I have a matrix with integers and I need to replace all appearances of 2 with -5. What is the most efficient way to do it? I made it the way below, but I am sure there is mo
Here's a trivial, unoptimised, probably slow implementation of changem
from the Mapping Toolbox.
function mapout = changem(Z, newcode, oldcode)
% Idential to the Mapping Toolbox's changem
% Note the weird order: newcode, oldcode. I left it unchanged from Matlab.
if numel(newcode) ~= numel(oldcode)
error('newcode and oldcode must be equal length');
end
mapout = Z;
for ii = 1:numel(oldcode)
mapout(Z == oldcode(ii)) = newcode(ii);
end
end
The Martin B's method is good if you are changing values in vector. However, to use it in matrix you need to get linear indices.
The easiest solution I found is to use changem
function. Very easy to use:
mapout = changem(Z,newcode,oldcode)
In your case: newA = changem(a, 5, -2)
More info: http://www.mathworks.com/help/map/ref/changem.html
Try this:
a(a==2) = -5;
The somewhat longer version would be
ind_plain = find(a == 2);
a(ind_plain) = -5;
In other words, you can index a matrix directly using linear indexes, no need to convert them using ind2sub
-- very useful! But as demonstrated above, you can get even shorter if you index the matrix using a boolean matrix.
By the way, you should put semicolons after your statements if (as is usually the case) you're not interested in getting the result of the statement dumped out to the console.
find
is not needed in this case.
Use logical indexing instead:
a(a == 2) = -5
In case of searching whether a matrix is equal to inf
you should use
a(isinf(a)) = -5
The general case is:
Mat(boolMask) = val
where Mat
is your matrix, boolMask
is another matrix of logical
values, and val
is the assignment value