在函数Python之间传递列表 [英] Passing a list between functions Python

问题描述

如果我在一个函数中创建列表并进行编辑，那么我希望能够将完成的列表传递给另一个函数，在该函数中它将用于执行更多操作.

If I make a list in one function, and edit it, then I want to be able to pass the finished list off to another function where it will be used to do more things.

def func1():
    numbers = [1, 2, 3, 4]
    if numbers.count(1) > 0:
        numbers.remove(1)
        func2()

def func2():
    two = numbers[0]

func1()

如何获取它，以便它不会提取未全局定义的错误号.我知道为什么会导致错误，但是经过研究，我仍然找不到解决此问题的好方法.

How could I get it so that it doesn't pull the error numbers not defined globally. I know why it is pulling the error, but after researching, I still can't find a good way to fix this problem.

推荐答案

根据您的里程和要求，选择一个

Based on your mileage and requirement, choose one

1. 将列表作为参数传递给被调用的函数

1. Pass the list as a parameter to the called function

def func2(numbers):
    two = numbers[0]


def func1():
    numbers = [1, 2, 3, 4]
    if numbers.count(1) > 0:
        numbers.remove(1)
        func2(numbers)

使两个类的功能都成为一个类的一部分，并使列表成为实例属性

Make both the functions part of a single Class and make the list an instance attribute

class Foo(object):
    def __init__(self, numbers):
        self.numbers = numbers
        pass
    def func1(self):
        if self.numbers.count(1) > 0:
            self.numbers.remove(1)
            self.func2()
    def func2(self):
        two = self.numbers[0]


Foo([1, 2, 3, 4]).func1()

重新设计，以便调用方装饰被调用的函数

Redesign so that the called function is decorated with the caller

def func1(func):
    def wraps(*argv):
        numbers = argv[0]
        if numbers.count(1) > 0:
        numbers.remove(1)
        func(*argv)
    return wraps

@func1
def func2(numbers):
    two = numbers[0]


func2([1,2,3,4])

使用嵌套函数关闭

Closure with nested function

def func1():
    def func2():
        two = numbers[0]
    numbers = [1, 2, 3, 4]
    if numbers.count(1) > 0:
        numbers.remove(1)
        func2()


func1()

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