double array of 0's and 1's conversion help

I have an array of 0's and 1's
Im trying to convert the 0's to nans and 1's to another value.
A(find(A==1)=256; %this part works
A(find(A==0)=NaN; %this part doesn't
when i try to replace the 0's, it replaces everything in the array with NaN, even though find(A==0) does return only the indices of where that array has a 0 value.
is there technical thing I'm missing here?

Respuestas (2)

Walter Roberson
Walter Roberson el 20 de Jun. de 2012
A(A==1) = 256;
A(A==0) = NaN;
If you want to live a life of confusion, and you only have 0 and 1s in the matrix,
A = A ./ A * 256;

4 comentarios

the cyclist
the cyclist el 20 de Jun. de 2012
Walter, I recognize that your syntax avoids the unnecessary find(), but what's the use case in which your syntax works, but David's doesn't? I'm puzzled.
Tom
Tom el 20 de Jun. de 2012
He may have left the bracket off like in the question?
Walter Roberson
Walter Roberson el 20 de Jun. de 2012
Reduces the steps that can go wrong ?
David C
David C el 20 de Jun. de 2012
sorry, i just forgot to put the brackets in while typing, i have them in my actual code

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David C
David C el 20 de Jun. de 2012
just tried
A(A==1) = 256;
A(A==0) = NaN;
and the same thing happened. The A==1 part worked, but as soon as it tried to logical index the ones that were 0, it replaced everything with NaN

2 comentarios

Sean de Wolski
Sean de Wolski el 20 de Jun. de 2012
What happens if you run this:
A = double(rand(10)>0.5);
A(A==1) = 256;
A(A==0) = NaN;
A
Walter Roberson
Walter Roberson el 20 de Jun. de 2012
Try breaking it down and do some experiments to see which step is going wrong:
A
T = A == 0
B = A;
A(T) = NaN
B(1) = NaN
C = rand(size(A));
C(T) = NaN

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