Python 初始化和填充 numpy 数组的最佳方法？
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Best way to initialize and fill an numpy array?
提问by tbc
I want to initialize and fill a numpyarray. What is the best way?
我想初始化并填充一个numpy数组。什么是最好的方法？
This works as I expect:
这按我的预期工作：
>>> import numpy as np
>>> np.empty(3)
array([ -1.28822975e-231, -1.73060252e-077, 2.23946712e-314])
But this doesn't:
但这不会：
>>> np.empty(3).fill(np.nan)
>>>
Nothing?
没有？
>>> type(np.empty(3))
<type 'numpy.ndarray'>
It seems to me that the np.empty()call is returning the correct type of object, so I don't understand why .fill()is not working?
在我看来，np.empty()调用正在返回正确类型的对象，所以我不明白为什么.fill()不起作用？
Assigning the result of np.empty()first works fine:
分配np.empty()first的结果工作正常：
>>> a = np.empty(3)
>>> a.fill(np.nan)
>>> a
array([ nan, nan, nan])
Why do I need to assign to a variable in order to use np.fill()? Am I missing a better alternative?
为什么我需要分配给一个变量才能使用np.fill()？我错过了更好的选择吗？
采纳答案by shx2
回答by JoshAdel
You could also try:
你也可以试试：
In [79]: np.full(3, np.nan)
Out[79]: array([ nan, nan, nan])
The pertinent doc:
相关文档：
Definition: np.full(shape, fill_value, dtype=None, order='C')
Docstring:
Return a new array of given shape and type, filled with `fill_value`.
Although I think this might be only available in numpy 1.8+
虽然我认为这可能仅在 numpy 1.8+ 中可用
回答by ryanjdillon
I find this easy to remember:
我觉得这很容易记住：
numpy.array([numpy.nan]*3)
Out of curiosity, I timed it, and both @JoshAdel's answerand @shx2's answerare far faster than mine with large arrays.
出于好奇，我给它计时，@JoshAdel 的回答和@shx2 的回答都比我的大数组快得多。
In [34]: %timeit -n10000 numpy.array([numpy.nan]*10000)
10000 loops, best of 3: 273 μs per loop
In [35]: %timeit -n10000 numpy.empty(10000)* numpy.nan
10000 loops, best of 3: 6.5 μs per loop
In [36]: %timeit -n10000 numpy.full(10000, numpy.nan)
10000 loops, best of 3: 5.42 μs per loop
回答by norok2
Just for future reference, the multiplication by np.nanonly works because of the mathematical properties of np.nan.
For a generic value N, one would need to use np.ones() * Nmimicking the accepted answer, however, speed-wise, this is not a terribly good choice.
仅供将来参考，乘以np.nan仅适用于 的数学特性np.nan。对于通用值N，需要使用np.ones() * N模仿已接受的答案，但是，就速度而言，这不是一个非常好的选择。
Best choice would be np.full()as already pointed out, and, if that is not available for you, np.zeros() + Nseems to be a better choice than np.ones() * N, while np.empty() + Nor np.empty() * Nare simply not suitable. Note that np.zeros() + Nwill also work when Nis np.nan.
最佳选择np.full()如已经指出的那样，如果这对您不可用，则np.zeros() + N似乎是比 更好的选择np.ones() * N，而np.empty() + N或np.empty() * N根本不适合。请注意，np.zeros() + N这也适用于Nis np.nan。
%timeit x = np.full((1000, 1000, 10), 432.4)
8.19 ms ± 97.8 μs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%timeit x = np.zeros((1000, 1000, 10)) + 432.4
9.86 ms ± 55.1 μs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%timeit x = np.ones((1000, 1000, 10)) * 432.4
17.3 ms ± 104 μs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%timeit x = np.array([432.4] * (1000 * 1000 * 10)).reshape((1000, 1000, 10))
316 ms ± 37.4 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
