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Replacing multiple values at specific indices in a matrix with a single value

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bsriv
bsriv 2019-9-8
编辑： Adam Danz 2019-9-8
Hi,
I have a 91282x1 (d) array with a lot of different values. I would like to replace certain values (all at multiple indices) with 1 (there is no value of 1 in the original matrix) and the rest with zeros- but in a much more efficient way. Here's what I have
d(d==7201 | d==7205 | d=7207 | d==7208 | d==7209 | d==7210 | d==7215)=1;
d.(d~=1)=0
I know this is really short but I have to do this for a lot and I'm wondering if there's a more efficient way to make the mask by using zeros and then ID'ing the appropriate indices... or something
Thanks

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采纳的回答

David Hill
David Hill 2019-9-8
values = [7201,7205,7207,7208,7209,7210,7215];
d=sum(d==values,2);
The above should be easier.
  2 个评论
Adam Danz
Adam Danz 2019-9-8
编辑：Adam Danz 2019-9-8
Just be aware that this method requires 'd' to be a column vector and 'values' to be a row vector. Also, if there are duplicate values in "values" then you will not have just 1s and 0s.
The ismember() approach avoids both of those potential issues.
One improvement to this method would be to force the inputs to be the correct shape (row or column).
d=sum(d(:)==values(:).',2);

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Adam Danz
Adam Danz 2019-9-8
编辑：Adam Danz 2019-9-8
ismember() will return a logical array the same size as 'd'. If you need ones and zeros rather than true and false, you can convert that to double.
d0 = ismember(d,[7201 7205 7207 7208 7209 7210 7215]);
% If you need double rather than logical
d0 = double(ismember(d,[7201 7205 7207 7208 7209 7210 7215]); );

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