How to remove zeros from an element in a cell array

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??

Answer 1

[Walter Roberson]

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]}

[Walter Roberson]

monthlengths = [28 30 31];
A = arrayfun(@(x) randi([0 9], 1, monthlengths(randi(3))), 1:12, 'uniform', 0)

A = 1×12 cell array
    {[7 1 4 3 2 6 0 3 9 5 9 2 6 9 7 7 7 4 1 7 5 5 6 0 5 4 8 4 2 7]}    {[7 9 3 2 2 0 7 1 6 9 9 2 1 3 2 5 1 4 1 2 1 1 3 1 3 8 2 8]}    {[5 7 7 4 8 4 3 5 5 3 5 0 6 5 6 0 6 3 0 7 7 3 7 3 4 5 5 7]}    {[6 3 3 8 3 1 2 3 4 1 8 9 9 5 6 9 4 8 5 3 9 5 8 6 9 1 2 0 0 9 6]}    {[5 2 4 9 3 0 3 1 7 9 0 2 4 4 4 2 9 5 9 1 1 8 5 6 5 5 4 9]}    {[1 3 4 6 2 8 9 1 1 0 6 9 4 1 5 9 1 5 1 9 3 1 3 3 4 6 5 5]}    {[2 2 7 2 7 1 3 5 7 2 0 0 4 6 1 6 6 5 6 0 9 6 4 7 4 0 6 4]}    {[3 3 4 6 4 9 9 0 8 5 4 9 2 0 7 3 9 8 1 7 0 4 4 0 7 3 4 0 8 9]}    {[5 6 4 8 2 5 1 0 6 1 7 0 7 9 9 9 4 0 8 9 6 2 0 2 1 2 6 5]}    {[5 0 6 1 3 1 1 9 2 8 4 8 2 5 3 1 1 8 0 5 2 0 2 9 2 3 9 4 5 5]}    {[5 6 1 9 9 0 4 5 4 2 8 3 8 3 5 4 7 6 1 3 8 3 0 3 4 7 4 8 7 7 7]}    {[6 8 1 6 4 2 0 5 8 9 3 0 7 3 4 7 7 8 4 9 1 7 6 5 2 7 8 8 8 6 1]}

B = cellfun(@(c) randi([0 9], 1, length(c)), A, 'uniform', 0)

B = 1×12 cell array
    {[3 5 0 6 9 5 9 8 7 4 0 8 1 8 9 3 0 4 8 5 6 5 9 0 0 9 3 4 4 8]}    {[1 2 6 3 4 0 9 3 2 6 4 1 0 5 0 9 6 0 9 9 1 6 1 3 8 1 0 2]}    {[9 7 7 3 5 1 7 3 9 7 1 9 5 1 8 8 9 7 7 2 7 6 9 4 0 0 3 3]}    {[3 6 7 3 2 1 4 0 4 7 0 2 4 3 4 8 3 8 8 0 2 3 1 7 7 1 4 5 4 9 8]}    {[0 2 6 4 6 0 8 2 5 8 8 9 2 5 9 2 2 0 8 2 1 2 8 6 9 2 7 2]}    {[9 7 7 9 6 1 8 8 5 0 8 8 5 3 6 6 4 0 7 0 4 7 2 6 8 1 9 2]}    {[3 7 8 6 3 0 9 8 1 4 1 7 0 4 2 5 4 3 7 4 9 4 4 5 7 7 8 7]}    {[2 6 4 9 5 7 3 4 0 9 6 0 6 6 7 5 7 0 3 3 3 2 8 6 9 6 6 5 7 0]}    {[1 3 2 0 5 5 0 3 9 8 7 2 9 3 6 5 2 8 2 2 3 9 7 1 8 8 3 9]}    {[4 9 2 0 3 0 7 6 9 6 1 0 7 0 5 9 2 4 3 2 1 3 2 5 4 5 5 4 0 9]}    {[5 9 8 9 4 7 4 5 2 5 3 6 3 1 6 9 3 2 1 8 0 3 5 5 4 9 7 8 2 8 9]}    {[0 8 9 4 0 8 2 6 2 1 4 2 9 0 0 0 8 0 6 4 0 0 2 3 3 8 7 1 9 6 4]}

mask_A = cellfun(@(c) c~=0, A, 'uniform', 0);
mask_B = cellfun(@(c) c~=0, B, 'uniform', 0);
mask = cellfun(@(mA, mB) mA & mB, mask_A, mask_B, 'uniform', 0)

mask = 1×12 cell array
    {[1 1 0 1 1 1 0 1 1 1 0 1 1 1 1 1 0 1 1 1 1 1 1 0 0 1 1 1 1 1]}    {[1 1 1 1 1 0 1 1 1 1 1 1 0 1 0 1 1 0 1 1 1 1 1 1 1 1 0 1]}    {[1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 0 1 1 0 1 1 1 1 1 0 0 1 1]}    {[1 1 1 1 1 1 1 0 1 1 0 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 0 0 1 1]}    {[0 1 1 1 1 0 1 1 1 1 0 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1]}    {[1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 0 1 0 1 1 1 1 1 1 1 1]}    {[1 1 1 1 1 0 1 1 1 1 0 0 0 1 1 1 1 1 1 0 1 1 1 1 1 0 1 1]}    {[1 1 1 1 1 1 1 0 0 1 1 0 1 0 1 1 1 0 1 1 0 1 1 0 1 1 1 0 1 0]}    {[1 1 1 0 1 1 0 0 1 1 1 0 1 1 1 1 1 0 1 1 1 1 0 1 1 1 1 1]}    {[1 0 1 0 1 0 1 1 1 1 1 0 1 0 1 1 1 1 0 1 1 0 1 1 1 1 1 1 0 1]}    {[1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 0 1 1 1 1 1 1 1 1]}    {[0 1 1 1 0 1 0 1 1 1 1 0 1 0 0 0 1 0 1 1 0 0 1 1 1 1 1 1 1 1 1]}

reduced_A = cellfun(@(C, m) C(m), A, mask, 'uniform', 0)

reduced_A = 1×12 cell array
    {[7 1 3 2 6 3 9 5 2 6 9 7 7 4 1 7 5 5 6 4 8 4 2 7]}    {[7 9 3 2 2 7 1 6 9 9 2 3 5 1 1 2 1 1 3 1 3 8 8]}    {[5 7 7 4 8 4 3 5 5 3 5 6 5 6 6 3 7 7 3 7 3 5 7]}    {[6 3 3 8 3 1 2 4 1 9 9 5 6 9 4 8 5 9 5 8 6 9 1 2 9 6]}    {[2 4 9 3 3 1 7 9 2 4 4 4 2 9 9 1 1 8 5 6 5 5 4 9]}    {[1 3 4 6 2 8 9 1 1 6 9 4 1 5 9 1 1 3 1 3 3 4 6 5 5]}    {[2 2 7 2 7 3 5 7 2 6 1 6 6 5 6 9 6 4 7 4 6 4]}    {[3 3 4 6 4 9 9 5 4 2 7 3 9 1 7 4 4 7 3 4 8]}    {[5 6 4 2 5 6 1 7 7 9 9 9 4 8 9 6 2 2 1 2 6 5]}    {[5 6 3 1 9 2 8 4 2 3 1 1 8 5 2 2 9 2 3 9 4 5]}    {[5 6 1 9 9 4 5 4 2 8 3 8 3 5 4 7 6 1 3 3 3 4 7 4 8 7 7 7]}    {[8 1 6 2 5 8 9 3 7 7 4 9 6 5 2 7 8 8 8 6 1]}

reduced_B = cellfun(@(C,m) C(m), B, mask, 'uniform', 0)

reduced_B = 1×12 cell array
    {[3 5 6 9 5 8 7 4 8 1 8 9 3 4 8 5 6 5 9 9 3 4 4 8]}    {[1 2 6 3 4 9 3 2 6 4 1 5 9 6 9 9 1 6 1 3 8 1 2]}    {[9 7 7 3 5 1 7 3 9 7 1 5 1 8 9 7 2 7 6 9 4 3 3]}    {[3 6 7 3 2 1 4 4 7 2 4 3 4 8 3 8 8 2 3 1 7 7 1 4 9 8]}    {[2 6 4 6 8 2 5 8 9 2 5 9 2 2 8 2 1 2 8 6 9 2 7 2]}    {[9 7 7 9 6 1 8 8 5 8 8 5 3 6 6 4 7 4 7 2 6 8 1 9 2]}    {[3 7 8 6 3 9 8 1 4 4 2 5 4 3 7 9 4 4 5 7 8 7]}    {[2 6 4 9 5 7 3 9 6 6 7 5 7 3 3 2 8 9 6 6 7]}    {[1 3 2 5 5 9 8 7 9 3 6 5 2 2 2 3 9 1 8 8 3 9]}    {[4 2 3 7 6 9 6 1 7 5 9 2 4 2 1 2 5 4 5 5 4 9]}    {[5 9 8 9 4 4 5 2 5 3 6 3 1 6 9 3 2 1 8 3 5 4 9 7 8 2 8 9]}    {[8 9 4 8 6 2 1 4 9 8 6 4 2 3 3 8 7 1 9 6 4]}

[Ebtesam Farid]

Hello @Walter Roberson

it works well with me for one cell, which was my previous question before editing. could you help me by telling me how to get the index of zeros in each cell, my idea for solving this issue, to get the index of zeros in each cell and combine them, then removing the adjacent indcies from both cells.

if there are a faster solution, I'll appreciate :)

[Walter Roberson]

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]}

[Ebtesam Farid]

it works well, Thanks @Walter Roberson so much :)

Answer 2

[dpb]

yourarray=cellfun(@(c)c(~-0),yourarray,'UniformOutput',false);

[Walter Roberson]

Pretty sure the ~-0 is a typo ;-)

[Ebtesam Farid]

Hi @dpb

unfortunately, idt doesn't give me the required result, it instead gives me one cell, instead of 12 cells removing the zeros from each one.

I edited the purpose and the required result in my main quesiton

[dpb]

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]

Yes, but this is different code; your posted code had c(~-0) not c(c~=0)