comparing matices that contain numbers and NaNs

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I have 2 cell matrices R and W and I want to check if they contain the same elements. Specifically, I want to compare R which is obtained from
[N,T,R] = xlsread( filename); %
with another matrix W.Both matrices are cell matrices and can also contain NaN entries.
My goal: If my calculations are correct then
kk=W(7 ,3:end); should contain the same elements with
zz1=R(7,3:end);
or zz2=R(8,3:end);
So to test this I do something like
if zz==xx1 | zz== xx2
'OK' else 'Not OK' EndThe problem is that
zz==xx1
??? Undefined function or method 'eq' for input arguments of type 'cell'.
So I use cell2mat(zz)==cell2mat(xx1) and it works.
But if zz and xx1 contain NaN then for some reason I cannot use the cell2mat() approach
I am looking for a general approach that will enable me to do that comparison whether I have NaN and numbers or only numbers’
thanks
  1 Comment
Jan
Jan on 5 Jul 2012
@Readers: It seems like anonet wants to compare cells, which contain numerical and string values.

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Accepted Answer

Andrei Bobrov
Andrei Bobrov on 5 Jul 2012
Edited: Andrei Bobrov on 5 Jul 2012
variant of comparison cell arrays (edit)
A = {...
[84] [ 93] [ 109]
[99] [NaN] [ 62]
[29] [123] 'dgdgdg'};
B = {...
[84] [ 93] [ 109]
[99] [NaN] [ 62]
[29] [123] 'dgdgdg'};
A1=A;
B1=B;
A1(cellfun(@(x)all(isnan(x)),A)) = {0};
B1(cellfun(@(x)all(isnan(x)),B)) = {0};
out = isequal(A1,B1);
  4 Comments
Andrei Bobrov
Andrei Bobrov on 5 Jul 2012
Hi Antonet! Thank you for "Accepted answer", but I mean that Jan's answer better.

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More Answers (2)

Jan
Jan on 5 Jul 2012
C1 = {'String1', 23, -17, NaN, 1:29);
C2 = {'String2', 24, -17.3, 21, 1:30);
C3 = {'String1', 23, -17, NaN, 1:29);
% |C1==C2| fails!
isequalwithequalnans(C1, C2)
isequalwithequalnans(C1, C3)
I cannot test this currently. When NaNs should be assumed to be different, isequal would do it.
  4 Comments

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Kevin Claytor
Kevin Claytor on 4 Jul 2012
My guess from searching the help docs for 'cell' gives;
m = cell2mat(c) converts a multidimensional cell array c with contents of the same data type into a single matrix, m.
  3 Comments
Jan
Jan on 5 Jul 2012
@antonet: Please provide the necessary information in the original question, because readers in this forum will read it at first. Therefore you can edit the question.
The code you have posted is hard to read. Please read the instructions about code formatting again.

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