Python列表切片语法使用没有明显的原因

Question

我偶尔会看到Python代码中使用的列表切片语法,如下所示:

newList = oldList[:]

当然这与以下相同:

newList = oldList

或者我错过了什么？

Answer 1

[:] Shallow复制列表,制作包含对原始列表成员的引用的列表结构的副本.这意味着副本上的操作不会影响原始结构.但是,如果您对列表成员执行某些操作,则两个列表仍会引用它们,因此如果通过原始组件访问成员,则会显示更新.

一个深复制会使所有的成员的拷贝.

下面的代码片段显示了一个浅层副本.

# ================================================================
# === ShallowCopy.py =============================================
# ================================================================
#
class Foo:
    def __init__(self, data):
        self._data = data

aa = Foo ('aaa')
bb = Foo ('bbb')

# The initial list has two elements containing 'aaa' and 'bbb'
OldList = [aa,bb]
print OldList[0]._data

# The shallow copy makes a new list pointing to the old elements
NewList = OldList[:]
print NewList[0]._data

# Updating one of the elements through the new list sees the
# change reflected when you access that element through the
# old list.
NewList[0]._data = 'xxx'
print OldList[0]._data

# Updating the new list to point to something new is not reflected
# in the old list.
NewList[0] = Foo ('ccc')
print NewList[0]._data
print OldList[0]._data

在python shell中运行它会得到以下脚本.我们可以看到使用旧对象的副本制作的列表.其中一个对象可以通过旧列表引用更新其状态,并且可以在通过旧列表访问对象时看到更新.最后,可以看到更改新列表中的引用不会反映在旧列表中,因为新列表现在指的是不同的对象.

>>> # ================================================================
... # === ShallowCopy.py =============================================
... # ================================================================
... #
... class Foo:
...     def __init__(self, data):
...         self._data = data
...
>>> aa = Foo ('aaa')
>>> bb = Foo ('bbb')
>>>
>>> # The initial list has two elements containing 'aaa' and 'bbb'
... OldList = [aa,bb]
>>> print OldList[0]._data
aaa
>>>
>>> # The shallow copy makes a new list pointing to the old elements
... NewList = OldList[:]
>>> print NewList[0]._data
aaa
>>>
>>> # Updating one of the elements through the new list sees the
... # change reflected when you access that element through the
... # old list.
... NewList[0]._data = 'xxx'
>>> print OldList[0]._data
xxx
>>>
>>> # Updating the new list to point to something new is not reflected
... # in the old list.
... NewList[0] = Foo ('ccc')
>>> print NewList[0]._data
ccc
>>> print OldList[0]._data
xxx

Answer 2

就像NXC所说的那样,Python变量名实际上指的是一个对象,而不是内存中的特定位置.

newList = oldList会创建指向同一对象的两个不同变量,因此,更改oldList也会发生变化newList.

但是,当您这样做时newList = oldList[:],它会"切片"列表,并创建一个新列表.默认值为[:]0和列表的结尾,因此它会复制所有内容.因此,它会创建一个新列表,其中包含第一个中包含的所有数据,但两者都可以在不更改另一个的情况下进行更改.

Answer 3

至于已经回答,我只需添加一个简单的演示:

>>> a = [1, 2, 3, 4]
>>> b = a
>>> c = a[:]
>>> b[2] = 10
>>> c[3] = 20
>>> a
[1, 2, 10, 4]
>>> b
[1, 2, 10, 4]
>>> c
[1, 2, 3, 20]