这种numpy选择行为的背后是什么? [英] What is going on behind this numpy selection behavior?

问题描述

回答此问题，其他人和我实际上认为是错误的，因为认为以下方法可行:

Answering this question, some others and I were actually wrong by considering that the following would work:

说一个有

test = [ [ [0], 1 ],
         [ [1], 1 ]
       ]
import numpy as np
nptest = np.array(test)

背后的原因是什么

>>> nptest[:,0]==[1]
array([False, False], dtype=bool)

一个人拥有

>>> nptest[0,0]==[1],nptest[1,0]==[1]
(False, True)

---

或

>>> nptest==[1]
array([[False,  True],
       [False,  True]], dtype=bool)

或

>>> nptest==1
array([[False,  True],
       [False,  True]], dtype=bool)

是导致尺寸下降的原因是简并性.

Is it the degeneracy in term of dimensions which causes this.

推荐答案

nptest是对象dtype的2D数组，每行的第一个元素是列表.

nptest is a 2D array of object dtype, and the first element of each row is a list.

nptest[:, 0]是对象dtype的一维数组，其每个元素都是列表.

nptest[:, 0] is a 1D array of object dtype, each of whose elements are lists.

当您执行nptest[:,0]==[1]时，NumPy不会对nptest[:,0]的每个元素与列表[1]进行逐元素比较.它从[1]创建尽可能高维的数组，生成一维数组np.array([1])，然后生成

When you do nptest[:,0]==[1], NumPy does not perform an elementwise comparison of each element of nptest[:,0] against the list [1]. It creates as high-dimensional an array as it can from [1], producing the 1D array np.array([1]), and then broadcasts the comparison, comparing each element of nptest[:,0] against the integer 1.

由于nptest[:, 0]中没有列表等于1，因此结果的所有元素均为False.

Since no list in nptest[:, 0] is equal to 1, all elements of the result are False.

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