使用Python测试一个列表是否包含另一个列表

以下是我的版本:

def contains(small, big):
    for i in xrange(len(big)-len(small)+1):
        for j in xrange(len(small)):
            if big[i+j] != small[j]:
                break
        else:
            return i, i+len(small)
    return False

因为在他的注释中,Andrew,think,is,in,the,start,of,the,and,think,as,the,start,of,and,因为他在评论的开头和结尾都是。它不切片任何子列表,因此应该是合理有效的。

这是新手的兴趣所在 else clause on the for statement

这与在字符串中查找子字符串相同,因此对于大型列表,实现类似于 Boyer-Moore algorithm .

如果所有项目都是唯一的,则可以使用集合。

>>> items = set([-1, 0, 1, 2])
>>> set([1, 2]).issubset(items)
True
>>> set([1, 3]).issubset(items)
False

all() 和 any() 函数来执行此操作。 big 包含中的所有元素 small

result = all(elem in big for elem in small)

检查是否 小的 包含中的任何元素

result = any(elem in big for elem in small)

变量结果将是布尔值(TRUE/FALSE)。

Rabin-Karp algorithm 如果 big 名单真的很大。这个链接甚至包含了几乎可以用Python编写的代码。

编辑后:

def contains(small, big):
    for i in xrange(1 + len(big) - len(small)):
        if small == big[i:i+len(small)]:
            return i, i + len(small) - 1
    return False

这是有效的,而且相当快,因为它使用内置的线性搜索 list.index() 方法和 == 操作员:

def contains(sub, pri):
    M, N = len(pri), len(sub)
    i, LAST = 0, M-N+1
    while True:
        try:
            found = pri.index(sub[0], i, LAST) # find first elem in sub
        except ValueError:
            return False
        if pri[found:found+N] == sub:
            return [found, found+N-1]
        else:
            i = found+1

#!/usr/bin/env python

def list_find(what, where):
    """Find `what` list in the `where` list.

    Return index in `where` where `what` starts
    or -1 if no such index.

    >>> f = list_find
    >>> f([2, 1], [-1, 0, 1, 2])
    -1
    >>> f([-1, 1, 2], [-1, 0, 1, 2])
    -1
    >>> f([0, 1, 2], [-1, 0, 1, 2])
    1
    >>> f([1,2], [-1, 0, 1, 2])
    2
    >>> f([1,3], [-1, 0, 1, 2])
    -1
    >>> f([1, 2], [[1, 2], 3])
    -1
    >>> f([[1, 2]], [[1, 2], 3])
    0
    """
    if not what: # empty list is always found
        return 0
    try:
        index = 0
        while True:
            index = where.index(what[0], index)
            if where[index:index+len(what)] == what:
                return index # found
            index += 1 # try next position
    except ValueError:
        return -1 # not found

def contains(what, where):
    """Return [start, end+1] if found else empty list."""
    i = list_find(what, where)
    return [i, i + len(what)] if i >= 0 else [] #NOTE: bool([]) == False

if __name__=="__main__":
    import doctest; doctest.testmod()

def contains(subseq, inseq):
    return any(inseq[pos:pos + len(subseq)] == subseq for pos in range(0, len(inseq) - len(subseq) + 1))

这里是我用来调整这一行程序的单元测试:

https://gist.github.com/anonymous/6910a85b4978daee137f

def contains(a,b):
    str(a)[1:-1].find(str(b)[1:-1])>=0

这是我的答案。这个函数将帮助您找出B是否是a的子列表。时间复杂度是O(n)。

`def does_A_contain_B(A, B): #remember now A is the larger list
    b_size = len(B)
    for a_index in range(0, len(A)):
        if A[a_index : a_index+b_size]==B:
            return True
    else:
        return False`

我试着尽可能地提高效率。

their documentation 以及 yield expressions .

基本上,它从子序列中创建一个值生成器,可以通过发送一个真值来重置这些值。如果发电机复位,它将从 sub .

sequence 与发电机产量,如果不匹配,重置发电机。

附属的 如果不重置,就意味着我们找到了匹配的对象。

str.find False 而不是 -1 .

进一步说明:我认为,与Python标准一致,返回的tuple的第二个值通常应该高一个。即 "string"[0:2] == "st"

这取决于这是一个通用例程,还是实现了某个特定的目标;在后一种情况下,最好实现一个通用例程,然后将其包装在一个函数中,该函数可以旋转返回值以符合规范。

def reiterator(sub):
    """Yield elements of a sequence, resetting if sent ``True``."""
    it = iter(sub)
    while True:
        if (yield it.next()):
            it = iter(sub)

def find_in_sequence(sub, sequence):
    """Find a subsequence in a sequence.

    >>> find_in_sequence([2, 1], [-1, 0, 1, 2])
    False
    >>> find_in_sequence([-1, 1, 2], [-1, 0, 1, 2])
    False
    >>> find_in_sequence([0, 1, 2], [-1, 0, 1, 2])
    (1, 3)
    >>> find_in_sequence("subsequence",
    ...                  "This sequence contains a subsequence.")
    (25, 35)
    >>> find_in_sequence("subsequence", "This one doesn't.")
    False

    """
    start = None
    sub_items = reiterator(sub)
    sub_item = sub_items.next()
    for index, item in enumerate(sequence):
        if item == sub_item:
            if start is None: start = index
        else:
            start = None
        try:
            sub_item = sub_items.send(start is None)
        except StopIteration:
            # If the subsequence is depleted, we win!
            return (start, index)
    return False

我觉得这个很快。。。

def issublist(subList, myList, start=0):
    if not subList: return 0
    lenList, lensubList = len(myList), len(subList)
    try:
        while lenList - start >= lensubList:
            start = myList.index(subList[0], start)
            for i in xrange(lensubList):
                if myList[start+i] != subList[i]:
                    break
            else:
                return start, start + lensubList - 1
            start += 1
        return False
    except:
        return False

大多数答案的问题是,它们适合于列表中唯一的项目。如果项目不是唯一的,并且您仍然想知道是否存在交叉点,则应该对项目进行计数:

from collections import Counter as count

def listContains(l1, l2):
  list1 = count(l1)
  list2 = count(l2)

  return list1&list2 == list1

print( listContains([1,1,2,5], [1,2,3,5,1,2,1]) ) # Returns True
print( listContains([1,1,2,8], [1,2,3,5,1,2,1]) ) # Returns False

也可以使用 ''.join(list1&list2)

a=[[1,2] , [3,4] , [0,5,4]]
print(a.__contains__([0,5,4]))

它提供真实的输出。

a=[[1,2] , [3,4] , [0,5,4]]
print(a.__contains__([1,3]))

它提供错误的输出。