Loading in png file that is black
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Dries Weytjens
il 5 Mag 2017
Commentato: Zhifei Deng
il 16 Nov 2018
Hello,
I am trying to load in a large png file in Matlab with purposes of cropping it (I have a bunch of them so am trying to automate the cropping). Unfortunately, when I load in the images, I first get the message that the image is too large to display, hence Matlab displays at 67% which does not seem to be a problem for the quality of the images. The bigger problem, however, is that the image shows up in black.
I using the following code:
ToCrop = imread('iteration 31.png','png');
imshow(ToCrop,[]);
Thanks!
3 Commenti
Guillaume
il 5 Mag 2017
There's definitively at least one non-black pixels in that image. Can you attach it to your question?
If not, what is the output of
imfinfo('iteration 31.png')
Note that you did not need the [] in your call to imshow. However, as it has no effect in your case, removing it won't solve the problem.
Risposta accettata
Santhana Raj
il 5 Mag 2017
By saying that the image appears in black, do you mean that the image is still visible but the image is in grey scale?? The below code works fine for me in that case:
[I,m]=imread('iteration 31.png');
imshow(I,'Colormap',m);
2 Commenti
Guillaume
il 5 Mag 2017
No! The image was not in greyscale. The problem was that you were displaying the indexes into a colour map instead of the colours themselves. Read up on indexed images (or convert the image to RGB as suggested by Image Analyst)
Più risposte (4)
Image Analyst
il 5 Mag 2017
This works just fine for your indexed image. You just have to apply the correct colormap:
filename = 'iteration 31.png'
[indexedImage, storedColorMap] = imread(filename);
imshow(indexedImage, storedColorMap)
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/175921/image.png)
3 Commenti
Guillaume
il 5 Mag 2017
"Don't use image as the name of a variable". I'm not sure where you've seen image used as a variable name. In the question, the OP uses ToCrop which is a good name in my opinion.
Image Analyst
il 5 Mag 2017
Guillaume, between my Answer and immediately below was a response by Dries with some code where he used my code but changed the variable names, including making image variable names of "image" and "I". He deleted that comment after I advised him to not use those names and use descriptive names instead. I agree - it's now confusing since he did that.
Santhana Raj
il 5 Mag 2017
The problem might be due to improper scaling or the type of image.
Use imagesc
1 Commento
KSSV
il 5 Mag 2017
ToCrop = imread('iteration 31.png','BackgroundColor',[1 1 1]);
imshow(ToCrop);
2 Commenti
Zhifei Deng
il 16 Nov 2018
I had the same problem with SW, but it seems that this code works for my case, great thanks!
Guillaume
il 5 Mag 2017
Your description was not very accurate. The image is not black, it simply is the wrong colour, with the background showing black instead of white.
That is because, this is an indexed image. The pixel values are indices into a colour map that you're not loading. The white colour in the colour map is at index 0, so if you use 0 as intensity instead of an index in the colour map it is going to look black.
To fix this you need to load the colour map and pass it to imshow:
[ToCrop, map] = imread('iteration 31.png');
imshow(ToCrop, map);
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