从开始和返回索引的Python子字符串（有重复的情况）[关闭]

Question

我有一个长字符串和相应长字符串的子字符串列表。

示例 1：

字符串：

" This is paragraph one "

子字符串列表：["This", "is paragraph", "one"]

我需要返回对应子串的索引

结果：[[0,4], [5, 17], [18, 21]]

示例 2： （可能有更多的空白，并且可能有重复的子字符串）

字符串：

"
This is a book       a book.
"

列表： 子串列表：["This", "is a", "book", "a", "book"]

结果：[[0,4], [5, 9], [10, 14], [21,22], [23, 27]]

Answer 1

您可以使用生成器函数：

def get_matches(s, sub):
   inds = []
   for i in sub:
      if (k:=[j for j in range(len(s)) if s[j:].startswith(i) and (not inds or j > max(inds))]):
         yield [k[0], k[0]+len(i)]
         inds.append(k[0])
         
s = 'This is a book       a book.'
subs = ['This', 'is a', 'book', 'a', 'book']
print(list(get_matches(s, subs)))

输出：

[[0, 4], [5, 9], [10, 14], [21, 22], [23, 27]]

Answer 2

您可以尝试以下方法：

s = "This is a book       a book."
subs = ["This", "is a", "book", "a", "book"]

bounds = []
end = 0
for sub in subs:
    bounds.append((start := s[end:].find(sub) + end, end := start + len(sub)))
print(bounds)

它给出：

[(0, 4), (5, 9), (10, 14), (21, 22), (23, 27)]

为了娱乐，同样使用re：

s = "This is a book       a book."
subs = ["This", "is a", "book", "a", "book"]

import re 
re.match(".*".join(f"({t})" for t in subs), s).regs[1:]

它给出：

((0, 4), (5, 9), (10, 14), (21, 22), (23, 27))