Python一行代码取出列表中符合条件的元素

文章目录

• 问题描述
• 解决方案
• 对比
• 参考文献

问题描述

Python一行代码找出列表中符合条件的元素

解决方案

循环

def for_loop(l, target):
    for i in l:
        if i == target:
            return i
    return None


l = [1, 2, 3, 4, 5]
print(for_loop(l, 0))
print(for_loop(l, 1))
# None
# 1

next

def _next(l, target):
    return next((i for i in l if i == target), None)


l = [1, 2, 3, 4, 5]
print(_next(l, 0))
print(_next(l, 1))
# None
# 1

more_itertools

安装

pip install more-itertools

或不安装

def first_true(iterable, default=None, pred=None):
    return next(filter(pred, iterable), default)

调用 more_itertools.first_true(iterable, default=None, pred=None)

from more_itertools import first_true

l = [1, 2, 3, 4, 5]
print(first_true(l, pred=lambda x: x == 0))
print(first_true(l, pred=lambda x: x == 1))
# None
# 1

对比

方法	耗时/s
循环	2.81
next()	2.85
more_itertools.first_true()	10.58

import timeit
import more_itertools


def for_loop():
    for i in range(10000000):
        if i == 9999999:
            return i
    return None


def _next():
    return next((i for i in range(10000000) if i == 9999999), None)


def first_true():
    return more_itertools.first_true(range(10000000), pred=lambda x: x == 9999999)


print(timeit.timeit(for_loop, number=10))
print(timeit.timeit(_next, number=10))
print(timeit.timeit(first_true, number=10))
# 2.8123628000000003
# 2.851581
# 10.5818328

参考文献

1. Python: Find in list
2. More Itertools Documentation