分配给numpy数组时避免np.where [英] Avoiding np.where when assigning to numpy array

问题描述

我希望以下(或类似内容)能够正常工作(不使用np.where)

I would like for the following (or similar) to work (without using np.where)

>>> A = np.arange(0,10)
>>> ind = np.logical_and(A>4, A%2)
>>> k = np.array([0,1,0],dtype=bool)
>>> A[ind][k] = np.pi # Doesn't actually assign to A

也就是说，我希望k是对ind值为true的附加布尔掩码.

That is, I want k to be an additional boolean mask on the values of ind that are true.

我知道我可以使用np.where(ind)[0][k]，但这比逻辑索引更昂贵.

I know that I can use np.where(ind)[0][k], but this is more expensive than logical indexing.

有没有办法引用A[ind]来引用A的基本内存?

Is there a way to reference A[ind] that will refer to the base memory of A?

推荐答案

从经常引用的numpy索引页面:

From the oft-referenced numpy indexing page:

....单个布尔索引数组实际上与x [obj.nonzero()]相同....但是，当obj.shape == x.shape时，它会更快.

.... A single boolean index array is practically identical to x[obj.nonzero()] .... However, it is faster when obj.shape == x.shape.

np.where(cond)是np.nonzero(cond).

但是让我们做一些简单的计时

But let's do some simple timing

In [239]: x = np.arange(10000)
In [240]: y = (x%2).astype(bool)
In [241]: x[y].shape
Out[241]: (5000,)
In [242]: idx = np.nonzero(y)
In [243]: x[idx].shape
Out[243]: (5000,)
In [244]: timeit x[y].shape
89.9 µs ± 726 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
In [245]: timeit x[idx].shape
13.3 µs ± 107 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
In [246]: timeit x[np.nonzero(y)].shape
34.2 µs ± 893 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)

因此，即使我们使用显式的where，数组索引也比布尔索引快.

So array indexing is faster than boolean indexing, even when we use an explicit where.

A[ind][k]=不起作用，因为A[ind]是副本而不是视图.

A[ind][k]= does not work because A[ind] is a copy, not a view.

In [251]: A = np.arange(100,110)
In [252]: ind = np.logical_and(A>104, A%2)
In [253]: ind
Out[253]: 
array([False, False, False, False, False,  True, False,  True, False,
        True])
In [254]: k = np.array([0,1,0], dtype=bool)
In [255]: A[ind]
Out[255]: array([105, 107, 109])
In [256]: A[ind][k]
Out[256]: array([107])
In [257]: A[ind][k] = 12
In [258]: A
Out[258]: array([100, 101, 102, 103, 104, 105, 106, 107, 108, 109])

但是使用k从np.where(ind)中选择索引是可行的:

But using the k to select indices from np.where(ind) works:

In [262]: A[np.where(ind)[0][k]]=12
In [263]: A
Out[263]: array([100, 101, 102, 103, 104, 105, 106,  12, 108, 109])

获取而不是集合的时间:

Timings for a fetch rather than a set:

In [264]: timeit A[np.where(ind)[0][k]]
1.94 µs ± 75.7 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
In [265]: timeit A[ind][k]
1.34 µs ± 13.3 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

所以是的，在这种情况下，双重掩膜要快一些，但是如果它不起作用并不重要.不要浪费时间来改善时间.

So yes, the double masking is a bit faster in this case, but that doesn't matter if it doesn't work. Don't sweat the small time improvements.

In [345]: ind1=ind.copy()
In [346]: ind1[ind] = k
In [348]: A[ind1]=3
In [349]: A
Out[349]: array([100, 101, 102, 103, 104, 105, 106,   3, 108, 109])

在这个小例子中，时间基本上与A[np.where(ind)[0][k]]=12相同.

In this small example timeit is basically the same as for A[np.where(ind)[0][k]]=12.

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