How to remove zeros from an element in a cell array

2021 7 11
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回答已采纳
更新时间：2021 7 12
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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??

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 采纳的回答

Walter Roberson
Walter Roberson 2021-7-11
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0 个投票

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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]}

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Walter Roberson
Walter Roberson 2021-7-11
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Ran in:
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
Ebtesam Farid 2021-7-11
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
Walter Roberson 2021-7-12
Ran in:
Ran in:
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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更多回答（1 个）

dpb
dpb 2021-7-11

0 个投票

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

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Ebtesam Farid
Ebtesam Farid 2021-7-11
编辑：Ebtesam Farid 2021-7-11
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
dpb 2021-7-12
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 2021-7-12
Yes, but this is different code; your posted code had c(~-0) not c(c~=0)

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