I just learned about Matlab's ability to use matrices as arguments to a for loop;
This displays the columns of the matrix A
A = magic(5)
for n = A
disp(n)
end
I was wondering if it is possible to obtain a sequential index into a loop like this. ie. if I want to print the column number along with the column itself, can I continue with the form above or would I need to revert to a more typical form like
A = magic(5)
for jj = 1:length(A)
disp(['Column #',num2str(jj)])
disp(A(:,jj))
end

 Akzeptierte Antwort

Walter Roberson
Walter Roberson am 24 Sep. 2012

0 Stimmen

You would need the more traditional form. There is no way to query which iteration number you are on.

Weitere Antworten (2)

Matt Fig
Matt Fig am 24 Sep. 2012

1 Stimme

A = magic(5)
cnt = 1;
for jj = A
disp(['Column #' num2str(cnt)])
disp(jj),
cnt = cnt+1;
end

4 Kommentare

Ravi
Ravi am 24 Sep. 2012
Thanks for this. Unfortunately, it looks like I can only mark one answer as accepted...
Matt Fig
Matt Fig am 24 Sep. 2012
Bearbeitet: Matt Fig am 24 Sep. 2012
Walter, did you look closely at the code? I ran it, and it does exactly what Ravi asked for...
Column #1
17
23
4
10
11
Column #2
24
5
6
12
18
Column #3
1
7
13
19
25
Column #4
8
14
20
21
2
Column #5
15
16
22
3
9
Walter Roberson
Walter Roberson am 24 Sep. 2012
Oh... yes, you are right, I had forgotten that it went by columns when matrices are used. blush
Ravi
Ravi am 24 Sep. 2012
I see... Backstory is that I was thinking about rewriting some existing code with for loops using the matrix approach for potential performance improvements. I did some quick tests and both methods take almost identical times. I can see the matrix approach being helpful down the line. Kind of like having a 'foreach' command.

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Shadow
Shadow am 2 Apr. 2024

0 Stimmen

enumer = @(my_array) cell2mat(arrayfun(@(x,idx) struct("cargo",x,"idx",idx), my_array(:).', 1:numel(my_array(:).'),UniformOutput=false));
data = rand(5,4)
s=size(data);
for elem = enumer(data)
[row,col] = ind2sub(s,elem.idx);
disp("Element " + string(elem.cargo) + " with index " + string(elem.idx) + " at location " + string(row) + "," + string(col))
end

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Gefragt:

am 24 Sep. 2012

Beantwortet:

am 2 Apr. 2024

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