l1是一个列表，l2 = l1；为什么不"l2 = l2 + [item]"?影响"l1"，而"l2 + = [item]"则影响l1.做?
可能重复:
在Python中加等于(+ =)有什么作用? >
Possible Duplicate:
What does plus equals (+=) do in Python?
我注意到一个奇怪的问题:
I noticed a strange problem:
l1 = ['1', '2', '3']
l2 = l1
item = l2.pop(0)
# the pop operation will effect l1
print l1
l2 = l2 + [item]
# why "l2 = l2 + [item]" does't effect l1 while "l2 += [item]" does.
print l2
print l1
输出为:
['2', '3']
['2', '3', '1']
['2', '3']
但是如果我将l2 = l2 + [item]更改为l2 += [item]，输出将是:
But if i change l2 = l2 + [item] into l2 += [item], the output will be:
['2', '3']
['2', '3', '1']
['2', '3', '1']
+ and += are different operators woth different internal meaning as described here.
l2 = l1 + x调用l2 = l1.__add__(x)，如果不起作用，则调用x.__radd__(l1).两者都应返回一个独立于旧对象的新对象，以形成运算结果，从而不会影响l1.
l2 = l1 + x calls l2 = l1.__add__(x), if that doesn't work it calls x.__radd__(l1). Both should return a new object forming the result of the operation, independent from the old one, thus not affecting l1.
l2 += x调用l2 = l2.__iadd__(x)(扩展分配")，只有在这种方法不起作用的情况下，才会如上所述退回到l2 = l2 + x.
l2 += x calls l2 = l2.__iadd__(x) ("augmented assignment"), and only if this doesn't work, falls back to l2 = l2 + x as described above.
在数字上，两者是相同的，因为它们是不可变的，因此不能用+=修改.在列表上，+返回一个新对象，而+=修改左侧的对象.
With numbers, both are the same, because they are immutable and thus cannot be modified with +=, while on lists, + returns a new object while += modifies the left hand side one.
由于修改了l2后面的对象，并且l1引用了同一对象，因此您也会注意到l1上的更改.
As the object behind l2 is modified and l1 refers the same object, you notice the change on l1 as well.