append（object）通过将对象添加到列表来更新列表。

x = [20]
# List passed to the append(object) method is treated as a single object.
x.append([21, 22, 23])
# Hence the resultant list length will be 2
print(x)
--> [20, [21, 22, 23]]

extend（list）本质上连接两个列表。

x = [20]
# The parameter passed to extend(list) method is treated as a list.
# Eventually it is two lists being concatenated.
x.extend([21, 22, 23])
# Here the resultant list's length is 4
print(x)
--> [20, 21, 22, 23]

append将元素添加到列表中。extend将第一个列表与另一个列表/可迭代列表连接起来。

>>> xs = ['A', 'B']
>>> xs
['A', 'B']

>>> xs.append("D")
>>> xs
['A', 'B', 'D']

>>> xs.append(["E", "F"])
>>> xs
['A', 'B', 'D', ['E', 'F']]

>>> xs.insert(2, "C")
>>> xs
['A', 'B', 'C', 'D', ['E', 'F']]

>>> xs.extend(["G", "H"])
>>> xs
['A', 'B', 'C', 'D', ['E', 'F'], 'G', 'H']

extend（）可以与迭代器参数一起使用。这里有一个例子。您希望通过以下方式从列表列表中列出一个列表：

From

list2d = [[1,2,3],[4,5,6], [7], [8,9]]

你想要的

>>>
[1, 2, 3, 4, 5, 6, 7, 8, 9]

您可以使用itertools.chain.from_iterable（）来执行此操作。此方法的输出是迭代器。它的实现相当于

def from_iterable(iterables):
    # chain.from_iterable(['ABC', 'DEF']) --> A B C D E F
    for it in iterables:
        for element in it:
            yield element

回到我们的例子，我们可以

import itertools
list2d = [[1,2,3],[4,5,6], [7], [8,9]]
merged = list(itertools.chain.from_iterable(list2d))

拿到通缉名单。

以下是如何将extend（）等效地用于迭代器参数：

merged = []
merged.extend(itertools.chain.from_iterable(list2d))
print(merged)
>>>
[1, 2, 3, 4, 5, 6, 7, 8, 9]

有一个有趣的点已经被暗示，但没有解释，那就是扩展比追加更快。对于任何内部有append的循环，都应该考虑用list.exextend（processed_elements）替换。

记住，附加新元素可能会导致整个列表重新分配到内存中更好的位置。如果因为一次添加一个元素而多次执行此操作，则会影响整体性能。在这个意义上，list.extend类似于“”.jjoin（stringlist）。

append（）方法将单个项添加到列表末尾。

x = [1, 2, 3]
x.append([4, 5])
x.append('abc')
print(x)
# gives you
[1, 2, 3, [4, 5], 'abc']

extend（）方法接受一个参数，一个列表，并将该参数的每个项附加到原始列表中。（列表被实现为类。“创建”列表实际上是实例化一个类。因此，列表具有对其进行操作的方法。）

x = [1, 2, 3]
x.extend([4, 5])
x.extend('abc')
print(x)
# gives you
[1, 2, 3, 4, 5, 'a', 'b', 'c']

从潜入Python。

append（object）通过将对象添加到列表来更新列表。

x = [20]
# List passed to the append(object) method is treated as a single object.
x.append([21, 22, 23])
# Hence the resultant list length will be 2
print(x)
--> [20, [21, 22, 23]]

extend（list）本质上连接两个列表。

x = [20]
# The parameter passed to extend(list) method is treated as a list.
# Eventually it is two lists being concatenated.
x.extend([21, 22, 23])
# Here the resultant list's length is 4
print(x)
--> [20, 21, 22, 23]