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声明numpy矩阵的简洁方法

numpy arrays python

riqitang · 技术社区 · 7 年前

什么是一种简短的、可读的方法来声明一个999x999 numpy矩阵,其中每一行是 [1,2,3,...,999] ? 最终矩阵应为:

[[1,2,3,...,999]
[1,2,3,...,999]
...
[1,2,3,...,999]]

2 回复 | 直到 7 年前

jpp 7 年前

你可以用 numpy.tile :

import numpy as np

res = np.tile(range(10), (5, 1))

print(res)

array([[0, 1, 2, 3, 4, 5, 6, 7, 8, 9],
       [0, 1, 2, 3, 4, 5, 6, 7, 8, 9],
       [0, 1, 2, 3, 4, 5, 6, 7, 8, 9],
       [0, 1, 2, 3, 4, 5, 6, 7, 8, 9],
       [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]])

或者,也可以添加到零数组中:

res = np.zeros((5, 10)) + range(10)

AGN Gazer 7 年前

@jpp的回答很优雅,但以下解决方案更有效:

res = np.empty((nrows, ncols))
res[:, :] = np.arange(ncols)

时间安排:

%timeit a = np.empty((1000,1000)); a[:, :] = np.arange(1000)
445 Âµs Â± 9.08 Âµs per loop (mean Â± std. dev. of 7 runs, 1000 loops each)

%timeit np.tile(range(1000), (1000, 1))
1.43 ms Â± 15.6 Âµs per loop (mean Â± std. dev. of 7 runs, 1000 loops each)

进一步的定时测试:

在@jpp注释之后,我又添加了一个直接在Python解释器中完成的测试(与Jupyter笔记本中运行的原始测试不同,因为它当时已经启动并运行):

>>> import sys
>>> print(sys.version)
3.6.5 |Anaconda, Inc.| (default, Apr 26 2018, 08:42:37) 
[GCC 4.2.1 Compatible Clang 4.0.1 (tags/RELEASE_401/final)]
>>> import numpy as np
>>> print(np.__version__)
1.13.3
>>> import timeit
>>> t = timeit.repeat('res = np.empty((nrows, ncols)); res[:, :] = np.arange(ncols)', setup='import numpy as np; nrows=ncols=1000', number=100, repeat=50)
>>> print(min(t), max(t), np.mean(t), np.std(t))
0.04336756598786451 0.053294404002372175 0.0459639201409 0.00240180447219
>>> t = timeit.repeat('res = np.tile(range(ncols), (nrows, 1))', setup='import numpy as np; nrows=ncols=1000', number=100, repeat=50)
>>> print(min(t), max(t), np.mean(t), np.std(t))
0.05032560401014052 0.05859642301220447 0.0530669655403 0.00225117881195

结果是 numpy 1.14.5 几乎相同:

>>> import sys
>>> print(sys.version)
3.6.6 |Anaconda, Inc.| (default, Jun 28 2018, 11:07:29) 
[GCC 4.2.1 Compatible Clang 4.0.1 (tags/RELEASE_401/final)]
>>> import numpy as np
>>> print(np.__version__)
1.14.5
>>> import timeit
>>> t = timeit.repeat('res = np.empty((nrows, ncols)); res[:, :] = np.arange(ncols)', setup='import numpy as np; nrows=ncols=1000', number=100, repeat=50)
>>> print(min(t), max(t), np.mean(t), np.std(t))
0.04360878499574028 0.05562149798788596 0.04657964294136036 0.0025253372244474614
>>> t = timeit.repeat('res = np.tile(range(ncols), (nrows, 1))', setup='import numpy as np; nrows=ncols=1000', number=100, repeat=50)
>>> print(min(t), max(t), np.mean(t), np.std(t))
0.05024543400213588 0.06169128899637144 0.05339125283906469 0.00276210097759817