测试列表是否包含另一个列表 [英] Test if list contains another list
问题描述
你好,
我正在尝试编写一些非常简单的东西来测试一个列表
是否包含另一个:
>
a = [1,2,3]
b = [3,2,1,4]
但是'' a in b''返回False。我怎样检查a确实包含了
b?
谢谢
mathieu写道:
你好,
我想写一些非常简单的东西来测试一下清单
包含另一个:
a = [1,2,3]
b = [3,2 ,1,4]
但是''in b''返回False。我怎样检查b确实包含了
?
使用套数:
>> a = [1,2,3]
b = [3,2,1,4]
设置(a).issubset(set(b))
True
Christian
>> a = [1,2,3]
b = [3,2,1,4]
a = set(a)
b = set(b)
a.intersection(b)
set([1,2,3])
这就是你想要的吗?
欢呼
詹姆斯
2008年9月8日,mathieu< ma ************* **@gmail.com写道:
你好,
我想写一些非常简单的东西来测试一下list
包含另一个:
a = [1,2,3]
b = [3,2,1,4]
但''在b'中'返回False。如何检查a确实包含
b?
谢谢
-
http://mail.python.org/mailman/listinfo/python -list
-
-
- "问题解决了方法
9月8日上午9:32,Bruno Desthuilliers
< bdesth.quelquech ... @ free.quelquepart.frwrote :
mathieuaécrit:
你好,
我正在尝试编写一些非常简单的东西来测试一个列表
是否包含另一个:
a = [1,2,3]
b = [3,2,1,4]
但''在b''返回False。
的确如此。列表不是集合,并且列表a
的所有元素恰好也是列表b的一部分这一事实并不会使列表本身成为
元素list b。
>> a = [1,2,3]
> > b = [3,2,1,4]
>> c = [b,a]
>> a in c
True
>> b in c
True
>> c
[[3,2,1,4],[1,2,3]]
>>>
如何检查a确实包含
in b?
但这就是你所做的 - 你*确认*是否包含在b中,并且
这不是案件。你想要的是检查一下
的*所有元素*是否包含在b中,这是另一个问题。现在回答
你的问题:使用套装。
>> set(a) .issubset(set(b))
True
>>>
谢谢大家!
Hi there,
I am trying to write something very simple to test if a list
contains another one:
a = [1,2,3]
b = [3,2,1,4]
but ''a in b'' returns False. How do I check that a is indeed contained
in b ?
thanks
mathieu wrote:Hi there,
I am trying to write something very simple to test if a list
contains another one:
a = [1,2,3]
b = [3,2,1,4]
but ''a in b'' returns False. How do I check that a is indeed contained
in b ?Use sets:
>>a = [1,2,3]
b = [3,2,1,4]
set(a).issubset(set(b))
True
Christian
Hi,
>>a = [1,2,3]
b = [3,2,1,4]
a = set(a)
b = set(b)
a.intersection(b)
set([1, 2, 3])
Is this what you want ?
cheers
James
On 9/8/08, mathieu <ma***************@gmail.comwrote:Hi there,
I am trying to write something very simple to test if a list
contains another one:
a = [1,2,3]
b = [3,2,1,4]
but ''a in b'' returns False. How do I check that a is indeed contained
in b ?
thanks
--
http://mail.python.org/mailman/listinfo/python-list
--
--
-- "Problems are solved by method"
On Sep 8, 9:32 am, Bruno Desthuilliers
<bdesth.quelquech...@free.quelquepart.frwrote:mathieu a écrit :
Hi there,
I am trying to write something very simple to test if a list
contains another one:
a = [1,2,3]
b = [3,2,1,4]
but ''a in b'' returns False.
Indeed. Lists are not sets, and the fact that all elements of list a
happens to also be part of list b doesn''t make the list a itself an
element of list b.
>>a = [1, 2, 3]
>>b = [3,2,1,4]
>>c = [b, a]
>>a in c
True
>>b in c
True
>>c
[[3, 2, 1, 4], [1, 2, 3]]
>>>
How do I check that a is indeed contained
in b ?
But that''s what you did - you *did* checked if a was contained in b, and
this is not the case. What you want is to check if *all elements* of a
are contained in b, which is quite another problem. Now the answer to
your question : use sets.
>>set(a).issubset(set(b))
True
>>>
thanks all !
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