How to replace cell values with consecutive values related to their index

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Dc215905
Dc215905 2021-5-28
评论： dpb 2021-5-29
For example:
a = 1x3 cell array
columns 1 through 3
{1x1} {1x5} {1x3}
I'd like to be able to replace the values in each cell array with consecutive values so that:
a = 1x3 cell array
{[1]} {[2 3 4 5 6]} {[7 8 9]}
Thanks!

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采纳的回答

dpb
dpb 2021-5-29
c=[{cell(1,1)} {cell(1,5)} {cell(1,3)}]; % original cell array
% engine
n=cellfun(@numel,c);
v=1:sum(n);
i1=1;
for i=1:numel(n)
i2=i1+n(i)-1;
c{i}=v(i1:i2);i1=i2+1;
end
results in
>> c
c =
1×3 cell array
{[1]} {1×5 double} {1×3 double}
>> c{:}
ans =
1
ans =
2 3 4 5 6
ans =
7 8 9
>>
There's probably a clever accumarray() or arrayfun() syntax to avoid the explicit loop, but nothing came to me quickly...
  1 个评论
Dc215905
Dc215905 2021-5-29
Thank you!
That's way less convoluted than what I cam up with:
b = 0;
in = 1;
for i = 1:length(CellMatrix)
c = b(end);
b = [];
for in = 1:length(CellMatrix{i})
length(CellMatrix{i});
b(end+1) = c + 1;
c = c +1;
end
in = 1;
CellMatrix(i) = {b};
end

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更多回答（1 个）

Stephen23
Stephen23 2021-5-29
编辑：Stephen23 2021-5-29
Ran in:
Simpler:
C = {cell(1,1),cell(1,5),cell(1,3)}
C = 1×3 cell array
{1×1 cell} {1×5 cell} {1×3 cell}
N = cellfun(@numel,C);
D = mat2cell(1:sum(N),1,N)
D = 1×3 cell array
{[1]} {[2 3 4 5 6]} {[7 8 9]}
  1 个评论
dpb
dpb 2021-5-29
That's what I intended, but kept muffing the mat2cell syntax...it's always thrown me for a loop for some reason -- rarely ever use it I guess is likely cause.

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