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python for 循环中使用星号(*) 实现分组展开列表

时间:2019-02-15 14:10:03

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python for 循环中使用星号(*) 实现分组展开列表

import numpy as npa = np.random.uniform(0, 1, (5, 6))print(a)for *b, c, d in a:print(b, c, d)for b, c, *d in a:print(b, c, d)

三个print的结果如下:

[[2.59082666e-01 9.33421608e-02 3.14394861e-01 5.21211733e-01

1.45397325e-01 9.50105984e-04]

[1.06471904e-01 2.08233001e-01 9.45687573e-01 4.77635986e-01

1.15698356e-01 8.77691012e-01]

[2.67165641e-01 9.69817760e-01 1.90024148e-02 4.34003087e-01

3.78529636e-01 7.91454564e-01]

[5.45704983e-01 5.27884108e-01 2.68554314e-01 2.31240267e-03

3.58575802e-02 3.29896612e-01]

[2.69259422e-01 6.62100724e-01 2.30178897e-01 4.78810788e-02

2.58792575e-02 9.19565646e-01]]

[0.259082665748668, 0.0933421607590641, 0.31439486129202887, 0.521211733137794]0.14539732546645678 0.0009501059838643577

[0.10647190379252569, 0.20823300141803736, 0.9456875734737963, 0.47763598619698655]0.11569835557059593 0.8776910120723697

[0.2671656406671584, 0.9698177601248633, 0.01900241482557319, 0.4340030873165195]0.3785296355936244 0.7914545642238436

[0.5457049827172322, 0.5278841077140354, 0.2685543140893826, 0.002312402674624514]0.03585758021822538 0.3298966119945801

[0.269259422244588, 0.6621007242460392, 0.2301788971051998, 0.047881078848715286]0.02587925750136999 0.9195656458937822

0.259082665748668 0.0933421607590641[0.31439486129202887, 0.521211733137794, 0.14539732546645678, 0.0009501059838643577]

0.10647190379252569 0.20823300141803736[0.9456875734737963, 0.47763598619698655, 0.11569835557059593, 0.8776910120723697]

0.2671656406671584 0.9698177601248633[0.01900241482557319, 0.4340030873165195, 0.3785296355936244, 0.7914545642238436]

0.5457049827172322 0.5278841077140354[0.2685543140893826, 0.002312402674624514, 0.03585758021822538, 0.3298966119945801]

0.269259422244588 0.6621007242460392[0.2301788971051998, 0.047881078848715286, 0.02587925750136999, 0.9195656458937822]

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