Find an index of a cell whose element satisfies a condition

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.

Answer 1

[Star Strider]

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

.

Answer 2

[DGM]

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

[Diaa]

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]

Ah yeah. I had read "second column" as "second element".