为什么python中的列表会有这种行为？

Question

如果我有以下功能和代码：

def do_something(a, b):
    a.insert(0, ’z’)
    b = [’z’] + b
a = [’a’, ’b’, ’c’]
a1 = a
a2 = a[:]
b = [’a’, ’b’, ’c’]
b1 = b
b2 = b[:]
do_something(a, b)

为什么print(a)产生['z','a','b','c']，而打印b仍然只打印['a','b','c']？你知道吗

我在函数中创建了b = b + ['z']，那么z不也应该在列表中吗？你知道吗

还有，为什么打印a[:]不打印新列表['z','a','b','c']，而是打印旧列表['a','b','c']？你知道吗

Answer 1

因为在do_something中，您正在修改具有标签a的列表，但是您正在创建一个新列表并将其重新分配给标签b，而不是修改具有标签b的列表

这意味着do_something之外的a列表已经更改，而不是b列表，因为您只是巧合地在func内部使用了相同的名称，您还可以对具有不同名称的func执行相同的操作，例如：

def do_something(x, y):
    x.insert(0, ’z’)
    y = [’z’] + y

外部的打印仍然会像您报告的那样工作，因为函数内部和外部对象的标签是不相关的，在您的示例中，它们恰好是相同的。你知道吗

Answer 2

从https://docs.python.org/2/library/copy.html

Shallow copies of dictionaries can be made using dict.copy(), and of lists by 
assigning a slice of the entire list, for example, copied_list = original_list[:].

好的

def do_something(a, b):
    a.insert(0, 'z') #this is still referencing a when executed. a changes.
    b = ['z'] + b #This is a shallow copy in which the b in this function, is now [’a’, ’b’, ’c’, 'z']. 

尽管上面说的是真的，但是你所想到的带有z的b与在程序的“结尾”处打印的b是不同的。但是，打印在第一行的b是函数def\u something（）中的b。你知道吗

代码：

def do_something(a, b):
    a.insert(0, 'z') #any changes made to this a changes the a that was passed in the function.
    b = ['z'] + b #This b is do_something() local only. LEGB scope: E. Link below about LEGB. 
print("a b in function: ", a, "|", b)
a = ['a', 'b', 'c']
a1 = a
a2 = a[:] #This is never touched following the rest of your code.
b = ['a', 'b', 'c']
b1 = b
b2 = b[:] #This is never touched following the rest of your code.
print("a b before function: ", a, "|", b)
do_something(a, b)
print("a b after function: ", a, "|", b) #This b is same thing it was after assignment.  

输出：

a b before function:  ['a', 'b', 'c'] | ['a', 'b', 'c']
a b in function:  ['z', 'a', 'b', 'c'] | ['z', 'a', 'b', 'c']
a b after function:  ['z', 'a', 'b', 'c'] | ['a', 'b', 'c']

有关LEGB的详细信息。你知道吗