Python列表切片语法没有明显的原因

[:] Shallow copies

A. Deep Copy 也会复制所有名单成员。

# ================================================================
# === 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pythonshell中运行它会得到以下脚本。我们可以看到 其状态通过旧列表通过引用更新,并且可以 通过旧列表访问对象时看到。最后,更改 新列表现在引用的是另一个对象。

>>> # ================================================================
... # === 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

正如NXC所说,Python变量名实际上指向一个对象,而不是内存中的特定位置。

newList = oldList oldList 也会改变 newList .

newList = oldList[:] ,它“切片”列表,并创建一个新列表。的默认值 [:] 是0,并且是列表的结尾,因此它复制所有内容。因此,它创建了一个新列表,其中包含第一个列表中包含的所有数据,但两个列表都可以在不更改另一个列表的情况下进行更改。

正如已经回答的那样,我将简单地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]

千万不要认为Python中的“a=b”意味着“将b复制到a”。如果两边都有变量,你就不可能真正知道这一点。相反,可以将其视为“给b额外的名称a”。

如果b是一个不可变的对象(如数字、元组或字符串),那么是的,其效果是得到一个副本。但那是因为当你处理不可变的(可能应该被称为 , 或 蠕虫 )你

如果b是可变的,那么 总是要做一些额外的事情,以确保你有一个真正的副本 . . 对于列表,它就像一个切片一样简单:a=b[:]。

可变性也是导致以下情况的原因:

def myfunction(mylist=[]): 
    pass

... 并不像你想象的那样。

http://docs.python.org/library/copy.html

浅拷贝: (将内存块从一个位置复制到另一个位置)

a = ['one','two','three']

b = a[:]

b[1] = 2

print id(a), a #Output: 1077248300 ['one', 'two', 'three']
print id(b), b #Output: 1077248908 ['one', 2, 'three']

深度复制: (复制对象引用)

a = ['one','two','three']

b = a

b[1] = 2


print id(a), a #Output: 1077248300 ['one', 2, 'three']
print id(b), b #Output: 1077248300 ['one', 2, 'three']