numpy布尔索引

Question

如何使用numpy广播重新编写这个python循环？你知道吗

>>> values.shape
(50000,)
>>> tests.shape  # booleans
(200, 50000)

>>> extracted = values[tests]
# FAILES

>>> extracted = values[test] for test in tests]
>>> extracted.shape
(200,)
>>> extracted[0].shape
(33,)
>>> extracted[1].shape
(468,)

花哨/布尔索引在这里不起作用。你知道吗

Answer 1

这样怎么样？ 如果您只想快速获得每行的大小。你知道吗

>>> values.shape
(50000,)
>>> tests.shape  # booleans
(200, 50000)

>>> extracted = np.prod((tests,values))
>>> extracted.shape
(200,50000)
>>> rows, cols = extracted.nonzero()
>>> cols[rows==0].shape
(33,)
>>> cols[rows==1].shape
(468,)