我想迭代CSR矩阵的行并将每个元素除以行的总和,类似于此处：
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？
解决方法
vstack是一个O(n)操作,因为它需要为结果分配内存,然后将作为参数传递的所有数组的内容复制到结果数组中.
您只需使用multiply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 ]])