了解numpy中奇怪的布尔二维数组索引行为 [英] Understanding weird boolean 2d-array indexing behavior in numpy

问题描述

为什么这样做:

a=np.random.rand(10,20)
x_range=np.arange(10)
y_range=np.arange(20)

a_tmp=a[x_range<5,:]
b=a_tmp[:,np.in1d(y_range,[3,4,8])]

而这不是:

a=np.random.rand(10,20)
x_range=np.arange(10)
y_range=np.arange(20)    

b=a[x_range<5,np.in1d(y_range,[3,4,8])]

推荐答案

Numpy参考文档的索引页面包含答案，但需要仔细阅读.

The Numpy reference documentation's page on indexing contains the answers, but requires a bit of careful reading.

此处的答案是，使用布尔值进行索引等效于使用首先通过np.nonzero转换布尔值数组而获得的整数数组进行索引.因此，对于布尔数组m1，m2

The answer here is that indexing with booleans is equivalent to indexing with integer arrays obtained by first transforming the boolean arrays with np.nonzero. Therefore, with boolean arrays m1, m2

a[m1, m2] == a[m1.nonzero(), m2.nonzero()]

(成功时，即m1.nonzero().shape == m2.nonzero().shape)等效于:

which (when it succeeds, i.e., m1.nonzero().shape == m2.nonzero().shape) is equivalent to:

[a[i, i] for i in range(a.shape[0]) if m1[i] and m2[i]]

我不确定为什么要设计成这样工作---通常，不是是您想要的.

I'm not sure why it was designed to work like this --- usually, this is not what you'd want.

要获得更直观的结果，您可以代替

To get the more intuitive result, you can instead do

a[np.ix_(m1, m2)]

产生的结果等同于

[[a[i,j] for j in range(a.shape[1]) if m2[j]] for i in range(a.shape[0]) if m1[i]]

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