Accessing columns of array elements in a cell array without for loops

I have a K-element cell array C1. Each element is an Nx3 array of doubles. How can I get a new cell array whose elements are the first column of each array in C1 using only logical indexing? For example, if C1 is {M1 M2 M3}, where M1=[1 2 3; 4 5 6; 7 8 9]=M2=M3; then I want a new cell array C2 that is {N1 N2 N3} where N1=[1;4;7]=N2=N3.

*Edit: My mistake, the K matrices do not all have the same number of rows (but they do all have 3 columns). See my comment below.

I can do this with a for loop:

for j=1:length(C1)
    mat=C1{j}(:,1);
    C2{j}=mat;
end

How can I do it without a loop?

[per isakson]

is = [true(3,1),false(3,2)]
cellfun( @(c) c(is), C1, 'uni',false)

outputs

is =
     1     0     0
     1     0     0
     1     0     0
ans = 
    [3x1 double]    [3x1 double]    [3x1 double]

However, I used array indexing to create the logical index. I give up.

[Ted]

Sorry, I should have specified that the K matrices are not necessarily all the same size; they each have 3 columns but an arbitrary number of rows. So they are not all Nx3 matrices, but rather nx3 matrices where n is different for each C{k}. My apologies.

Answer 1

[the cyclist]

C2 = cellfun(@(x)x(:,1),C1,'UniformOutput',false);

[per isakson]

This solution violates "using only logical indexing" :-)

[Image Analyst]

Clever (+1 vote) but I wonder why, since the array sizes are all the same and not variable, he doesn't simply use a 2D double array instead of a cell array, which is always more complicated.

[the cyclist]

I agree with the above comments. :-)

[Ted]

I guess this is impossible using only logical indexing... disappointing. But this cellfun does the trick, thanks!

[Image Analyst]

Did you see my solution below? It does it, at least for a known, fixed value of K. And you didn't explain why you are even using cells in the first place.

Answer 2

[Image Analyst]

Perhaps you want something like this:

% Setup:
% Define K random 3x3 matrices for the K M arrays. 
% K = 3 in this example.
M1 = [1 2 3; 4 5 6; 7 8 9]
M2 = randi(9, 3, 3)
M3 = randi(9, 3, 3)
% Create C1 from the M's.
C1 = {M1 M2 M3}

% Solution:
% Use logical indexing to get Ns as the first column of the M's.
N1 = M1(logical([1, 0, 0, 1, 0, 0, 1, 0, 0]))'
N2 = M2(logical([1, 0, 0, 1, 0, 0, 1, 0, 0]))'
N3 = M3(logical([1, 0, 0, 1, 0, 0, 1, 0, 0]))'
% Create C2, a 1-by-3 cell array, that is the 3 column vectors
% (first columns of the M's) each in their own cell.
C2 = {N1 N2 N3}
celldisp(C2)