Python “Fast normalized cross correlation”
I am trying to use the implementation of "Fast Normalized cross correlation" ( http://pastebin.com/x1NJqWWm ) in python to do some template matching. However, even for simple test-images, it produces values out of the [-1:1] range. I can't see what I am doing wrong. Is there a limitation of this algorithm? Help would be much appreciated!
example:
# Template creation
t = np.zeros((4,4))
t[:,0:3] = 1
# Image creation
image = np.zeros((15,15)) # image
k=1
for i in range(len(image)):
if k==1:
image[i,:]=1
image[:,i]=1
k=0
else:
k=1
# Fast Normalized Cross Correlation
TM = norm_xcorr.TemplateMatch(t, method='both')
result2, ssd = TM(image)
This gives values out of the [-1:1] range:
>>> np.max(result2)
Out[6]: 8.4913641920299376
>>> np.min(result2)
Out[7]: -3.1869961773458306
python performance correlation
add a comment |
I am trying to use the implementation of "Fast Normalized cross correlation" ( http://pastebin.com/x1NJqWWm ) in python to do some template matching. However, even for simple test-images, it produces values out of the [-1:1] range. I can't see what I am doing wrong. Is there a limitation of this algorithm? Help would be much appreciated!
example:
# Template creation
t = np.zeros((4,4))
t[:,0:3] = 1
# Image creation
image = np.zeros((15,15)) # image
k=1
for i in range(len(image)):
if k==1:
image[i,:]=1
image[:,i]=1
k=0
else:
k=1
# Fast Normalized Cross Correlation
TM = norm_xcorr.TemplateMatch(t, method='both')
result2, ssd = TM(image)
This gives values out of the [-1:1] range:
>>> np.max(result2)
Out[6]: 8.4913641920299376
>>> np.min(result2)
Out[7]: -3.1869961773458306
python performance correlation
add a comment |
I am trying to use the implementation of "Fast Normalized cross correlation" ( http://pastebin.com/x1NJqWWm ) in python to do some template matching. However, even for simple test-images, it produces values out of the [-1:1] range. I can't see what I am doing wrong. Is there a limitation of this algorithm? Help would be much appreciated!
example:
# Template creation
t = np.zeros((4,4))
t[:,0:3] = 1
# Image creation
image = np.zeros((15,15)) # image
k=1
for i in range(len(image)):
if k==1:
image[i,:]=1
image[:,i]=1
k=0
else:
k=1
# Fast Normalized Cross Correlation
TM = norm_xcorr.TemplateMatch(t, method='both')
result2, ssd = TM(image)
This gives values out of the [-1:1] range:
>>> np.max(result2)
Out[6]: 8.4913641920299376
>>> np.min(result2)
Out[7]: -3.1869961773458306
python performance correlation
I am trying to use the implementation of "Fast Normalized cross correlation" ( http://pastebin.com/x1NJqWWm ) in python to do some template matching. However, even for simple test-images, it produces values out of the [-1:1] range. I can't see what I am doing wrong. Is there a limitation of this algorithm? Help would be much appreciated!
example:
# Template creation
t = np.zeros((4,4))
t[:,0:3] = 1
# Image creation
image = np.zeros((15,15)) # image
k=1
for i in range(len(image)):
if k==1:
image[i,:]=1
image[:,i]=1
k=0
else:
k=1
# Fast Normalized Cross Correlation
TM = norm_xcorr.TemplateMatch(t, method='both')
result2, ssd = TM(image)
This gives values out of the [-1:1] range:
>>> np.max(result2)
Out[6]: 8.4913641920299376
>>> np.min(result2)
Out[7]: -3.1869961773458306
python performance correlation
python performance correlation
asked Feb 2 '15 at 16:39
sb9911
62
62
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
It's already available in skimage
source code
Also see this: https://dsp.stackexchange.com/questions/28322/python-normalized-cross-correlation-to-measure-similarites-in-2-images
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%2f28282415%2fpython-fast-normalized-cross-correlation%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
It's already available in skimage
source code
Also see this: https://dsp.stackexchange.com/questions/28322/python-normalized-cross-correlation-to-measure-similarites-in-2-images
add a comment |
It's already available in skimage
source code
Also see this: https://dsp.stackexchange.com/questions/28322/python-normalized-cross-correlation-to-measure-similarites-in-2-images
add a comment |
It's already available in skimage
source code
Also see this: https://dsp.stackexchange.com/questions/28322/python-normalized-cross-correlation-to-measure-similarites-in-2-images
It's already available in skimage
source code
Also see this: https://dsp.stackexchange.com/questions/28322/python-normalized-cross-correlation-to-measure-similarites-in-2-images
answered Nov 22 '18 at 21:37
seralouk
5,66522338
5,66522338
add a comment |
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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%2f28282415%2fpython-fast-normalized-cross-correlation%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