Possible Duplicate:
Unexpected feature in a Python list of lists可能重复:Python列表中的意外功能
So I am relatively new to Python and I am having trouble working with 2D Lists.
所以我对Python比较新,我在处理2D列表时遇到了麻烦。
Here's my code:
这是我的代码:
data = [[None]*5]*5
data[0][0] = 'Cell A1'
print data
and here is the output (formatted for readability):
这是输出(为便于阅读而格式化):
[['Cell A1', None, None, None, None],
['Cell A1', None, None, None, None],
['Cell A1', None, None, None, None],
['Cell A1', None, None, None, None],
['Cell A1', None, None, None, None]]
Why does every row get assigned the value?
为什么每一行都被分配了值?
3 个解决方案
#1
50
This makes a list with five references to the same list:
这使得列表具有对同一列表的五个引用:
data = [[None]*5]*5
Use something like this instead which creates five separate lists:
使用类似这样的东西,创建五个单独的列表:
>>> data = [[None]*5 for _ in range(5)]
Now it does what you expect:
现在它符合您的期望:
>>> data[0][0] = 'Cell A1'
>>> print data
[['Cell A1', None, None, None, None],
[None, None, None, None, None],
[None, None, None, None, None],
[None, None, None, None, None],
[None, None, None, None, None]]
#2
10
As the python library reference for sequence types, which includes lists, says
作为序列类型的python库引用,包括列表,说
Note also that the copies are shallow; nested structures are not copied. This often haunts new Python programmers; consider:
还要注意副本很浅;嵌套结构不会被复制。这常常困扰着新的Python程序员;考虑:
>>> lists = [[]] * 3
>>> lists
[[], [], []]
>>> lists[0].append(3)
>>> lists
[[3], [3], [3]]
What has happened is that [[]] is a one-element list containing an empty list, so all three elements of [[]] * 3 are (pointers to) this single empty list. Modifying any of the elements of lists modifies this single list.
发生的事情是[[]]是一个包含空列表的单元素列表,因此[[]] * 3的所有三个元素都是(指向)这个空列表。修改列表的任何元素都会修改此单个列表。
You can create a list of different lists this way:
您可以通过以下方式创建不同列表的列表:
>>> lists = [[] for i in range(3)]
>>> lists[0].append(3)
>>> lists[1].append(5)
>>> lists[2].append(7)
>>> lists
[[3], [5], [7]]
#3
2
In python every variable is an object, and so a reference. You first created an array of 5 Nones, and then you build an array with 5 times the same object.
在python中,每个变量都是一个对象,因此是一个引用。您首先创建了一个包含5个Nones的数组,然后构建一个具有相同对象5倍的数组。
#1
50
This makes a list with five references to the same list:
这使得列表具有对同一列表的五个引用:
data = [[None]*5]*5
Use something like this instead which creates five separate lists:
使用类似这样的东西,创建五个单独的列表:
>>> data = [[None]*5 for _ in range(5)]
Now it does what you expect:
现在它符合您的期望:
>>> data[0][0] = 'Cell A1'
>>> print data
[['Cell A1', None, None, None, None],
[None, None, None, None, None],
[None, None, None, None, None],
[None, None, None, None, None],
[None, None, None, None, None]]
#2
10
As the python library reference for sequence types, which includes lists, says
作为序列类型的python库引用,包括列表,说
Note also that the copies are shallow; nested structures are not copied. This often haunts new Python programmers; consider:
还要注意副本很浅;嵌套结构不会被复制。这常常困扰着新的Python程序员;考虑:
>>> lists = [[]] * 3
>>> lists
[[], [], []]
>>> lists[0].append(3)
>>> lists
[[3], [3], [3]]
What has happened is that [[]] is a one-element list containing an empty list, so all three elements of [[]] * 3 are (pointers to) this single empty list. Modifying any of the elements of lists modifies this single list.
发生的事情是[[]]是一个包含空列表的单元素列表,因此[[]] * 3的所有三个元素都是(指向)这个空列表。修改列表的任何元素都会修改此单个列表。
You can create a list of different lists this way:
您可以通过以下方式创建不同列表的列表:
>>> lists = [[] for i in range(3)]
>>> lists[0].append(3)
>>> lists[1].append(5)
>>> lists[2].append(7)
>>> lists
[[3], [5], [7]]
#3
2
In python every variable is an object, and so a reference. You first created an array of 5 Nones, and then you build an array with 5 times the same object.
在python中,每个变量都是一个对象,因此是一个引用。您首先创建了一个包含5个Nones的数组,然后构建一个具有相同对象5倍的数组。