在Python中查找子列表在列表中出现的位置（起始:end）。

Question

如果有一份很长的数字列表：

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

在列表中还有子列表，如下：

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

如何生成一个函数，输入这些子列表，然后输出它们在原始字符串中出现的位置？ 获取结果的方法：

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

我在尝试类似于以下的东西

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

但是那个方法不起作用了吗？

- O.rka

---

我理解results=[[0-2],[1-2],[5-8]]只是您自己的表示法，以说明您想要返回的数字，如果您真的想将其作为字符串获取，您应该使用字符串替换。类似于return "result=[[%d-%d], [%d-%d..." % (<sequence of values here>)。 - mac

可能是重复的问题：Python：在列表中查找子列表的起始和结束索引 - Nikana Reklawyks

Answer 1

这应该可以处理几乎所有情况，包括子列表出现多次的情况：

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]

- mac

Answer 2

这个方法可以奏效，但前提是子列表完整存在。

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]]

- Pengman