What's the right way to normalize a convolution filter?
I'm following these instructions. But my results aren't coming out right. I'm using this kernel:
-1 -1 -1
-1 8 -1
-1 -1 -1
So that means the sum for a given convolution will be between -8 and 8, assuming I have already normalized my input (0-255 -> 0-1). Then I do the convolution. Then I find the percent my value is between the minimum and maximum values. For example, with this kernel my min is -8 and max is 8. So if the value is 0 that's 50% which works out to 255 * .5 = 127.5. But that's clearly not right and what it gives is a mostly gray image. The parts that aren't gray are still monochrome even though I'm running the kernel on each channel individually.
static int EvaluateKernelAtPoint(Bitmap bitmap, Matrix<double> kernel, int x, int y, Func<int, int, int> onGetIntensity)
{
double sum = 0;
for (int a = 0; a < kernel.ColumnCount; a++)
{
for (int b = 0; b < kernel.RowCount; b++)
{
var xn = x + a - 1;
var yn = y + b - 1;
var intensity = (double)onGetIntensity(xn, yn); // returns R,G, or B color channel at that pixel
intensity /= 255; // intensity is 0-1
sum += intensity * kernel.At(a, b);
}
}
var result = (sum - (-8d)) / (8d - (-8d)); // find the % between the min and max of -8 and 8
result *= 255; // bring it back to 0-255
}
image-processing edge-detection
asked Nov 26 '18 at 1:42
user875234user875234
6752820
add a comment |
I'm following these instructions. But my results aren't coming out right. I'm using this kernel:
-1 -1 -1
-1 8 -1
-1 -1 -1
So that means the sum for a given convolution will be between -8 and 8, assuming I have already normalized my input (0-255 -> 0-1). Then I do the convolution. Then I find the percent my value is between the minimum and maximum values. For example, with this kernel my min is -8 and max is 8. So if the value is 0 that's 50% which works out to 255 * .5 = 127.5. But that's clearly not right and what it gives is a mostly gray image. The parts that aren't gray are still monochrome even though I'm running the kernel on each channel individually.
static int EvaluateKernelAtPoint(Bitmap bitmap, Matrix<double> kernel, int x, int y, Func<int, int, int> onGetIntensity)
{
double sum = 0;
for (int a = 0; a < kernel.ColumnCount; a++)
{
for (int b = 0; b < kernel.RowCount; b++)
{
var xn = x + a - 1;
var yn = y + b - 1;
var intensity = (double)onGetIntensity(xn, yn); // returns R,G, or B color channel at that pixel
intensity /= 255; // intensity is 0-1
sum += intensity * kernel.At(a, b);
}
}
var result = (sum - (-8d)) / (8d - (-8d)); // find the % between the min and max of -8 and 8
result *= 255; // bring it back to 0-255
}
image-processing edge-detection
asked Nov 26 '18 at 1:42
user875234user875234
6752820
1
Those extremes are extremely unlikely. You indeed want to set the zero to 127 or 128, then you can choose a scaling that will make your image look good. One approach is to find the min and max in the output image.
– Cris Luengo
Nov 26 '18 at 1:50
1
"The parts that aren't gray are still monochrome": don't expect the Laplacian output to show meaningful colors. Also, applying different gain coefficients to the three channels is not recommended as it introduces a bias.
– Yves Daoust
Nov 26 '18 at 7:07
add a comment |
I'm following these instructions. But my results aren't coming out right. I'm using this kernel:
-1 -1 -1
-1 8 -1
-1 -1 -1
So that means the sum for a given convolution will be between -8 and 8, assuming I have already normalized my input (0-255 -> 0-1). Then I do the convolution. Then I find the percent my value is between the minimum and maximum values. For example, with this kernel my min is -8 and max is 8. So if the value is 0 that's 50% which works out to 255 * .5 = 127.5. But that's clearly not right and what it gives is a mostly gray image. The parts that aren't gray are still monochrome even though I'm running the kernel on each channel individually.
static int EvaluateKernelAtPoint(Bitmap bitmap, Matrix<double> kernel, int x, int y, Func<int, int, int> onGetIntensity)
{
double sum = 0;
for (int a = 0; a < kernel.ColumnCount; a++)
{
for (int b = 0; b < kernel.RowCount; b++)
{
var xn = x + a - 1;
var yn = y + b - 1;
var intensity = (double)onGetIntensity(xn, yn); // returns R,G, or B color channel at that pixel
intensity /= 255; // intensity is 0-1
sum += intensity * kernel.At(a, b);
}
}
var result = (sum - (-8d)) / (8d - (-8d)); // find the % between the min and max of -8 and 8
result *= 255; // bring it back to 0-255
}
image-processing edge-detection
asked Nov 26 '18 at 1:42
user875234user875234
6752820
I'm following these instructions. But my results aren't coming out right. I'm using this kernel:
-1 -1 -1
-1 8 -1
-1 -1 -1
So that means the sum for a given convolution will be between -8 and 8, assuming I have already normalized my input (0-255 -> 0-1). Then I do the convolution. Then I find the percent my value is between the minimum and maximum values. For example, with this kernel my min is -8 and max is 8. So if the value is 0 that's 50% which works out to 255 * .5 = 127.5. But that's clearly not right and what it gives is a mostly gray image. The parts that aren't gray are still monochrome even though I'm running the kernel on each channel individually.
static int EvaluateKernelAtPoint(Bitmap bitmap, Matrix<double> kernel, int x, int y, Func<int, int, int> onGetIntensity)
{
double sum = 0;
for (int a = 0; a < kernel.ColumnCount; a++)
{
for (int b = 0; b < kernel.RowCount; b++)
{
var xn = x + a - 1;
var yn = y + b - 1;
var intensity = (double)onGetIntensity(xn, yn); // returns R,G, or B color channel at that pixel
intensity /= 255; // intensity is 0-1
sum += intensity * kernel.At(a, b);
}
}
var result = (sum - (-8d)) / (8d - (-8d)); // find the % between the min and max of -8 and 8
result *= 255; // bring it back to 0-255
}
image-processing edge-detection
image-processing edge-detection
asked Nov 26 '18 at 1:42
user875234user875234
6752820
asked Nov 26 '18 at 1:42
user875234user875234
6752820
asked Nov 26 '18 at 1:42
user875234user875234
6752820
asked Nov 26 '18 at 1:42
user875234user875234
6752820
asked Nov 26 '18 at 1:42
user875234user875234
6752820
6752820
1
Those extremes are extremely unlikely. You indeed want to set the zero to 127 or 128, then you can choose a scaling that will make your image look good. One approach is to find the min and max in the output image.
– Cris Luengo
Nov 26 '18 at 1:50
1
"The parts that aren't gray are still monochrome": don't expect the Laplacian output to show meaningful colors. Also, applying different gain coefficients to the three channels is not recommended as it introduces a bias.
– Yves Daoust
Nov 26 '18 at 7:07
add a comment |
1
Those extremes are extremely unlikely. You indeed want to set the zero to 127 or 128, then you can choose a scaling that will make your image look good. One approach is to find the min and max in the output image.
– Cris Luengo
Nov 26 '18 at 1:50
1
"The parts that aren't gray are still monochrome": don't expect the Laplacian output to show meaningful colors. Also, applying different gain coefficients to the three channels is not recommended as it introduces a bias.
– Yves Daoust
Nov 26 '18 at 7:07
1
1
Those extremes are extremely unlikely. You indeed want to set the zero to 127 or 128, then you can choose a scaling that will make your image look good. One approach is to find the min and max in the output image.
– Cris Luengo
Nov 26 '18 at 1:50
Those extremes are extremely unlikely. You indeed want to set the zero to 127 or 128, then you can choose a scaling that will make your image look good. One approach is to find the min and max in the output image.
– Cris Luengo
Nov 26 '18 at 1:50
1
1
"The parts that aren't gray are still monochrome": don't expect the Laplacian output to show meaningful colors. Also, applying different gain coefficients to the three channels is not recommended as it introduces a bias.
– Yves Daoust
Nov 26 '18 at 7:07
"The parts that aren't gray are still monochrome": don't expect the Laplacian output to show meaningful colors. Also, applying different gain coefficients to the three channels is not recommended as it introduces a bias.
– Yves Daoust
Nov 26 '18 at 7:07
add a comment |
1 Answer
1
active
oldest
votes
There is no right way to normalize, because the range is content-dependent.
In good RGB images, the range of values is usually [0, 255], provided the dynamic range is well adjusted.
But the output of this Laplacian filter, which can be seen as the difference between the original image and a smoothed version of it, usually has much smaller amplitude. But this depends on the local variations: an already smooth image will give no response, while noise (especially salt & pepper) can yield huge values.
You also need to decide what to do with negative values: shift so that zero appears as mid-gray, clamp to zero or take the absolute value.
Taking the min-max range and mapping it to 0-255 is an option, but leads to a "floating" zero and uncontrolled gain. I would prefer to set a constant gain for a set of images of the same origin.
Last but not least, histogram equalization is another option.
answered Nov 26 '18 at 7:15
Yves DaoustYves Daoust
38.3k72760
Allright, I will try fiddling with those values. It's on the back burner right now though. Quick question, the reason I got confused is because I was comparing my image to what you get in the browser with a -1 -1 -1 -1 8 -1 -1 -1 -1 convolution filter. Any guesses on what the browser might be doing here? jsfiddle.net/5zcro4hf
– user875234
Nov 27 '18 at 3:04
@user875234: it does the right thing, no ?
– Yves Daoust
Nov 27 '18 at 8:16
Yeah, it looks good to me. I just can't figure out how to get my own code to do it. If I follow the steps on mozilla (which point to the w3c spec which I have only read a couple parts of) I don't get the same result. I think the code in my question is a faithful interpretation of their pseudo code (bias would be 0 and divisor would be 1 so I've left them out of this question). That's why I'm confused about the results not matching up.
– user875234
Nov 27 '18 at 12:28
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "1"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53473765%2fwhats-the-right-way-to-normalize-a-convolution-filter%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
There is no right way to normalize, because the range is content-dependent.
In good RGB images, the range of values is usually [0, 255], provided the dynamic range is well adjusted.
But the output of this Laplacian filter, which can be seen as the difference between the original image and a smoothed version of it, usually has much smaller amplitude. But this depends on the local variations: an already smooth image will give no response, while noise (especially salt & pepper) can yield huge values.
You also need to decide what to do with negative values: shift so that zero appears as mid-gray, clamp to zero or take the absolute value.
Taking the min-max range and mapping it to 0-255 is an option, but leads to a "floating" zero and uncontrolled gain. I would prefer to set a constant gain for a set of images of the same origin.
Last but not least, histogram equalization is another option.
answered Nov 26 '18 at 7:15
Yves DaoustYves Daoust
38.3k72760
Allright, I will try fiddling with those values. It's on the back burner right now though. Quick question, the reason I got confused is because I was comparing my image to what you get in the browser with a -1 -1 -1 -1 8 -1 -1 -1 -1 convolution filter. Any guesses on what the browser might be doing here? jsfiddle.net/5zcro4hf
– user875234
Nov 27 '18 at 3:04
@user875234: it does the right thing, no ?
– Yves Daoust
Nov 27 '18 at 8:16
Yeah, it looks good to me. I just can't figure out how to get my own code to do it. If I follow the steps on mozilla (which point to the w3c spec which I have only read a couple parts of) I don't get the same result. I think the code in my question is a faithful interpretation of their pseudo code (bias would be 0 and divisor would be 1 so I've left them out of this question). That's why I'm confused about the results not matching up.
– user875234
Nov 27 '18 at 12:28
add a comment |
There is no right way to normalize, because the range is content-dependent.
In good RGB images, the range of values is usually [0, 255], provided the dynamic range is well adjusted.
But the output of this Laplacian filter, which can be seen as the difference between the original image and a smoothed version of it, usually has much smaller amplitude. But this depends on the local variations: an already smooth image will give no response, while noise (especially salt & pepper) can yield huge values.
You also need to decide what to do with negative values: shift so that zero appears as mid-gray, clamp to zero or take the absolute value.
Taking the min-max range and mapping it to 0-255 is an option, but leads to a "floating" zero and uncontrolled gain. I would prefer to set a constant gain for a set of images of the same origin.
Last but not least, histogram equalization is another option.
answered Nov 26 '18 at 7:15
Yves DaoustYves Daoust
38.3k72760
Allright, I will try fiddling with those values. It's on the back burner right now though. Quick question, the reason I got confused is because I was comparing my image to what you get in the browser with a -1 -1 -1 -1 8 -1 -1 -1 -1 convolution filter. Any guesses on what the browser might be doing here? jsfiddle.net/5zcro4hf
– user875234
Nov 27 '18 at 3:04
@user875234: it does the right thing, no ?
– Yves Daoust
Nov 27 '18 at 8:16
Yeah, it looks good to me. I just can't figure out how to get my own code to do it. If I follow the steps on mozilla (which point to the w3c spec which I have only read a couple parts of) I don't get the same result. I think the code in my question is a faithful interpretation of their pseudo code (bias would be 0 and divisor would be 1 so I've left them out of this question). That's why I'm confused about the results not matching up.
– user875234
Nov 27 '18 at 12:28
add a comment |
There is no right way to normalize, because the range is content-dependent.
In good RGB images, the range of values is usually [0, 255], provided the dynamic range is well adjusted.
But the output of this Laplacian filter, which can be seen as the difference between the original image and a smoothed version of it, usually has much smaller amplitude. But this depends on the local variations: an already smooth image will give no response, while noise (especially salt & pepper) can yield huge values.
You also need to decide what to do with negative values: shift so that zero appears as mid-gray, clamp to zero or take the absolute value.
Taking the min-max range and mapping it to 0-255 is an option, but leads to a "floating" zero and uncontrolled gain. I would prefer to set a constant gain for a set of images of the same origin.
Last but not least, histogram equalization is another option.
answered Nov 26 '18 at 7:15
Yves DaoustYves Daoust
38.3k72760
There is no right way to normalize, because the range is content-dependent.
In good RGB images, the range of values is usually [0, 255], provided the dynamic range is well adjusted.
But the output of this Laplacian filter, which can be seen as the difference between the original image and a smoothed version of it, usually has much smaller amplitude. But this depends on the local variations: an already smooth image will give no response, while noise (especially salt & pepper) can yield huge values.
You also need to decide what to do with negative values: shift so that zero appears as mid-gray, clamp to zero or take the absolute value.
Taking the min-max range and mapping it to 0-255 is an option, but leads to a "floating" zero and uncontrolled gain. I would prefer to set a constant gain for a set of images of the same origin.
Last but not least, histogram equalization is another option.
answered Nov 26 '18 at 7:15
Yves DaoustYves Daoust
38.3k72760
answered Nov 26 '18 at 7:15
Yves DaoustYves Daoust
38.3k72760
answered Nov 26 '18 at 7:15
Yves DaoustYves Daoust
38.3k72760
answered Nov 26 '18 at 7:15
Yves DaoustYves Daoust
38.3k72760
38.3k72760
Allright, I will try fiddling with those values. It's on the back burner right now though. Quick question, the reason I got confused is because I was comparing my image to what you get in the browser with a -1 -1 -1 -1 8 -1 -1 -1 -1 convolution filter. Any guesses on what the browser might be doing here? jsfiddle.net/5zcro4hf
– user875234
Nov 27 '18 at 3:04
@user875234: it does the right thing, no ?
– Yves Daoust
Nov 27 '18 at 8:16
Yeah, it looks good to me. I just can't figure out how to get my own code to do it. If I follow the steps on mozilla (which point to the w3c spec which I have only read a couple parts of) I don't get the same result. I think the code in my question is a faithful interpretation of their pseudo code (bias would be 0 and divisor would be 1 so I've left them out of this question). That's why I'm confused about the results not matching up.
– user875234
Nov 27 '18 at 12:28
add a comment |
Allright, I will try fiddling with those values. It's on the back burner right now though. Quick question, the reason I got confused is because I was comparing my image to what you get in the browser with a -1 -1 -1 -1 8 -1 -1 -1 -1 convolution filter. Any guesses on what the browser might be doing here? jsfiddle.net/5zcro4hf
– user875234
Nov 27 '18 at 3:04
@user875234: it does the right thing, no ?
– Yves Daoust
Nov 27 '18 at 8:16
Yeah, it looks good to me. I just can't figure out how to get my own code to do it. If I follow the steps on mozilla (which point to the w3c spec which I have only read a couple parts of) I don't get the same result. I think the code in my question is a faithful interpretation of their pseudo code (bias would be 0 and divisor would be 1 so I've left them out of this question). That's why I'm confused about the results not matching up.
– user875234
Nov 27 '18 at 12:28
Allright, I will try fiddling with those values. It's on the back burner right now though. Quick question, the reason I got confused is because I was comparing my image to what you get in the browser with a -1 -1 -1 -1 8 -1 -1 -1 -1 convolution filter. Any guesses on what the browser might be doing here? jsfiddle.net/5zcro4hf
– user875234
Nov 27 '18 at 3:04
Allright, I will try fiddling with those values. It's on the back burner right now though. Quick question, the reason I got confused is because I was comparing my image to what you get in the browser with a -1 -1 -1 -1 8 -1 -1 -1 -1 convolution filter. Any guesses on what the browser might be doing here? jsfiddle.net/5zcro4hf
– user875234
Nov 27 '18 at 3:04
@user875234: it does the right thing, no ?
– Yves Daoust
Nov 27 '18 at 8:16
@user875234: it does the right thing, no ?
– Yves Daoust
Nov 27 '18 at 8:16
Yeah, it looks good to me. I just can't figure out how to get my own code to do it. If I follow the steps on mozilla (which point to the w3c spec which I have only read a couple parts of) I don't get the same result. I think the code in my question is a faithful interpretation of their pseudo code (bias would be 0 and divisor would be 1 so I've left them out of this question). That's why I'm confused about the results not matching up.
– user875234
Nov 27 '18 at 12:28
Yeah, it looks good to me. I just can't figure out how to get my own code to do it. If I follow the steps on mozilla (which point to the w3c spec which I have only read a couple parts of) I don't get the same result. I think the code in my question is a faithful interpretation of their pseudo code (bias would be 0 and divisor would be 1 so I've left them out of this question). That's why I'm confused about the results not matching up.
– user875234
Nov 27 '18 at 12:28
add a comment |
Thanks for contributing an answer to Stack Overflow!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53473765%2fwhats-the-right-way-to-normalize-a-convolution-filter%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
Those extremes are extremely unlikely. You indeed want to set the zero to 127 or 128, then you can choose a scaling that will make your image look good. One approach is to find the min and max in the output image.
– Cris Luengo
Nov 26 '18 at 1:50
1
"The parts that aren't gray are still monochrome": don't expect the Laplacian output to show meaningful colors. Also, applying different gain coefficients to the three channels is not recommended as it introduces a bias.
– Yves Daoust
Nov 26 '18 at 7:07