# bsxfun - @minus - how to divide by maximum

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Ko Fa on 28 Dec 2020
Commented: Ko Fa on 28 Dec 2020
Hello everyone,
I have two arrays:
a = [1; 2; 3]
b = [1; 2; 3; 4]
out = abs(bsxfun(@minus,a,b')./a)
I am getting this:
out = [0 1 2 3
0.5 0 0.5 1.0
0.67 0.33 0 0.33]
I am looking for a way to divide not simply by ./a, but to divide by the maximum number in the numerator. So instead of calculating abs(1-3)/1=2 for out(1,3) I would like the calculation to be abs((1-3)/3)=0.67.
Any help is greatly appreciated.

Bruno Luong on 28 Dec 2020
Edited: Bruno Luong on 28 Dec 2020
Do do poor job to explain, up to everyone guess
>> a = [3; 5; 1;]
a =
3
5
1
>> b = [3; 2; 3; 4]
b =
3
2
3
4
>> abs(a-b')./max(a,b')
ans =
0 0.3333 0 0.2500
0.4000 0.6000 0.4000 0.2000
0.6667 0.5000 0.6667 0.7500
>> abs(bsxfun(@minus,a,b'))./bsxfun(@max,a,b')
ans =
0 0.3333 0 0.2500
0.4000 0.6000 0.4000 0.2000
0.6667 0.5000 0.6667 0.7500
Ko Fa on 28 Dec 2020
My bad, I just didnt have the apostrophe in max(a,b') after b and was wondering about dimension problems in my original data. Thanks.

Geoff Hayes on 28 Dec 2020
Ko - if you want to divide by the maximum of a, couldn't you just do
out = abs(bsxfun(@minus,a,b')./max(a))
?
Ko Fa on 28 Dec 2020
4 is the maximum numerator, so the number I want to divide with could be from either a or b. whatever the maximum numerator is. hope that helps
Thanks, I'll be taking bsxfun away then.