot be done in-place. When you update data[i][j]
, you'll be reading the updated value to compute data[i][j+1]
. You need to allocate a new image, and write the result there.
I would suggest not adding zeros for out-of-bounds pixels at all, because it introduces a bias to the output pixels near the boundary. The clearest example is for the pixels close to any of the corners. At the corner pixel, with a 3x3 kernel, you'll have 4 image pixels covered by the kernel. Adding 5 zeros for the out-of-bounds pixels guarantees that the output will be 0. For larger kernels this happens in more pixels of course. Instead, it is easy to simply remove the temp.append(0)
statements, leading to a non-biased result. Other options are to read values from elsewhere in the image, for example mirroring the image at the boundary or extending the image by extrapolation. For median filtering this has little effect, IMO.
You set temp =
at the very beginning of your function, then reset it when you're done using it, in preparation for the next loop. Instead, initialize it once inside the main double-loop over the pixels:
for i in range(len(data)):
for j in range(len(data[0])):
temp =
# ...
You're looping over i
and j
as image indices, then over z
and c
or k
for filter kernel indices. c
and k
have the same function in two different loops, I would suggest using the same variable for that. z
doesn't really fit in with either c
or k
. I would pick two names that are related in the way that i
and j
are, such as m
and n
. The choice of variable names is always very limited if it's just one letter. Using longer names would make this code clearer: for example img_column
, img_row
, kernel_column
, kernel_row
.
Out-of-bounds checking
This concludes my comments on your code. Now I'd like to offer some alternatives for out-of-bounds checking. These tests are rather expensive when performed for every pixel -- it's a test that is done $n k$ times (with $n$ pixels in the image and $k$ pixels in the kernel). Maybe in Python the added cost is relatively small, it's an interpreted language after all, but for a compiled language these tests can easily amount to doubling processing time. There are 3 common alternatives that I know of. I will use border = filter_size // 2
, and presume filter_size
is odd. It is possible to adjust all 3 methods to even-sized filters.
Separate loops for image border pixels
The idea here is that the loop over the first and last border
pixels along each dimension are handled separately from the loop over the core of the image. This avoids all tests. But it does require some code duplication (all in the name of speed!).
for i in range(border):
# here we loop over the kernel from -i to border+1
for i in range(border, len(data)-border):
# here we loop over the full kernel
for i in range(len(data)-border, len(data)):
# here we loop over the kernel from -border to len(data)-i
Of course, within each of those loops, a similar set of 3 loops is necessary to loop over j
. The filter logic is thus repeated 9 times. In a compiled language, where this is the most efficient method, code duplication can be avoided with inlined functions or macros. I don't know how a Python function call compares to a bunch of tests for out-of-bounds access, so can't comment on the usefulness of this method in Python.
A separate code path for border pixels
The idea here is to do out-of-bounds checking only for those pixels that are close to the image boundary. For pixels within the border, you use a version of the filtering logic with out-of-bounds checking. For the pixels in the core of the image (which is the big majority of pixels), you use a second version of the logic without out-of-bounds checking.
for i in range(len(data)):
i_border = i < border or i >= len(data)-border
for j in range(len(data[0])):
j_border = j < border or j >= len(data)-border
if i_border or j_border:
# filtering with bounds checking
else:
# filtering without bounds checking
Padding the image
The simplest solution, and also the most flexible one, is to create a temporary image that is larger than the input image by 2*border
along each dimension, and copy the input image into it. The "new" pixels can be filled with zeros (to replicate what OP intended to do), or with values taken from the input image (for example by mirroring the image at the boundary or extrapolating in some other way).
The filter now never needs to check for out-of-bounds reads. When the filtering kernel is placed over any of the input image pixels, all samples fall within the padded image.
Since for this type of filtering it is necessary to create a new output image anyway (it is not possible to compute it in-place, as I mentioned before), this is not a huge cost: the original input image can now be re-used as output image.
This solution leads to the simplest code, allows for all sorts of boundary extension methods without complicating the filtering code, and often results in the fastest code too.
answered Apr 14 at 6:35
Cris Luengo
1,877215
1,877215
add a comment |Â
add a comment |Â
up vote
3
down vote
You seem to have a few bugs.
if i + z - indexer < 0 or i + z - indexer > len(data) - 1:
If
i
andz
are0
, whereindexer
is 1, then you'll have0 + 0 - 1 < 0
. This would mean that you'd replace the data in(-1, j)
,(0, j)
and(1, j)
to 0. Since 0 and 1 probably do contain data this is just plain wrong.if j + z - indexer < 0 or j + indexer > len(data[0]) - 1:
temp.append(0)This removes some data, meaning that the median is shifted. Say you should have
(0, 0, 0, 1, 2, 3)
, however you removed the first three because of this you'd have(0, 1, 2, 3)
. Now the median is1
rather than0
.
Your code would be simpler if you:
- Made a window list, that contained all the indexes that you want to move to.
- Have an if to check if the data in that index is out of bounds.
- If it's out of bounds default to 0.
- If it's not out of bounds use the data.
This could become:
def median_filter(data, filter_size):
temp =
indexer = filter_size // 2
window = [
(i, j)
for i in range(-indexer, filter_size-indexer)
for j in range(-indexer, filter_size-indexer)
]
index = len(window) // 2
for i in range(len(data)):
for j in range(len(data[0])):
data[i][j] = sorted(
0 if (
min(i+a, j+b) < 0
or len(data) <= i+a
or len(data[0]) <= j+b
) else data[i+a][j+b]
for a, b in window
)[index]
return data
That is the whole point, replacing those values with 0. As I said adding zeros, it is very common approach. Probably I couldnt make myself clear to you in the comment section, sorry about that Thank you for your comments
â MeteHan
Apr 13 at 15:15
@M.Han Please read my answer again. You're changing too many values to black. If changing valid data to black was 'the whole point', then you'd want to just change the code todata[i][j] = 0
...
â Peilonrayz
Apr 13 at 15:17
@M.Han Is(0, 0)
meant to be replaced to black?
â Peilonrayz
Apr 13 at 15:23
The window list not only makes the code simpler, it also allows for arbitrary window shapes. A circular window has much nicer properties than a square one. It is easy to implement that with this code. Nice!
â Cris Luengo
Apr 14 at 6:36
add a comment |Â
up vote
3
down vote
You seem to have a few bugs.
if i + z - indexer < 0 or i + z - indexer > len(data) - 1:
If
i
andz
are0
, whereindexer
is 1, then you'll have0 + 0 - 1 < 0
. This would mean that you'd replace the data in(-1, j)
,(0, j)
and(1, j)
to 0. Since 0 and 1 probably do contain data this is just plain wrong.if j + z - indexer < 0 or j + indexer > len(data[0]) - 1:
temp.append(0)This removes some data, meaning that the median is shifted. Say you should have
(0, 0, 0, 1, 2, 3)
, however you removed the first three because of this you'd have(0, 1, 2, 3)
. Now the median is1
rather than0
.
Your code would be simpler if you:
- Made a window list, that contained all the indexes that you want to move to.
- Have an if to check if the data in that index is out of bounds.
- If it's out of bounds default to 0.
- If it's not out of bounds use the data.
This could become:
def median_filter(data, filter_size):
temp =
indexer = filter_size // 2
window = [
(i, j)
for i in range(-indexer, filter_size-indexer)
for j in range(-indexer, filter_size-indexer)
]
index = len(window) // 2
for i in range(len(data)):
for j in range(len(data[0])):
data[i][j] = sorted(
0 if (
min(i+a, j+b) < 0
or len(data) <= i+a
or len(data[0]) <= j+b
) else data[i+a][j+b]
for a, b in window
)[index]
return data
That is the whole point, replacing those values with 0. As I said adding zeros, it is very common approach. Probably I couldnt make myself clear to you in the comment section, sorry about that Thank you for your comments
â MeteHan
Apr 13 at 15:15
@M.Han Please read my answer again. You're changing too many values to black. If changing valid data to black was 'the whole point', then you'd want to just change the code todata[i][j] = 0
...
â Peilonrayz
Apr 13 at 15:17
@M.Han Is(0, 0)
meant to be replaced to black?
â Peilonrayz
Apr 13 at 15:23
The window list not only makes the code simpler, it also allows for arbitrary window shapes. A circular window has much nicer properties than a square one. It is easy to implement that with this code. Nice!
â Cris Luengo
Apr 14 at 6:36
add a comment |Â
up vote
3
down vote
up vote
3
down vote
You seem to have a few bugs.
if i + z - indexer < 0 or i + z - indexer > len(data) - 1:
If
i
andz
are0
, whereindexer
is 1, then you'll have0 + 0 - 1 < 0
. This would mean that you'd replace the data in(-1, j)
,(0, j)
and(1, j)
to 0. Since 0 and 1 probably do contain data this is just plain wrong.if j + z - indexer < 0 or j + indexer > len(data[0]) - 1:
temp.append(0)This removes some data, meaning that the median is shifted. Say you should have
(0, 0, 0, 1, 2, 3)
, however you removed the first three because of this you'd have(0, 1, 2, 3)
. Now the median is1
rather than0
.
Your code would be simpler if you:
- Made a window list, that contained all the indexes that you want to move to.
- Have an if to check if the data in that index is out of bounds.
- If it's out of bounds default to 0.
- If it's not out of bounds use the data.
This could become:
def median_filter(data, filter_size):
temp =
indexer = filter_size // 2
window = [
(i, j)
for i in range(-indexer, filter_size-indexer)
for j in range(-indexer, filter_size-indexer)
]
index = len(window) // 2
for i in range(len(data)):
for j in range(len(data[0])):
data[i][j] = sorted(
0 if (
min(i+a, j+b) < 0
or len(data) <= i+a
or len(data[0]) <= j+b
) else data[i+a][j+b]
for a, b in window
)[index]
return data
You seem to have a few bugs.
if i + z - indexer < 0 or i + z - indexer > len(data) - 1:
If
i
andz
are0
, whereindexer
is 1, then you'll have0 + 0 - 1 < 0
. This would mean that you'd replace the data in(-1, j)
,(0, j)
and(1, j)
to 0. Since 0 and 1 probably do contain data this is just plain wrong.if j + z - indexer < 0 or j + indexer > len(data[0]) - 1:
temp.append(0)This removes some data, meaning that the median is shifted. Say you should have
(0, 0, 0, 1, 2, 3)
, however you removed the first three because of this you'd have(0, 1, 2, 3)
. Now the median is1
rather than0
.
Your code would be simpler if you:
- Made a window list, that contained all the indexes that you want to move to.
- Have an if to check if the data in that index is out of bounds.
- If it's out of bounds default to 0.
- If it's not out of bounds use the data.
This could become:
def median_filter(data, filter_size):
temp =
indexer = filter_size // 2
window = [
(i, j)
for i in range(-indexer, filter_size-indexer)
for j in range(-indexer, filter_size-indexer)
]
index = len(window) // 2
for i in range(len(data)):
for j in range(len(data[0])):
data[i][j] = sorted(
0 if (
min(i+a, j+b) < 0
or len(data) <= i+a
or len(data[0]) <= j+b
) else data[i+a][j+b]
for a, b in window
)[index]
return data
answered Apr 13 at 15:11
Peilonrayz
24.3k336102
24.3k336102
That is the whole point, replacing those values with 0. As I said adding zeros, it is very common approach. Probably I couldnt make myself clear to you in the comment section, sorry about that Thank you for your comments
â MeteHan
Apr 13 at 15:15
@M.Han Please read my answer again. You're changing too many values to black. If changing valid data to black was 'the whole point', then you'd want to just change the code todata[i][j] = 0
...
â Peilonrayz
Apr 13 at 15:17
@M.Han Is(0, 0)
meant to be replaced to black?
â Peilonrayz
Apr 13 at 15:23
The window list not only makes the code simpler, it also allows for arbitrary window shapes. A circular window has much nicer properties than a square one. It is easy to implement that with this code. Nice!
â Cris Luengo
Apr 14 at 6:36
add a comment |Â
That is the whole point, replacing those values with 0. As I said adding zeros, it is very common approach. Probably I couldnt make myself clear to you in the comment section, sorry about that Thank you for your comments
â MeteHan
Apr 13 at 15:15
@M.Han Please read my answer again. You're changing too many values to black. If changing valid data to black was 'the whole point', then you'd want to just change the code todata[i][j] = 0
...
â Peilonrayz
Apr 13 at 15:17
@M.Han Is(0, 0)
meant to be replaced to black?
â Peilonrayz
Apr 13 at 15:23
The window list not only makes the code simpler, it also allows for arbitrary window shapes. A circular window has much nicer properties than a square one. It is easy to implement that with this code. Nice!
â Cris Luengo
Apr 14 at 6:36
That is the whole point, replacing those values with 0. As I said adding zeros, it is very common approach. Probably I couldnt make myself clear to you in the comment section, sorry about that Thank you for your comments
â MeteHan
Apr 13 at 15:15
That is the whole point, replacing those values with 0. As I said adding zeros, it is very common approach. Probably I couldnt make myself clear to you in the comment section, sorry about that Thank you for your comments
â MeteHan
Apr 13 at 15:15
@M.Han Please read my answer again. You're changing too many values to black. If changing valid data to black was 'the whole point', then you'd want to just change the code to
data[i][j] = 0
...â Peilonrayz
Apr 13 at 15:17
@M.Han Please read my answer again. You're changing too many values to black. If changing valid data to black was 'the whole point', then you'd want to just change the code to
data[i][j] = 0
...â Peilonrayz
Apr 13 at 15:17
@M.Han Is
(0, 0)
meant to be replaced to black?â Peilonrayz
Apr 13 at 15:23
@M.Han Is
(0, 0)
meant to be replaced to black?â Peilonrayz
Apr 13 at 15:23
The window list not only makes the code simpler, it also allows for arbitrary window shapes. A circular window has much nicer properties than a square one. It is easy to implement that with this code. Nice!
â Cris Luengo
Apr 14 at 6:36
The window list not only makes the code simpler, it also allows for arbitrary window shapes. A circular window has much nicer properties than a square one. It is easy to implement that with this code. Nice!
â Cris Luengo
Apr 14 at 6:36
add a comment |Â
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1
Your code doesn't make the complete sense to me. Could you explain a little what it does? Why do you loop through
filter_size
on two of the three options in thefor z in range(filter_size)
loop?â Peilonrayz
Apr 13 at 14:19
Sure, Median filter is usually used to reduce noise in an image. My code basically takes the array of the image which is corrupted by salt and pepper noise and remove the noise. I loop through "filter_size" because there are different sized median filters, like 3x3, 5x5. So there is more pixels that need to be considered. You can take a look at the GitHub repo, github.com/MeteHanC/Python-Median-Filter
â MeteHan
Apr 13 at 14:37
could you answer my second question? I don't really understand how you handle those median filters going out of bounds.
â Peilonrayz
Apr 13 at 14:40
I check the rows and columns going out of bounds with different if statements. So if a columns value is negative, I append zeros to the temp array. Adding zeros in the situation of an out bound, is a common approach. And if row value is not negative , I check the column value. And append zeros if out of bounds again. And about "i+z-indexer", with this way I put the current pixel in the hotspot(basically center of the array) and I check its surrounding pixels. For 3x3 median filter, lets say we got the [0][0] pixel. So program will check [-1][-1], [-1][0], [-1][1], [0][-1], [0][0], [0][1] so on
â MeteHan
Apr 13 at 14:47