# Find an index of a cell whose element satisfies a condition

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Diaa on 18 Jun 2021
Commented: DGM on 18 Jun 2021
Is there a way to economically fix the code below to make it work so that it returns the index of the cell element whose array has the number 1 in its second column?
c = {[1,2,3], [2,3,1], [3,1,2]};
find(c{:}(:,2)==1) % expected result is the cell index 3
The code above gets me this error:
Intermediate brace {} indexing produced a comma-separated list with 3 values, but it must produce a single value to perform subsequent indexing operations.
I came up with this one
find(~cellfun(@isempty,(cellfun(@(x) find(x(:,2)==1,1),c,'un',0))))
but I would like to find some other efficient ways of doing it.
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### Accepted Answer

Star Strider on 18 Jun 2021
Try this —
c = {[1,2,3], [2,3,1], [3,1,2]};
idx = cellfun(@(x)x(:,2)==1, c, 'Unif',0)
idx = 1×3 cell array
{[0]} {[0]} {[1]}
Out = find([idx{:}])
Out = 3
.
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### More Answers (1)

DGM on 18 Jun 2021
This is one way:
c = {[1,2,3], [2,3,1], [3,1,2], [1,2,3], [2,3,1], [3,1,2]};
idx = find(cellfun(@(x) x(2)==1,c))
idx = 1×2
3 6
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Diaa on 18 Jun 2021
I think your code will fail in this example
c = {[1,2,3;1,2,3], [2,3,1], [3,1,2;3,6,9]}; find(cellfun(@(x) x(2)==1,c))
DGM on 18 Jun 2021
Ah yeah. I had read "second column" as "second element".

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