Stack Overflow Asked by user13984029 on December 16, 2021
I have an array and need to normalize it in a way that the results will be numbers between 0 and 1. I already normalized the entire array as follows:
C = A / A.max(axis=0)
print(C)
____________________________________________________________________
[[0. 0.05263158 0.1 0.14285714 0.18181818 0.2173913 ]
[0.33333333 0.36842105 0.4 0.42857143 0.45454545 0.47826087]
[0.66666667 0.68421053 0.7 0.71428571 0.72727273 0.73913043]
[1. 1. 1. 1. 1. 1. ]]
But now I have to normalize by column and by line. How can I do that with axis reduction? If there is a better way to what I did, suggest me alterations.
My expected result is two arrays with the values normalized. One considering the columns and the other by the lines
This is my data
A = [[ 0 1 2 3 4 5]
[ 6 7 8 9 10 11]
[12 13 14 15 16 17]
[18 19 20 21 22 23]]
You skip the minimum part. Normally a 0-1 normalization demanding subtracting a minimum value from denominator and numerator. https://stats.stackexchange.com/questions/70801/how-to-normalize-data-to-0-1-range
import numpy as np
A = np.matrix([[ 0, 1, 2, 3, 4, 5],
[ 6, 7, 8, 9, 10, 11],
[12, 13, 14, 15, 16, 17],
[18, 19, 20, 21, 22, 23]])
(A-A.min(axis=1))/(A.max(axis=1)-A.min(axis=1))
(A-A.min(axis=0))/(A.max(axis=0)-A.min(axis=0))
Answered by polkas on December 16, 2021
My expected result is two arrays with the values normalized. One considering the columns and the other by the lines
a = np.array([[ 0, 1, 2, 3, 4, 5],
[ 6, 7, 8, 9, 10, 11],
[12, 13, 14, 15, 16, 17],
[18, 19, 20, 21, 22, 23]])
If
c = a / a.max(axis=0)
gives you what you want for the columns then
d = a / a.max(axis=1)[:,None]
will suffice for the rows.
>>> d.round(4)
array([[0. , 0.2 , 0.4 , 0.6 , 0.8 , 1. ],
[0.5455, 0.6364, 0.7273, 0.8182, 0.9091, 1. ],
[0.7059, 0.7647, 0.8235, 0.8824, 0.9412, 1. ],
[0.7826, 0.8261, 0.8696, 0.913 , 0.9565, 1. ]])
Answered by wwii on December 16, 2021
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