如何提取Python列表中一定比例的均匀分布的元素? [英] How should a certain percentage of evenly-distributed elements of a Python list be extracted?

问题描述

我有一个数据点列表.对于程序的完整运行，我将使用所有数据点，但是对于代码测试，我只想使用其中的一小部分，以便程序在短时间内运行.但是，我不想简单地获取列表的前n个元素.我想从列表中选择元素的均匀分布.因此，如果我使用50％的数据点，则可能需要每隔两个数据点从数据点列表中进行选择.

I have a list of data points. For the full run of my program, I'll use all of the data points, but for testing of the code, I want to use only a small percentage of them in order that the program run in a short time. I do not want simply to take the first n elements of the list, though; I want to select an even distribution of the elements from the list. So, if I'm using 50% of the data points, I might want to select from the list of data points every second data points.

基本上，我希望有一个函数，该函数将一个列表和一个百分比作为参数，并返回一个列表，该列表由输入列表中元素的均匀分布组成，其数量与请求的百分比尽可能接近.

Basically, I want to have a function that takes as arguments a list and a percentage and returns a list consisting of an even distribution of elements from the input list, the number of which corresponds as closely as possible to the percentage requested.

什么是做到这一点的好方法?

What would be a good way to do this?

推荐答案

这可以通过设置带台阶的切片来简单地实现:

This can trivially be achieved by setting a slice with a step:

def select_elements(seq, perc):
    """Select a defined percentage of the elements of seq."""
    return seq[::int(100.0/perc)]

使用中:

>>> select_elements(range(10), 50)
[0, 2, 4, 6, 8]
>>> select_elements(range(10), 33)
[0, 3, 6, 9]
>>> select_elements(range(10), 25)
[0, 4, 8]

您还可以添加round，因为int会截断:

You could also add round, as int will truncate:

>>> int(3.6)
3
>>> int(round(3.6))
4

如果要使用比例而不是百分比(例如，用0.5代替50)，只需将100.0替换为1.

If you want to use a proportion rather than a percentage (e.g. 0.5 instead of 50), simply replace 100.0 with 1.

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