How can I make each cell array consistent in length?

In a cell array each element is 1Xlength. This length is not equal. I want to get the maximum length and then want to make each cell with same length by padding zeros in it.

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Jos (10584)
Jos (10584) el 18 de Mzo. de 2019
A final attempt to answer this question :-)
C = {1:4 1:2 ; 1:5 1:6 ; 1 1:3} % a m-by-n cell array
N = cellfun(@numel, C) % old lengths of cell elements
M = 3 ; % new length should be multiple of M
newN = M * ceil(N / M) % new lengths of cell elements
padfun = @(k) [C{k} zeros(1, newN(k) - N(k))] ;
C2 = arrayfun(padfun, 1:numel(C) , 'un', 0) ; % apply padding to all elements of C
C2 = reshape(C2, size(C)) % reshape (if needed)

5 comentarios

Zara Khan
Zara Khan el 18 de Mzo. de 2019
Jos (10584):
Thank you so much. You are really great. You dont know how you have saved me. I was trying this from long . Thanks again.
But I apologise for mixing up two questions here. This was my another question where I asked how to divide a cell by32.
You're welcome, and thank you. I am glad we sorted it out :-)
Zara Khan
Zara Khan el 19 de Mzo. de 2019
I am asking something more:
How can I put a condition here to check whether cell length is less than M or not . If it is less than M then only there will be padding otherwise not ?
After you calculated the new N you can use this:
tf = ~(N < M) % true for the large cells
newN(tf) = N(tf) % reset to the old lengths
Zara Khan
Zara Khan el 19 de Mzo. de 2019
Thank you . It works.

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Más respuestas (3)

tmarske
tmarske el 7 de Mzo. de 2019
Editada: tmarske el 7 de Mzo. de 2019
%set up a dummy example
tst = {[1 1], [1 1 1 1 1], [1 1 1], [1]}
%get the maximum length
maxlen = max(cellfun(@length, tst))
%pad zeros
tstPadded = cellfun(@(x)([x zeros(1, maxlen - length(x))]), tst, 'UniformOutput', false)

13 comentarios

Zara Khan
Zara Khan el 8 de Mzo. de 2019
tmarske:
I am working with mXn cell array. so I am getting this error "Size inputs must be scalar" while implementing your code.
madhan ravi
madhan ravi el 8 de Mzo. de 2019
Editada: madhan ravi el 8 de Mzo. de 2019
maxlen = max(cellfun('prodofsize', tst),[],[1 2]); % tst is m x n cell array
tstPadded = cellfun(@(x)([x zeros(1, maxlen - numel(x))]), tst, 'un', 0)
Zara Khan
Zara Khan el 8 de Mzo. de 2019
Now I am getting this : Dimension argument must be a positive integer scalar within indexing range.
madhan ravi
madhan ravi el 8 de Mzo. de 2019
Editada: madhan ravi el 8 de Mzo. de 2019
>> tst = {[1 3 2] 7 6 ; 1:6 [11 2] 6}
maxlen = max(cellfun('prodofsize', tst),[],[1 2]); % tst is m x n cell array
tstPadded = cellfun(@(x)([x zeros(1, maxlen - numel(x))]), tst, 'un', 0)
tst =
2×3 cell array
{1×3 double} {[ 7]} {[6]}
{1×6 double} {1×2 double} {[6]}
tstPadded =
2×3 cell array
{1×6 double} {1×6 double} {1×6 double}
{1×6 double} {1×6 double} {1×6 double}
>> tstPadded{:}
ans =
1 3 2 0 0 0
ans =
1 2 3 4 5 6
ans =
7 0 0 0 0 0
ans =
11 2 0 0 0 0
ans =
6 0 0 0 0 0
ans =
6 0 0 0 0 0
>>
Zara Khan
Zara Khan el 17 de Mzo. de 2019
madhan ravi :
maxlen = max(cellfun('prodofsize', tst),[],[1 2]);
Error using max
Dimension argument must be a positive integer scalar within indexing range.
M=reshape(cellfun('prodofsize',tst),1,[]);
maxlen=max(M);
Zara Khan
Zara Khan el 18 de Mzo. de 2019
Its easy. Thanks for the response. But I am always facing the same problem when using cellfun
Error using cellfun
Input #2 expected to be a cell array, was double instead.
Error in prog_32 (line 66)
tstPadded = cellfun(@(x)([x zeros(1, maxlen - numel(x))]), M, 'un', 0)
Each cell consists of consequtives 0's and 1's. Whenever I am doing padding I am getting this error.
madhan ravi
madhan ravi el 18 de Mzo. de 2019
Editada: madhan ravi el 18 de Mzo. de 2019
I have no problem with the code , I have no idea why you get an error upload the code & data(as .mat file) that your trying.
A=cell(numImages,8);
B{k,j}=profile;
A=B;
len = cellfun(@length, B);
len1 = 32 * ceil(len/32);
width_needed=cell(numImages,8);
width_needed=len1-len;
I am making each cell a multiple of 32. Now my task is to pad extra zero after the multiplication. And I want to keep them as cell format , no merging will be done. Here the profile is varying in length. Basically it consists of consequtives 0's and 1's. I am attaching few pictures here too.
Your code makes little sense. You overwrite A and width_needed.
I thought you needed the maximum of the lengths, so why don't you apply max?
Zara Khan
Zara Khan el 18 de Mzo. de 2019
Editada: Zara Khan el 18 de Mzo. de 2019
I dont want maxlength . I am making each cell a multiple of 32 . so i just want to add the extra places by zeros. maxlength will only find out maximum length among all.
Your question reads otherwise ... " I want to get the maximum length and then want to make each cell with same length by padding zeros in it.

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Jos (10584)
Jos (10584) el 7 de Mzo. de 2019
Editada: Jos (10584) el 8 de Mzo. de 2019
If you make them the same length, you can also store them in a matrix. In that case, my PADCAT function is your friend :-)
C = {[1 2] ; 1 ; [1 2 3]}
[M, tf] = padcat(C{:}) % pads with NaNs
M(~tf) = 0
PADCAT can be found on the File Exchange, for free:
(edited answer)

6 comentarios

Zara Khan
Zara Khan el 7 de Mzo. de 2019
what it will be if there is C{i,j} like thing ?
You mean a M-by-N cell array? I suggest you try it! :-)
C = {[1 2] 1 ; [1 2 3] [1 2]} % a 2-by-2 cell array with row vectors
Zara Khan
Zara Khan el 7 de Mzo. de 2019
yes yes M×N cell array .
Zara Khan
Zara Khan el 8 de Mzo. de 2019
This is shwoing wrong syntax : (M,tf) = padcat(C{:})
My mistake, should be square brackets of course ... (I edited my answer).
(assuming you also downloaded and installed the function)
Zara Khan
Zara Khan el 18 de Mzo. de 2019
Jos (10584):
by using padcat I am getting a matrix where all coloumns are merged. That I dont want. I want to keep each cell as it is, just want want to add extra zeros. Like , first cell is 1X16, second is 1X31 and so on. I want to work on each indivisually. By adding extra zeros the first cell will be suppose 1X32 and so on. Remember I am dealing with mXn cell array.

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C = {1:3 4 ; 5:9 10:12 ; 13:14 15} % a m-by-n cell array
N = cellfun(@numel, C)
maxN = max(N(:))
padfun = @(v) [v zeros(1, maxN - numel(v))] ;
C2 = cellfun(padfun, C , 'un', 0)

3 comentarios

Zara Khan
Zara Khan el 18 de Mzo. de 2019
Each cell array consists of consequtives 0's and 1's. Padding with extra zero showing me error
size input must be scalar.
I do not get this error in the above code for a cell array like this
C = {[1 0 0 1], [0 1] ; [1 0 1 0], [0 0 1]}
You should give more details about the error and the input ...
A=cell(numImages,8);
B{k,j}=profile;
A=B;
len = cellfun(@length, B);
len1 = 32 * ceil(len/32);
width_needed=cell(numImages,8);
width_needed=len1-len;
I am making each cell a multiple of 32. Now my task is to pad extra zero after the multiplication. And I want to keep them as cell format , no merging will be done. Here the profile is varying in length. Basically it consists of consequtives 0's and 1's. I am attaching few pictures here too.

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el 7 de Mzo. de 2019

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el 19 de Mzo. de 2019

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