python - 查找子列表出现在列表中的（开始：结束）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，例如

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

您将如何生成一个函数来获取这些子列表并为您提供它们在原始字符串中出现的位置？得到结果：

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

我正在尝试类似的东西

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

但这不起作用？

Answer 1

这应该可以处理几乎所有情况，包括不止一次出现的子列表：

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

这行得通，但只是因为我依赖于子列表存在于它们的完整性中的事实

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]]