I have a cell type variable with 5000 rows and 8 cells. For example:
a={182 1999 63,8 229 [] 30 [] 1
185 1999 44,5 123 19,7 51 [] []
194 1999 50,7 273 [] 44 [] 1
195 1999 53,2 [] [] [] [] []}
And I would like to substitute the blank entries in the last column only by zero so I would get:
a={182 1999 63,8 229 [] 30 [] 1
185 1999 44,5 123 19,7 51 [] 0
194 1999 50,7 273 [] 44 [] 1
195 1999 53,2 [] [] [] [] 0}
I tried something like this, but is not working:
a(cellfun(@isempty,a(:,8))) = {0};
Can someone help me? Thank you

1 commentaire

dpb
dpb le 14 Août 2014
Ahhh....my old eyes missed the missing column addressing index...couldn't see the difference at first.

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 Réponse acceptée

Adam
Adam le 14 Août 2014

2 votes

a( cellfun(@isempty,a(:,8)),8 ) = {0}

Plus de réponses (1)

dpb
dpb le 14 Août 2014

1 vote

a(cellfun(@isempty,a(:,end)),end)={0}
worked here...had to convert the ',' decimal points to '.' to enter the array, however...

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le 14 Août 2014

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dpb
le 14 Août 2014

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