python中的搜索排序是如何工作的? [英] how does searchsort in python work?

问题描述

为了清楚我的问题说如果我有一个数组 a 作为出[123]:[1, 3, 4, 6, 9, 10, 54]当我尝试搜索列表中的数字时，searchsort 返回正确的值，但是当我尝试不在列表中的内容时，它返回一个荒谬的值

To make my question clear say if I have an array a as Out[123]: [1, 3, 4, 6, 9, 10, 54] When I try to search the numbers in the list, searchsort returns correct value but when I try something not in the list, it returns an absurd value

这是一些结果

In [131]: a
Out[131]: [1, 3, 4, 6, 9, 10, 54]

In [132]: searchsorted(a,1)
Out[132]: 0

In [133]: searchsorted(a,6)
Out[133]: 3

In [134]: searchsorted(a,[9,54,1])
Out[134]: array([4, 6, 0])

In [135]: searchsorted(a,[9,54,1,0])
Out[135]: array([4, 6, 0, 0])
***> # here 0 is not in the list, but turns up @ position 0***

In [136]: searchsorted(a,740)
Out[136]: 7
***> # here 0 is not in the list, but turns up @ position 7***

为什么会这样?

推荐答案

searchsorted 告诉你哪里元素属于保证排序:

searchsorted tells you where the element belongs to guarantee ordering:

在已排序的数组 a 中查找索引 ，这样，如果 v 中的相应元素插入在索引之前，则 a 的顺序将被保留.

Find the indices into a sorted array a such that, if the corresponding elements in v were inserted before the indices, the order of a would be preserved.

在位置 7 插入 740 将保持排序，就像在位置 0 插入 0 一样.

inserting 740 at position 7 would preserve ordering, as would inserting 0 at position 0.

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