查找(start:end)子列表在列表中出现的位置。Python

Question

如果一个人有一长串数字：

example=['130','90','150','123','133','120','160','45','67','55','34']

和列表中的子列表，如

sub_lists=[['130','90','150'],['90','150'],['120','160','45','67']]

如何生成一个函数来获取这些子列表，并给出它们在原始字符串中出现的位置？要获得结果：

results=[[0-2],[1-2],[5-8]]

我在尝试一些类似的东西

example=['130','90','150','123','133','120','160','45','67','55','34']

sub_lists=[['130','90','150'],['90','150'],['120','160','45','67']]

for p in range(len(example)):
    for lists in sub_lists:
        if lists in example:
            print p

但那不管用吗？

Answer 1

这应该可以处理几乎所有的情况，包括一个子列表出现多次：

example=['130','90','150','123','133','120','160','45','67','55','34']
sub_lists=[['130','90','150'],['90','150'],['120','160','45','67']]

for i in range(len(example)):
    for li in sub_lists:
        length = len(li)
        if example[i:i+length] == li:
            print 'List %s has been matched at index [%d, %d]' % (li, i, i+length-1)

输出：

List ['130', '90', '150'] has been matched at index [0, 2]
List ['90', '150'] has been matched at index [1, 2]
List ['120', '160', '45', '67'] has been matched at index [5, 8]

Answer 2

这是可行的，但这只是因为我依赖于子列表存在于其内部的事实。

example=['130','90','150','123','133','120','160','45','67','55','34']

sub_lists=[['130','90','150'],['90','150'],['120','160','45','67']]

def f(example, sub_lists) :
    for l in sub_lists:
        yield [example.index(l[0]),example.index(l[-1])]

print [x for x in f(example,sub_lists)]

>>> [[0, 2], [1, 2], [5, 8]]