How to remove zeros from an element in a cell array

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Hi all,
I have two cell arrays each contain 12 cells, each cell contains different matrix size (31,1) or (30,1) or (28,1), each cell in the array contains a bunch of zeros, and I need to remove these zeros, because it indicates that there is no data at this day
here are an example of my data,
A = {[1 0 1], [2 0 2], ......., [12 0 12]}
B = {[0 2 1], [1 2 0], ........, [12 0 12]}
my purpose is to compare A with B, and to do that I have to remove the zeros and the coincident index in the other cell
A(1,1){2} and B(1,1){2} = []; aslo A(1,1){1} and B(1,1){1}=[];
any suggestions, please??

Accepted Answer

Walter Roberson
Walter Roberson on 11 Jul 2021
monthlengths = [28 30 31];
C = arrayfun(@(x) randi([0 9], 1, monthlengths(randi(3))), 1:12, 'uniform', 0)
C = 1×12 cell array
{[5 3 1 4 6 4 6 8 7 0 9 1 5 7 9 3 7 3 4 9 5 2 0 3 5 6 1 0 5 0]} {[2 9 1 7 5 2 6 8 1 6 2 6 1 3 3 3 0 1 7 1 4 3 6 0 2 7 1 8 9 4]} {[4 8 6 4 8 7 1 0 5 7 3 1 9 1 1 8 2 0 7 9 8 6 1 9 6 5 5 9 4 0 8]} {[4 1 0 6 7 0 0 7 9 9 3 1 3 5 6 5 2 3 7 1 4 0 7 4 7 1 9 5]} {[7 0 5 7 7 2 3 3 9 8 7 3 9 6 5 3 0 0 0 3 5 3 9 2 2 9 6 9]} {[7 3 3 1 2 5 0 0 8 4 9 9 8 0 7 5 0 7 0 2 6 1 4 8 0 2 1 0 6 7 4]} {[1 8 7 3 9 7 3 1 1 6 7 7 6 9 4 3 4 2 9 8 7 6 0 6 1 8 6 7 4 3]} {[1 9 4 2 2 9 8 9 3 9 9 6 3 7 0 3 8 1 7 5 0 9 0 3 9 6 8 7 9 9]} {[7 1 5 9 9 9 8 7 0 5 6 0 7 2 0 1 1 8 5 2 1 2 4 0 5 1 7 9]} {[8 5 2 2 7 3 5 9 1 5 7 2 1 2 9 6 0 9 3 9 9 2 1 2 8 2 0 1 8 5 0]} {[9 5 2 0 8 1 3 5 6 5 3 4 9 2 6 9 7 0 5 2 6 7 0 6 2 1 4 3 0 9]} {[7 1 3 2 3 7 8 4 0 4 0 9 2 8 9 3 7 6 4 9 6 9 4 5 7 7 5 2 0 0 9]}
reduced_C = cellfun(@(c) c(c~=0), C, 'uniform', 0)
reduced_C = 1×12 cell array
{[5 3 1 4 6 4 6 8 7 9 1 5 7 9 3 7 3 4 9 5 2 3 5 6 1 5]} {[2 9 1 7 5 2 6 8 1 6 2 6 1 3 3 3 1 7 1 4 3 6 2 7 1 8 9 4]} {[4 8 6 4 8 7 1 5 7 3 1 9 1 1 8 2 7 9 8 6 1 9 6 5 5 9 4 8]} {[4 1 6 7 7 9 9 3 1 3 5 6 5 2 3 7 1 4 7 4 7 1 9 5]} {[7 5 7 7 2 3 3 9 8 7 3 9 6 5 3 3 5 3 9 2 2 9 6 9]} {[7 3 3 1 2 5 8 4 9 9 8 7 5 7 2 6 1 4 8 2 1 6 7 4]} {[1 8 7 3 9 7 3 1 1 6 7 7 6 9 4 3 4 2 9 8 7 6 6 1 8 6 7 4 3]} {[1 9 4 2 2 9 8 9 3 9 9 6 3 7 3 8 1 7 5 9 3 9 6 8 7 9 9]} {[7 1 5 9 9 9 8 7 5 6 7 2 1 1 8 5 2 1 2 4 5 1 7 9]} {[8 5 2 2 7 3 5 9 1 5 7 2 1 2 9 6 9 3 9 9 2 1 2 8 2 1 8 5]} {[9 5 2 8 1 3 5 6 5 3 4 9 2 6 9 7 5 2 6 7 6 2 1 4 3 9]} {[7 1 3 2 3 7 8 4 4 9 2 8 9 3 7 6 4 9 6 9 4 5 7 7 5 2 9]}
  4 Comments
Walter Roberson
Walter Roberson on 12 Jul 2021
Well, if you really want... but I do not see what advantages that would have over the code I already posted using masks to select elements. The setdiff(1:end,m) to remove the zero elements whose index you know is pretty clumsy: easier to use logical variables, which you can use to index or use the negation to index.
monthlengths = [28 30 31];
A = arrayfun(@(x) randi([0 9], 1, monthlengths(randi(3))), 1:12, 'uniform', 0)
A = 1×12 cell array
{[9 8 1 9 3 1 9 9 3 8 1 6 1 7 7 0 4 9 0 2 4 2 0 9 1 4 2 3 4 6 9]} {[7 5 1 0 7 5 6 0 4 4 6 9 9 9 4 4 8 6 2 5 7 3 1 4 6 6 2 9]} {[4 4 2 7 3 4 2 8 5 1 8 3 2 5 1 2 3 3 6 2 5 9 0 8 0 2 5 5]} {[0 7 6 7 7 2 5 8 2 4 0 4 4 3 2 2 9 5 3 2 1 8 1 9 1 4 2 3]} {[9 9 9 6 3 4 2 5 7 6 1 3 0 7 7 0 7 1 1 4 3 2 2 7 7 9 1 4 3 3]} {[5 8 6 2 1 7 2 3 0 1 1 4 5 1 4 7 6 8 6 7 7 9 8 8 3 0 1 2]} {[3 8 7 2 4 8 9 1 3 2 9 5 1 0 9 7 9 9 0 0 8 0 7 8 5 7 4 4 0 0 2]} {[8 8 9 9 7 9 7 7 7 0 4 3 1 8 6 0 9 0 2 1 2 9 4 0 9 8 6 9 1 9 8]} {[8 7 2 8 3 3 3 6 6 9 7 6 9 2 2 3 7 6 6 1 3 3 3 7 3 6 8 2 5 7]} {[1 1 1 5 8 0 6 1 9 6 8 4 2 8 7 7 3 0 6 6 2 6 8 8 3 3 2 7 6 2 9]} {[6 8 7 7 8 4 8 1 4 7 2 4 9 2 9 8 0 6 8 7 7 8 3 6 3 0 5 3 5 6]} {[8 1 7 8 3 2 9 1 7 7 8 0 3 4 9 6 3 0 8 0 4 5 3 8 6 2 9 2 5 5]}
B = cellfun(@(c) randi([0 9], 1, length(c)), A, 'uniform', 0)
B = 1×12 cell array
{[1 1 8 3 2 1 6 9 0 8 5 8 8 9 0 5 9 6 0 4 9 6 1 4 0 8 5 5 9 4 7]} {[5 0 4 2 8 5 9 1 2 2 6 0 1 6 3 5 0 3 5 5 5 4 3 9 8 5 6 6]} {[1 6 5 2 2 1 3 1 5 8 0 1 0 5 0 5 0 7 3 5 2 3 5 8 7 0 9 4]} {[4 5 5 2 9 9 0 3 1 6 4 9 8 8 0 1 5 8 0 7 5 8 1 8 4 0 1 6]} {[5 7 0 4 7 0 0 2 1 5 1 4 1 8 2 3 7 7 7 1 6 9 1 6 8 2 2 5 8 5]} {[9 6 3 6 3 1 6 7 1 5 5 7 4 6 8 2 4 8 9 1 7 1 5 5 6 4 8 8]} {[7 9 8 7 9 1 6 3 0 1 7 6 6 6 9 0 9 4 3 3 9 2 1 5 1 2 7 3 8 7 6]} {[7 1 0 1 9 9 4 7 6 7 3 3 0 4 7 7 6 9 0 6 3 7 9 3 3 6 3 1 6 3 0]} {[8 8 4 6 1 6 4 7 5 3 8 7 7 8 8 6 1 8 2 9 2 0 2 4 5 1 1 0 9 9]} {[2 7 7 2 2 1 5 8 3 0 3 1 8 1 5 2 5 3 5 8 8 8 3 2 7 2 9 6 0 9 6]} {[2 9 5 4 8 0 5 8 8 8 3 2 7 1 7 3 5 2 6 2 6 8 4 8 7 8 0 4 1 2]} {[0 0 6 9 7 5 9 0 5 9 9 4 2 4 7 7 5 4 0 8 9 2 6 1 5 5 2 8 9 7]}
z_A = cellfun(@(c) find(c==0), A, 'uniform', 0);
z_B = cellfun(@(c) find(c==0), B, 'uniform', 0);
z = cellfun(@(zA, zB) union(zA, zB), z_A, z_B, 'uniform', 0)
z = 1×12 cell array
{[9 15 16 19 23 25]} {[2 4 8 12 17]} {[11 13 15 17 23 25 26]} {[1 7 11 15 19 26]} {[3 6 7 13 16]} {[9 26]} {[9 14 16 19 20 22 29 30]} {[3 10 13 16 18 19 24 31]} {[22 28]} {[6 10 18 29]} {[6 17 26 27]} {[1 2 8 12 18 19 20]}
reduced_A = cellfun(@(C, m) C(setdiff(1:end,m)), A, z, 'uniform', 0)
reduced_A = 1×12 cell array
{[9 8 1 9 3 1 9 9 8 1 6 1 7 4 9 2 4 2 9 4 2 3 4 6 9]} {[7 1 7 5 6 4 4 6 9 9 4 4 6 2 5 7 3 1 4 6 6 2 9]} {[4 4 2 7 3 4 2 8 5 1 3 5 2 3 6 2 5 9 8 5 5]} {[7 6 7 7 2 8 2 4 4 4 3 2 9 5 2 1 8 1 9 1 2 3]} {[9 9 6 3 5 7 6 1 3 7 7 7 1 1 4 3 2 2 7 7 9 1 4 3 3]} {[5 8 6 2 1 7 2 3 1 1 4 5 1 4 7 6 8 6 7 7 9 8 8 3 1 2]} {[3 8 7 2 4 8 9 1 2 9 5 1 9 9 9 8 7 8 5 7 4 4 2]} {[8 8 9 7 9 7 7 7 4 3 8 6 9 1 2 9 4 9 8 6 9 1 9]} {[8 7 2 8 3 3 3 6 6 9 7 6 9 2 2 3 7 6 6 1 3 3 7 3 6 8 5 7]} {[1 1 1 5 8 6 1 9 8 4 2 8 7 7 3 6 6 2 6 8 8 3 3 2 7 2 9]} {[6 8 7 7 8 8 1 4 7 2 4 9 2 9 8 6 8 7 7 8 3 6 3 3 5 6]} {[7 8 3 2 9 7 7 8 3 4 9 6 3 4 5 3 8 6 2 9 2 5 5]}
reduced_B = cellfun(@(C,m) C(setdiff(1:end,m)), B, z, 'uniform', 0)
reduced_B = 1×12 cell array
{[1 1 8 3 2 1 6 9 8 5 8 8 9 9 6 4 9 6 4 8 5 5 9 4 7]} {[5 4 8 5 9 2 2 6 1 6 3 5 3 5 5 5 4 3 9 8 5 6 6]} {[1 6 5 2 2 1 3 1 5 8 1 5 5 7 3 5 2 3 8 9 4]} {[5 5 2 9 9 3 1 6 9 8 8 1 5 8 7 5 8 1 8 4 1 6]} {[5 7 4 7 2 1 5 1 4 8 2 7 7 7 1 6 9 1 6 8 2 2 5 8 5]} {[9 6 3 6 3 1 6 7 5 5 7 4 6 8 2 4 8 9 1 7 1 5 5 6 8 8]} {[7 9 8 7 9 1 6 3 1 7 6 6 9 9 4 9 1 5 1 2 7 3 6]} {[7 1 1 9 9 4 7 6 3 3 4 7 6 6 3 7 9 3 6 3 1 6 3]} {[8 8 4 6 1 6 4 7 5 3 8 7 7 8 8 6 1 8 2 9 2 2 4 5 1 1 9 9]} {[2 7 7 2 2 5 8 3 3 1 8 1 5 2 5 5 8 8 8 3 2 7 2 9 6 9 6]} {[2 9 5 4 8 5 8 8 8 3 2 7 1 7 3 2 6 2 6 8 4 8 7 4 1 2]} {[6 9 7 5 9 5 9 9 2 4 7 7 5 9 2 6 1 5 5 2 8 9 7]}

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More Answers (1)

dpb
dpb on 11 Jul 2021
yourarray=cellfun(@(c)c(~-0),yourarray,'UniformOutput',false);
  4 Comments
dpb
dpb on 12 Jul 2021
Au contraire --
>> c{:}
ans =
5
3
1
3
0
ans =
2
5
2
1
1
>>
>> c=cellfun(@(c)c(c~=0),c,'UniformOutput',false)
c =
1×2 cell array
{4×1 double} {5×1 double}
>> c{:}
ans =
5
3
1
3
ans =
2
5
2
1
1
>>
Walter Roberson
Walter Roberson on 12 Jul 2021
Yes, but this is different code; your posted code had c(~-0) not c(c~=0)

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