replacing the NaN with specific values
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Dear all,
I have the following matrix
A={'country' 'values'
'E' '1'
'E' [NaN]
'E' [NaN]
'E' '2'
'E' [NaN]
'E' [NaN]
'E' [NaN]
'E' [NaN]
'E' [NaN]
'E' '3'
'E' [NaN]
'E' [NaN]
'E' [NaN]
'E' [NaN]
'EE' [NaN]
'E' '4'
'I' '1'
'I' [NaN]
'I' [NaN]
'I' '2'
'I' [NaN]
'I' [NaN]
'I' [NaN]
'I' [NaN]
'I' [NaN]
'I' '3'
'I' [NaN]
'I' [NaN]
'I' [NaN]
'I' [NaN]
'I' [NaN]
'I' '4'
'I' [NaN]
'I' [NaN]
'I' [NaN]
'I' [NaN]
'I' [NaN]
'I' '5'
'K' '1'
'K' [NaN]
'K' [NaN]
'K' '2'
'K' [NaN]
'K' [NaN]
'K' [NaN]
'K' [NaN]
'K' [NaN]
'K' [NaN]
'K' [NaN]
'K' '3'
'K' [NaN]
'K' [NaN]
'K' [NaN]
'K' '4'
'K' [NaN]
'K' [NaN]
'K' [NaN]
'K' [NaN]
'K' [NaN]
'K' '5'
};
The first column describes the countries and the second the values corresponding to each country.
For each country the initial value is always '1'. For each country I want to replace the NaNs with the value that is before them.
So for the first country I want to have
'E' '1'
'E' '1'
'E' '1'
'E' '2'
'E' '2'
'E' '2'
'E' '2'
'E' '2'
'E' '2'
'E' '3'
'E' '3'
'E' '3'
'E' '3'
'E' '3'
'EE' '3'
'E' '4'
for country I I want to have
'I' '1'
'I' '1'
'I' '1'
'I' '2'
'I' '2'
'I' '2'
'I' '2'
'I' '2'
'I' '2'
'I' '3'
'I' '3'
'I' '3'
'I' '3'
'I' '3'
'I' '3'
'I' '4'
'I' '4'
'I' '4'
'I' '4'
'I' '4'
'I' '4'
'I' '5'
and so so..
I am trying to come up with a code but I got stuck
Is there any way of doing that
My real vector contains 300 countries, so doing manual is quite tedious
thanks in advance
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采纳的回答
Andrei Bobrov
2013-3-14
编辑:Andrei Bobrov
2013-3-14
A1 = A(2:end,:);
t = ~cellfun(@isnan,A1(:,2));
k = A1(t,2);
A(2:end,2) = k(cumsum(t));
added on comment
A1 = A(2:end,2);
t = cellfun(@ischar,A1);
A1(t) = cellfun(@str2double,A1(t),'un',0);
t1 = ~cellfun(@isnan,A1);
k = A1(t1);
A(2:end,2) = k(cumsum(t1));
5 个评论
Khaing Zin Htwe
2016-5-10
To replace integer zeros to NaN vaules in Array ,how can I do it ,sir.I have 141x3084 array that include NaN values unfortunately. After reducing dimensions to give input to ANFIS classifier, i get the result as NaN,seriously. Please help me ,sir , Andrei Bobrov.
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