By the way, the recommended syntax these days for that operation would be
[min_val1, idx1] = min(chi_coeff, [], "all", "linear");
but personally it feels to me that surely that must require more processing than what you are using.
Error "Array indices must be positive integers or logical values" occurs at the line [row1, col1, s1] = ind2sub(size(chi_coeff), idx1);
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I am working on a MATLAB script to find the minimum value in a 3D matrix chi_coeff and convert the linear index to subscripts using ind2sub. Despite idx1 being a valid scalar positive integer index, and chi_coeff confirmed as a 3D double array, I get the error:
Array indices must be positive integers or logical values.
on the line:
[row1, col1, s1] = ind2sub(size(chi_coeff), idx1);
Diagnostic checks I performed:
- idx1 is scalar, positive, and integer (assertion passed).
- chi_coeff is a valid 3D numeric array with size 5x10x4 and double class.
- which ind2sub -all confirms the built-in function is used.
- The error is not caused by passing a non-integer or invalid index to ind2sub.
The error still persists and might be coming from subsequent lines where I use (row1, col1, s1) to index other arrays of the same size, e.g.:
matlab
found_Scat_Calc(id,:) = Scat_Coeff_Calc(row1, col1, s1);
I suspect there might be a mismatch in array dimensions or an unexpected indexing issue.
Key code snippet:
[min_val1, idx1] = min(chi_coeff(:));
[row1, col1, s1] = ind2sub(size(chi_coeff), idx1);
found_Scat_Calc(id,:) = Scat_Coeff_Calc(row1, col1, s1);
Additional info:
- Confirmed sizes with whos and class commands; all arrays involved have compatible dimensions.
- The arrays involved in indexing are initialized with zeros and filled inside nested loops with parfor.
Would appreciate guidance on why this error occurs in the context of ind2sub and further indexing, and how to properly diagnose/fix dimension or indexing mismatch issues.
回答 (1 件)
Star Strider
2025 年 7 月 26 日
My guess is that you have a variable names 'size' and that is throwing the error.
Run this to test that hypothesis:
which size -all
You should get something like this result.
If the first entry is 'size is a variable' you have found the problem. The solution is to rename your variable to something meaningful in the context of your code that does not conflict with any MATLAB function names.
To illustrate --
size = 42;
idx1 = 84;
chi_coeff = rand(5,10,4);
which size -all
[row1, col1, s1] = ind2sub(size(chi_coeff), idx1);
.
4 件のコメント
dpb
2025 年 7 月 26 日
編集済み: dpb
2025 年 7 月 26 日
Good catch, @Star Strider. OP did well to have checked about ind2sub, but forgot or didn't think about size (nor did I after noticed he had checked the one, but it is a fairly common mistake).
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