Python将列表作为参数传递

Python passing list as argument(Python将列表作为参数传递)

问题描述

如果我要运行这段代码:

If i were to run this code:

def function(y):
    y.append('yes')
    return y

example = list()
function(example)
print(example)

为什么即使我没有直接更改变量example"，它也会返回 ['yes']，以及如何修改代码以使example"不受函数影响?

Why would it return ['yes'] even though i am not directly changing the variable 'example', and how could I modify the code so that 'example' is not effected by the function?

推荐答案

Python 中的一切都是参考.如果您希望避免这种行为，您必须使用 list() 创建原始副本的新副本.如果列表包含更多引用，您需要使用 deepcopy()

Everything is a reference in Python. If you wish to avoid that behavior you would have to create a new copy of the original with list(). If the list contains more references, you'd need to use deepcopy()

def modify(l):
 l.append('HI')
 return l

def preserve(l):
 t = list(l)
 t.append('HI')
 return t

example = list()
modify(example)
print(example)

example = list()
preserve(example)
print(example)

输出

['HI']
[]

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