Fastest way to replace values in an array if they equal a certain value?
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Eric Chadwick
on 26 May 2019
Commented: Eric Chadwick
on 28 May 2019
Hello,
I have a very large array (4 billion x 2). I frequently need to update the values in column two based on a value from another list. This is how I currently have it:
for i = 1:size(net,1)
imcoords(imcoords(:,2)==net{i,7},2) = i;
end
Where net is a cell array and imcoords is the Nx2 array. Just as it is written, I need to replace all values of the second column of imcoords that equal the value in the 7th column of the cell array with the index of the current row of the cell array we are in. With the method I have right now this takes way too long and is the current bottleneck of my code since I need to do this frequently.
Does anyone have any ideas to make this quicker?
Thanks you!
Eric
2 Comments
John D'Errico
on 26 May 2019
Don't forget, to NEVER test for exct equality of floating point numbers. If you do not learn to use a tolerance, then your next plaintive question will be why does my code not work some of the time?
Accepted Answer
John D'Errico
on 26 May 2019
Since the numbers are all integers, just do it as a lookup table. That makes the replacement a simple index into a vector. Done in milliseconds.
3 Comments
Walter Roberson
on 26 May 2019
Your net{i,7} appear to be scalar . If so then
net7 = [net{:,7}];
look7(net7) = 1:length(net7);
imcoords(:,2) = look7(imcoords(:,2));
This depends upon:
- that the net{:,7} are scalars
- that there are no duplicate values (if there are then the replacement is order dependent)
- that the existing values are all positive integers
- that all of the existing values are being replaced
- that the existing values all exceed the number of rows in net (otherwise you could get multiple replacements with your code
If not all of the existing values are being replaced, then you can initialize
look7 = 1 : maximum_expected_value
look7(net7) = 1:length(net7);
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