# replacing NaN with zeros in a cell column of strings

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antonet on 24 Apr 2013
Edited: Shane Hagen on 22 Mar 2015
Dear all,
I have this column
A={[NaN]
[NaN]
[NaN]
[NaN]
[NaN]
'ka'
[NaN]
[NaN]
'jdfksjkg'
'ldsiu'
[NaN]
[NaN]
'jdfkkg'
'jdfkkg'
};
and I want to replace the NaN with zeros that is ,
A={[0]
[0]
[0]
[0]
[0]
'ka'
[0]
[0]
'jdfksjkg'
'ldsiu'
[0]
[0]
'jdfkkg'
'jdfkkg'
};
Is there a way of doing that in matlab?

Jan on 25 Apr 2013
Are the NaNs the only non-CHAR elements? Then:
A(~cellfun('isclass', A, 'char')) = {0};

#### 1 Comment

antonet on 25 Apr 2013
thanks Jan. I have columns of dimension 200000 by 1. so I want to avoid using loops

Wayne King on 24 Apr 2013
One way is just with a for loop:
for ii = 1:length(A)
if isnan(A{ii})
A{ii} = 0;
end
end

Show 1 older comment
Image Analyst on 25 Apr 2013
Wow, 5 lines is too complicated?!? Any shorter way is likely to be more obtuse, not as straightforward and easy to follow, and harder to maintain because it'd be more cryptic. What's wrong with a loop, especially for only 14 variables? I mean, it's not like you have tens of millions or anything, so speed is definitely NOT going to be an issue.
Jan on 25 Apr 2013
This is the most simplest method. +1
Shane Hagen on 22 Mar 2015
isnan doesnt work for cells...can anyone help?
If i separate out the individual cell with NaN it works but not in the loop
ix=cellfun(@isempty,spiketimes);
spiketimes(ix)={0} ; %which I wish would just put 0's
for ii = 1:length(spiketimes)
if isnan(spiketimes{ii})
spiketimes{ii} = 0;
end end
results = spiketimes{j};
%n = length(results(j,:));
plot([results{:} ; results{:}], [ones(1,n)*j-1;ones(1,n)*j],'k-'); end; end;
Undefined function 'isnan' for input arguments of type 'cell'.
this is at the end of my code..would appreciate any help

Cedric Wannaz on 25 Apr 2013
Edited: Cedric Wannaz on 25 Apr 2013
You can go for something like:
A(cellfun(@(x)all(x~=x), A)) = {0} ;
but it is just "cosmetic" (in some sense) and the loop proposed by Wayne (with a little update so it manages well non-scalar content) is likely to be more efficient.