概述

numpy divide row by row sum

我的问题是我正在处理一个大矩阵：(96582,350138)

当从链接的帖子应用操作时,它会膨胀我的记忆,因为返回的矩阵是密集的.

所以这是我的第一次尝试：

for row in counts:
    row = row / row.sum()

不幸的是,这根本不会影响矩阵,所以我想出了第二个想法来创建一个新的csr矩阵并使用vstack连接行：

from scipy import sparse
import time

start_time = curr_time = time.time()
mtx = sparse.csr_matrix((0,counts.shape[1]))
for i,row in enumerate(counts):
   prob_row = row / row.sum()
   mtx = sparse.vstack([mtx,prob_row])
   if i % 1000 == 0:
      delta_time = time.time() - curr_time
      total_time = time.time() - start_time
      curr_time = time.time()
      print('step: %i,total time: %i,delta_time: %i' % (i,total_time,delta_time))

这很好用,但经过一些迭代后,它变得越来越慢：

step: 0,total time: 0,delta_time: 0
step: 1000,total time: 1,delta_time: 1
step: 2000,total time: 5,delta_time: 4
step: 3000,total time: 12,delta_time: 6
step: 4000,total time: 23,delta_time: 11
step: 5000,total time: 38,delta_time: 14
step: 6000,total time: 55,delta_time: 17
step: 7000,total time: 88,delta_time: 32
step: 8000,total time: 136,delta_time: 47
step: 9000,total time: 190,delta_time: 53
step: 10000,total time: 250,delta_time: 59
step: 11000,total time: 315,delta_time: 65
step: 12000,total time: 386,delta_time: 70
step: 13000,total time: 462,delta_time: 76
step: 14000,total time: 543,delta_time: 81
step: 15000,total time: 630,delta_time: 86
step: 16000,total time: 722,delta_time: 92
step: 17000,total time: 820,delta_time: 97

有什么建议？知道为什么vstack越来越慢？

您只需使用@L_403_1@即可完成操作：

>>> res = counts.multiply(1 / counts.sum(1))  # multiply with inverse
>>> res.todense()
matrix([[ 0.33333333,0.,0.66666667],[ 0.,1.        ],[ 0.26666667,0.33333333,0.4       ]])

但是使用np.lib.stride_tricks.as_strided进行你想要的操作(相对高效)也很容易.这个as_strided函数还允许对数组执行更复杂的操作(如果您的情况没有方法或函数).

例如,使用scipy documentation的示例csr：

>>> from scipy.sparse import csr_matrix
>>> import numpy as np
>>> row = np.array([0,1,2,2])
>>> col = np.array([0,2])
>>> data = np.array([1.,3,4,5,6])
>>> counts = csr_matrix( (data,(row,col)),shape=(3,3) )
>>> counts.todense()
matrix([[ 1.,2.],3.],[ 4.,5.,6.]])

您可以将每行除以它的总和,如下所示：

>>> row_start_stop = np.lib.stride_tricks.as_strided(counts.indptr,shape=(counts.shape[0],2),strides=2*counts.indptr.strides)
>>> for start,stop in row_start_stop:   
...    row = counts.data[start:stop]
...    row /= row.sum()
>>> counts.todense()
matrix([[ 0.33333333,0.4       ]])

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