Replace values in one column in multiple cells in array

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mzaya
mzaya 2016-2-23
评论： Amos 2016-2-23
I'm trying to replace the 5th column in each cell in a cell array with the 5th column of each cell from another cell array. I made the following function, which does this but also replaces the values in all other columns with 0. How do I do this without deleting all other values from the other columns. The function is:
function [X]=replace_cells(cell)
X={};
for i=1:length(cell)
X{i}(:,[5])=cell{i}(:,[5]);
end
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Amos
Amos 2016-2-23
编辑：Amos 2016-2-23
then I can't see why your code should not work ... if you post a full example (including test cell arrays), people can have a look at the issue
edit: wait, you shouldn't initialize X as empty cell array, but rather give it as an input to your function...
mzaya
mzaya 2016-2-23
Yes that was the main issue, thanks! New code is as follows and it works:
function X=replace_cells(c,X)
for i=1:length(c)
X{i}(:,5)=c{i}(:,5);
end
end

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Amos
Amos 2016-2-23
Maybe try:
function [X]=replace_cells(cell,X)
for i=1:length(cell)
X{i}(:,[5])=cell{i}(:,[5]);
end
  2 个评论
Guillaume
Guillaume 2016-2-23
编辑：Guillaume 2016-2-23
Please replace the variable name cell with something that is not a matlab function. While it does not cause any problem in this short function, it is still confusing to the casual reader.
For that matter, using meaningful names for all variables would be much better. How about:
function destination = replace_cell(destination, source, column)
assert(size(destination) == size(source));
for cidx = 1:numel(source)
destination{cix}(:, column) = source{cidx}(:, column);
end
end
Also note that I'm using numel instead of length. My version works regardless of the shape of source and destination (vector, 2D, 3D, ND) as long as they're the same size. The original code with length would utterly fail on 2D or ND cell arrays.

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