Python numpy逐行除以总和
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/16202348/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
numpy divide row by row sum
提问by Stefan Profanter
How can I divide a numpy array row by the sum of all values in this row?
如何将 numpy 数组行除以该行中所有值的总和？
This is one example. But I'm pretty sure there is a fancy and much more efficient way of doing this:
这是一个例子。但我很确定有一种奇特且更有效的方法可以做到这一点：
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:
结果：
array([[ 0. , 1. ],
[ 0.33333333, 0.66666667],
[ 0.16666667, 0.83333333]])
采纳答案by DSM
Method #1: use None(or np.newaxis) to add an extra dimension so that broadcasting will behave:
方法 #1：使用None(或np.newaxis) 添加一个额外的维度，以便广播的行为：
>>> 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:
方法#2：转置快乐：
>>> (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.)
（axis=为了简洁起见，您可以删除该部分，如果您愿意。）
Method #3: (promoted from Jaime's comment)
方法 #3：（从 Jaime 的评论中提升）
Use the keepdimsargument on sumto preserve the dimension:
使用keepdims参数 onsum保留维度：
>>> e/e.sum(axis=1, keepdims=True)
array([[ 0. , 1. ],
[ 0.33333333, 0.66666667],
[ 0.16666667, 0.83333333]])
回答by Ali
You can do it mathematically as 
.
您可以在数学上将其作为.
Here, Eis your original matrix and Dis a diagonal matrix where each entry is the sum of the corresponding row in E. If you're lucky enough to have an invertible D, this is a pretty mathematically convenient way to do things.
这里，E是您的原始矩阵，D是一个对角矩阵，其中每个条目是 中相应行的总和E。如果你有幸拥有一个 invertible D，这是一种在数学上非常方便的做事方式。
In numpy:
在 numpy 中：
import numpy as np
diagonal_entries = [sum(e[row]) for row in range(e.shape[0])]
D = np.diag(diagonal_entries)
D_inv = np.linalg.inv(D)
e = np.dot(e, D_inv)
