在列表中查找子列表出现的(start:end)位置。Python

Question

如果你有一长串数字：

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

而且在这串数字中还有一些子列表，比如：

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

那么你该如何写一个函数，让它能找出这些子列表在原始数字串中出现的位置呢？这样你就能得到结果：

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

我试着做了一些类似的事情：

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

但那并没有成功？

Answer 1

这个方法有效，但仅仅是因为我依赖于子列表完整存在的事实。

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]]

Answer 2

这段代码应该能处理几乎所有情况，包括一个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)

输出结果：

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]