先来看列表中的append和extend的帮助文档
app
>>> help(list.append) # 往列表末尾添加对象ide
Help on method_descriptor:函数
append(...)性能
L.append(object) -> None -- append object to end对象
>>> help(list.extend) # 往经过添加迭代对象来扩充列表ip
Help on method_descriptor:element
extend(...)文档
L.extend(iterable) -> None -- extend list by appending elements from the iterableit
>>>io
再来看二者的示例
# append日后添加单独一个元素
>>> A = ["a", "b"]
>>> A.append(1)
>>> A
['a', 'b']
# append日后添加一整个列表
>>> A = ["a", "b"]
>>> A.append(["c", "d"])
>>> A
['a', 'b', ['c', 'd']]
# append日后添加一个迭代对象
>>> A = ["a", "b"]
>>> A.append(range(10))
>>> A
['a', 'b', range(0, 10)]
# append日后添加函数和类对象
>>> a = [1, 2]
>>> class A:
... pass
>>> def B():
... pass
>>> a.append(A)
>>> a.append(B)
>>> a
[1, 2, <class '__main__.A'>, <function B at 0x000000000404B598>]
# 往列表A,扩充一个列表对象
>>> A = ["a", "b"]
>>> A.extend(["c", "d"])
>>> A
['a', 'b', 'c', 'd']
# 往列表A,扩充一个迭代对象range(4)
>>> A = ["a", "b"]
>>> A.extend(range(4))
>>> A
['a', 'b', 0, 1, 2, 3]
# 往列表A,扩充一个迭代对象"ABCD"
>>> A = [1, 2]
>>> A.extend("ABCD")
>>> A
[1, 2, 'A', 'B', 'C', 'D']
再来看往extend后添加单独元素会报错,由于单独元素是不可迭代的。
>>> A = ["a", "b"]
>>> A.extend(1)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: 'int' object is not iterable
'int' object is not iterable
综合上面的示例及报错信息,来总结extend, append方法:
extend只能添加可迭代对象,而append便可添加可迭代对象,也能够添加单独元素
append后面添加的对象,无论是单独元素,列表,可迭代对象,通通算一个元素往列表最后加入
extend日后扩充的对象,是拿后面可迭代的全部元素往里面添加
再来看+,+=的示例
# +=操做
>>> A = ["a", "b"]
>>> B = ["c", "d"]
>>> A += B
>>> A
['a', 'b', 'c', 'd']
# +操做
>>> A = ["a", "b"]
>>> B = ["c", "d"]
>>> A = A + B
>>> A
['a', 'b', 'c', 'd']
从上面的例子能看看出不管是+仍是+=,都只能在列表之间操做
>>> A = ["a", "b"]
>>> B = 1
>>> A += B
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: 'int' object is not iterable
'int' object is not iterable
>>> A = A + B
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: can only concatenate list (not "int") to list
can only concatenate list (not "int") to list
>>> A + range(10)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: can only concatenate list (not "range") to list
can only concatenate list (not "range") to list
>>>
综合上面的示例及报错,来总结+,+=:
不管是+仍是+=只能在两个列表之间操做
都不能添加非列表对象
再来从id来看操做对象的变化
# 执行extend后,A的id不变
>>> A = [1, 2]
>>> B = [3, 4]
>>> id(A)
61099656
>>> A.extend(B)
>>> A
[1, 2, 3, 4]
>>> id(A)
61099656
# 执行append后,A的id不变
>>> A = [1, 2]
>>> B = [3, 4]
>>> id(A)
65986184
>>> A.append(B)
>>> A
[1, 2, [3, 4]]
>>> id(A)
65986184
# 执行+=后,A的id不变
>>> A = [1, 2]
>>> B = [3, 4]
>>> id(A)
63943112
>>> A += B
>>> A
[1, 2, 3, 4]
>>> id(A)
63943112
# 执行+后,A的id发生变化
>>> A = [1, 2]
>>> B = [3, 4]
>>> id(A)
61026568
>>> A = A + B
>>> A
[1, 2, 3, 4]
>>> id(A)
61026952
从上面的id来看,彷佛又可总结出:
执行extend,append,+=这三个操做都是原地修改列表,都不会引发id的变化
执行+,会拷贝出一个新的对象,会引发id的变化
若是执行+操做,一旦操做对象不少的话,常常须要拷贝新对象,性能是四者中最差的
最后总结:
extend原地修改列表,只可添加可迭代对象,待添加的对象有多少个就往原对象中添加多少个
append原地修改列表,可添加任何对象,但不管是什么对象,在末尾添加,并且只算一个元素
+=原地修改列表,只可执行列表之间的操做,效果上至关于extend
+会拷贝成一个新对象,并且只能执行列表之间的操做,性能问题要考虑