python 中的函数和全局 scope/nested 函数混淆

Functions and global scope/nested function confusion in python

fridge={"cheese":10, "milk":11, "feta":12, "cream":21, "onion":32, "pepper":25} 
def fridge_function (fridge):
    del fridge["feta"]

    return

print (fridge)

我很困惑。

1) 我认为字典具有全局范围。

2) 我想做的是从字典中删除项目,然后根据需要更新修改后的字典,即删除值 "feta"。

3) 当我打印 fridge 时,我得到了未经编辑的原始列表。我什至尝试制作冰箱的副本,然后尝试在函数之外打印该值,但是,最终结果是相同的:原始冰箱字典没有任何修改。

你还没有用 fridge

调用函数 fridge_function

这应该有效:

fridge_function(fridge)
print (fridge)

您将 fridge 作为参数传入,因此它将成为局部变量。要使用全局 fridge 你会这样做:

FRIDGE = {"cheese": 10, "milk": 11, "feta": 12, "cream": 21, "onion": 32, "pepper": 25}

def fridge_function():
    del FRIDGE["feta"]

fridge_function()   # you need to call it
print(FRIDGE)

globals 通常是大写的,所以我将变量命名为 FRIDGE.

如果你需要赋值到全局变量(并且只有这样),你需要添加global语句:

def myfunction():
    global MYGLOBAL
    MYGLOBAL = 42

然而,使用全局变量和改变全局变量的函数通常不是一个好的解决方案。也许面向对象的东西会更好? (Python 2.7 语法):

class Fridge(dict):
    def __init__(self, **contents):
        super(Fridge, self).__init__(contents)

def main():
    fridge = Fridge(
        cheese=10,
        milk=11,
        feta=12,
        cream=21,
        onion=32,
        pepper=25
    )
    print fridge
    del fridge['feta']
    print fridge

if __name__ == "__main__":
    main()

输出是

{'cheese': 10, 'pepper': 25, 'feta': 12, 'onion': 32, 'milk': 11, 'cream': 21}
{'cheese': 10, 'pepper': 25, 'onion': 32, 'milk': 11, 'cream': 21}

为了进一步说明,请注意以下任一方法都有效:

fridge作为函数参数:

fridge={"cheese":10, "milk":11, "feta":12, "cream":21, "onion":32, "pepper":25} 
def fridge_function(fridge):
    del fridge["feta"]  # reference to fridge passed as a parameter

fridge_function(fridge)
print(fridge)

fridge作为全局变量

fridge={"cheese":10, "milk":11, "feta":12, "cream":21, "onion":32, "pepper":25} 
def fridge_function():
    del fridge["feta"]  # refers to the global fridge 

fridge_function()
print(fridge)

在大多数情况下,第一种被认为是更好的编程风格,如 global variables are usually best avoided