error in edge function line 213
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error description:error in line 213 [a,method,thresh,sigma,thinning,H,kx,ky] = parse_inputs(args{:});
1 件のコメント
DGM
2023 年 12 月 15 日
Nobody can know what your code is, how you called edge() or what the actual error message was unless you actually tell us.
回答 (1 件)
Maneet Kaur Bagga
2023 年 12 月 26 日
編集済み: Maneet Kaur Bagga
2023 年 12 月 26 日
Hi Rengasri,
As per my understanding the edge function admits only GrayScale Images. Therefore assuming that the image used here is in rgb format, convert the image to gray scale using the "rgb2gray" function, and then apply the edge function to obtain the desired result.
Please refer to the following MATLAB Answer that address comparable issues to achieve the intended outcome.
For better understanding, refer to the MATLAB documentation below:
edge function: https://www.mathworks.com/help/images/ref/edge.html
Hope this helps!
2 件のコメント
Dyuman Joshi
2023 年 12 月 26 日
How are you sure that OP has supplied a rgb image?
DGM
2023 年 12 月 27 日
編集済み: DGM
2023 年 12 月 27 日
The given lines will be in the error backtrace for any error in any of the input arguments.
- It could be because the image has more than one channel, but not all multichannel images are RGB.
- Or maybe it's not a strictly real-valued numeric array.
- Maybe the method string is invalid.
- Maybe the threshold parameters aren't valid.
Given that OP doesn't need an answer anymore, I think it's fair to presume the likely case that it's RGB. After all, that has a good chance of helping future readers.
That said, there are plenty of times where I need to do edge finding on multichannel images in whole, without converting them to grayscale. An answer to that variant of the question creates new problems.
Answering dead questions gives us the latitude to choose our assumptions about the intent and context, but that's also why I sometimes like to address multiple possible interpretations of things in an answer. This is especially the case when the question itself is ill-defined. After all, we're not answering it for the OP anymore; we're answering it for everyone in the present and future who might hope to find an answer to their own interpretation of the question.
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