LIST.append(1) 和 LIST = LIST + [1] (Python) 有什么区别 [英] What is the difference between LIST.append(1) and LIST = LIST + [1] (Python)
问题描述
当我执行(我使用交互式 shell)这些语句时，我得到了这个:
L=[1,2,3]K=LL.append(4)升[1,2,3,4]钾[1,2,3,4]但是当我用 L=L+[4] 替换 L.append(4) 做同样的事情时我得到:
L[1,2,3,4]钾[1,2,3]这是某种参考吗?为什么会发生这种情况?
我注意到的另一件有趣的事情是 L+=[4] 的作用类似于 .append，这很奇怪，因为我认为它的作用类似于 L = L + [4].
对所有这些进行澄清将不胜感激.
谢谢
L.append(4)这会在现有列表L的末尾添加一个元素.
L += [4]+= 运算符调用神奇的 __iadd__() 方法.结果证明 list 覆盖了 __iadd__() 方法并使其等效于 extend() ，类似于 append()code>，将元素直接添加到现有列表中.
L = L + [4]L + [4] 生成一个新列表，它等于 L 的末尾添加 4.这个新列表然后被分配回L.因为你已经创建了一个新的列表对象，K 不会被这个赋值改变.
我们可以使用 id() 来识别何时创建了新的对象引用:
When I execute (I'm using the interactive shell) these statements I get this:
L=[1,2,3]
K=L
L.append(4)
L
[1,2,3,4]
K
[1,2,3,4]
But when I do exactly the same thing replacing L.append(4) with L=L+[4] I get:
L
[1,2,3,4]
K
[1,2,3]
Is this some sort of reference thing? Why does this happen?
Another funny thing I noticed is that L+=[4] acts like .append, which is odd as I thought it would act like L = L + [4].
Clarification to all of this would be greatly appreciated.
Thanks
L.append(4)
This adds an element on to the end of the existing list L.
L += [4]
The += operator invokes the magic __iadd__() method. It turns out list overrides the __iadd__() method and makes it equivalent to extend() which, like append(), adds elements directly onto an existing list.
L = L + [4]
L + [4] generates a new list which is equal to L with 4 added to the end. This new list is then assigned back to L. Because you've created a new list object, K is unchanged by this assignment.
We can use id() to identify when a new object reference is created:
>>> L = [1, 2, 3]
>>> id(L)
152678284
>>> L.append(4)
>>> id(L)
152678284
>>> L = [1, 2, 3]
>>> id(L)
152680524
>>> L = L + [4]
>>> id(L)
152678316
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