python - 列表的可变性，因为它与函数的参数相关

Question

我正在使用 Lutz 的书学习 Python。我正在使用 Anaconda 发行版的 Python 3.6.5。我确实研究了这个问题，但没有找到任何可以回答我的问题的线程。 Mutability of lists in python讨论 append 而不是我们如何将可变对象传递给函数。

我的问题是，当我使用函数内的索引对列表进行就地更改时，更改确实会如预期的那样得到反射(reflect)，因为可变对象是通过引用传递的。但是，当我直接分配列表时，更改不会得到反射(reflect)。

具体来说，我创建了两个列表 L1 和 L2。对于L1，我会使用索引进行赋值，但对于L2，我会在函数内部进行直接赋值。

L1=[2]
L2=['a']
print("Before, L1:",L1)
print("Before, L2:",L2)
def f(a,b):
    a[0] =[3] #Using index-based assignment
    b = ['b'] #Direct assignment

#Pass L to f
f(L1,L2)
print("After, L1:",L1)
print("After, L2:",L2)

输出为:

Before, L1: [2]
Before, L2: ['a']
After, L1: [[3]]
After, L2: ['a']

正如我们所见，L1 发生了变化，但 L2 没有变化。

问题:有人可以解释一下为什么L2的值没有更改为'b'吗？

如果您认为这篇文章是重复的，那么如果您为相关文章添加标签那就太好了。

---

顺便说一句，我进行了一个小实验，看看是否与基于索引的赋值或直接赋值有关。

l=[2]
id(l)
l[0] = 3 #Index assignment
id(l) # Memory location doesn't change

l = 3 # direct assignment
id(l) #Memory location changes.

因此，我似乎缺少一个概念，这意味着我不确定为什么直接赋值会改变内存位置。

Answer 1

如果我们稍微更改您的代码，我们可以使用 id 来查看引用如何更改(或不更改):

L1=[2]
L2=['a']
print("Before, L1:", L1, id(L1))
print("Before, L2:", L2, id(L2))
def f(a,b):
    print("Inside, Before, a:", id(a))
    print("Inside, Before, b:", id(b))
    a[0] =[3] #Using index-based assignment
    b = ['b'] #Direct assignment
    print("Inside, After, a:", id(a))
    print("Inside, After, b:", id(b))

#Pass L to f
f(L1,L2)
print("After, L1:", L1, id(L1))
print("After, L2:", L2, id(L2))

输出:

Before, L1: [2]     1870498294152  # L1
Before, L2: ['a']   1870498294280  # L2
Inside, Before, a:  1870498294152  # L1
Inside, Before, b:  1870498294280  # L2
Inside, After, a:   1870498294152  # L1
Inside, After, b:   1870498239496  # Something different, not L2
After, L1: [[3]]    1870498294152  # L1
After, L2: ['a']    1870498294280  # L2

Note, the numbers aren't significant in themselves other than to help distinguish references to different objects. Running this yourself (or if I ran it again), would cause the ids to change.

With a, you're modifying/mutating a but not attempting to re-assign the reference. That's fine.

With b, you're re-assigning the reference. This will work inside the function (as the "Inside, After, b:" print call shows), but this change will not be reflected outside of the function. b will be restored to reference the original object, ['a'].

As to why...

meaning I am unsure why direct assignment changes the memory location.

Inside your function, a and b are just references to objects. Initially, they reference (the objects referenced by) L1 and L2 respectively because by calling f, you're passing references to those objects.

a[0] = [3] first dereferences a (or L1 in this case), then the [0] index, and sets that value.

In fact, if you look at id(a[0]) before and after that call then that would change. a is a list of references. Try it:

print(id(a[0]))   # One thing
a[0] =[3] #Using index-based assignment
print(id(a[0]))   # Something different

这很好。当您退出函数时，L1 仍将引用该函数使用 a 引用的对象，并且它在 0 索引处的突变将保留。

使用 b = ['b']，您可以将 b 重新分配或重新绑定(bind)到新对象。旧对象仍然存在(供以后在函数外部使用)。

最后，我经常使用“引用”这个术语，但 Python 并不完全是一种“按引用传递”语言，rather variable names are bound to objects 。在第二种情况下，您将重新绑定(bind) b，从而永远失去与最初引用的对象 L2 的关联。