给出一个列索引 i 和一个稀疏的csr矩阵 X ，我想获得在第i列上所有值为1的所有行的均值向量。 / p>
这是我想出的解决方案，但我认为它的速度很慢：
# a) create a mask of rows containing True if this column was > 0 or False otherwise
mask = (X[:,i] > 0).transpose().toarray()[0]
# b) now get the indices of these rows as list
indices_of_row = list(np.where(mask > 0)[0])
if len(indices_of_row) == 0:
return
# c) use the indices of these rows to create the mean vector
mean_vector = X[indices_of_row,].mean(axis=0)
有什么想法可以使它更有效或更可读？
编辑：我想避免在整个矩阵上调用toarray（）
以下是三个相对较快的解决方案：
from scipy import sparse
import numpy as np
def pp():
m = np.maximum.reduceat(a.indices==i,a.indptr[:-1])
cnt = np.count_nonzero(m)
m = m.repeat(np.diff(a.indptr))
return np.bincount(a.indices[m],a.data[m],a.shape[1])/cnt
def qq():
idx = a.indptr.searchsorted(*(a.indices==i).nonzero(),"right")-1
return np.bincount(
np.concatenate([a.indices[a.indptr[i]:a.indptr[i+1]] for i in idx]),np.concatenate([a.data[a.indptr[i]:a.indptr[i+1]] for i in idx]),a.shape[1]) / len(idx)
def mm():
idx = (a@(np.arange(a.shape[1])==i))!=0
return idx/np.count_nonzero(idx)@a
def OP():
# a) create a mask of rows containing True if this column was > 0 or False otherwise
mask = (a[:,i] > 0).transpose().toarray()[0]
# b) now get the indices of these rows as list
indices_of_row = list(np.where(mask > 0)[0])
if len(indices_of_row) == 0:
return
# c) use the indices of these rows to create the mean vector
return a[indices_of_row,].mean(axis=0)
from timeit import timeit
n = 1000
a = sparse.random(n,n,format="csr")
i = np.random.randint(0,n)
print("mask ",timeit(pp,number=1000),"ms")
print("concat",timeit(qq,"ms")
print("matmul",timeit(mm,"ms")
print("OP ",timeit(OP,"ms")
assert np.allclose(pp(),OP())
assert np.allclose(qq(),OP())
assert np.allclose(mm(),OP())
样品运行：
mask 0.08981675305403769 ms
concat 0.04179211403243244 ms
matmul 0.14177833893336356 ms
OP 0.9761617160402238 ms
我认为就足够了。
Y = X.toarray()
MeanVec = Y[Y[:,i] > 0].mean(axis=1)
已编辑
X.mean(axis=1)[(X.getcol(i) > 0).toarray()]