python 如何在没有循环的情况下检查另一个列表包含的列表？
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How to check a list contained by another list without a loop?
提问by Young
As the title mentions,is there any builtins to do this job?I looked for that in dir(list)but got no usable one.Thanks.
正如标题所提到的，是否有任何内置函数可以完成这项工作？我在里面找过，dir(list)但没有可用的。谢谢。
回答by nosklo
Depends on what you mean by "contained". Maybe this:
取决于你所说的“包含”是什么意思。也许这个：
if set(a) <= set(b):
print "a is in b"
回答by Etaoin
Assuming that you want to see if all elements of sublistare also elements of superlist:
假设您想查看 的所有元素sublist是否也是 的元素superlist：
all(x in superlist for x in sublist)
all(x in superlist for x in sublist)
回答by Adrien Plisson
the solution depends on what values you expect from your lists.
解决方案取决于您对列表的期望值。
if there is the possiblity of a repetition of a value, and you need to check that there is enough values in the tested container, then here is a time-inefficient solution:
如果存在重复值的可能性，并且您需要检查测试容器中是否有足够的值，那么这是一个时间效率低的解决方案：
def contained(candidate, container):
temp = container[:]
try:
for v in candidate:
temp.remove(v)
return True
except ValueError:
return False
test this function with:
使用以下命令测试此功能：
>>> a = [1,1,2,3]
>>> b = [1,2,3,4,5]
>>> contained(a,b)
False
>>> a = [1,2,3]
>>> contained(a,b)
True
>>> a = [1,1,2,4,4]
>>> b = [1,1,2,2,2,3,4,4,5]
>>> contained(a,b)
True
of course this solution can be greatly improved: list.remove() is potentially time consuming and can be avoided using clever sorting and indexing. but i don't see how to avoid a loop here...
当然，这个解决方案可以大大改进：list.remove() 可能很耗时，可以使用巧妙的排序和索引来避免。但我不知道如何在这里避免循环......
(anyway, any other solution will be implemented using sets or list-comprehensions, which are using loops internally...)
（无论如何，任何其他解决方案都将使用集合或列表推导来实现，它们在内部使用循环......）
