Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
824 views
in Technique[技术] by (71.8m points)

python - numpy divide row by row sum

How can I divide a numpy array row by the sum of all values in this row?

This is one example. But I'm pretty sure there is a fancy and much more efficient way of doing this:

import numpy as np
e = np.array([[0., 1.],[2., 4.],[1., 5.]])
for row in xrange(e.shape[0]):
    e[row] /= np.sum(e[row])

Result:

array([[ 0.        ,  1.        ],
       [ 0.33333333,  0.66666667],
       [ 0.16666667,  0.83333333]])
See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

Method #1: use None (or np.newaxis) to add an extra dimension so that broadcasting will behave:

>>> e
array([[ 0.,  1.],
       [ 2.,  4.],
       [ 1.,  5.]])
>>> e/e.sum(axis=1)[:,None]
array([[ 0.        ,  1.        ],
       [ 0.33333333,  0.66666667],
       [ 0.16666667,  0.83333333]])

Method #2: go transpose-happy:

>>> (e.T/e.sum(axis=1)).T
array([[ 0.        ,  1.        ],
       [ 0.33333333,  0.66666667],
       [ 0.16666667,  0.83333333]])

(You can drop the axis= part for conciseness, if you want.)

Method #3: (promoted from Jaime's comment)

Use the keepdims argument on sum to preserve the dimension:

>>> e/e.sum(axis=1, keepdims=True)
array([[ 0.        ,  1.        ],
       [ 0.33333333,  0.66666667],
       [ 0.16666667,  0.83333333]])

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...