如何从字符串中找到子串列表的位置？

Question

如何从字符串中找到子串列表的位置？

给定一个字符串：

“这架飞往圣彼得堡的飞机周六从沙姆沙伊赫起飞后仅 23 分钟就坠毁在埃及的西奈沙漠。”

和子字符串列表：

['The', 'plane', ',', 'bound', 'for', 'St', 'Petersburg', ',', 'crashed', 'in', 'Egypt', "'s" , 'Sinai', 'desert', 'just', '23', 'minutes', 'after', 'take-off', 'from', 'Sharm', 'el-Sheikh', 'on', '星期六'，'。']

期望的输出：

>>> s = "The plane, bound for St Petersburg, crashed in Egypt's Sinai desert just 23 minutes after take-off from Sharm el-Sheikh on Saturday."
>>> tokens = ['The', 'plane', ',', 'bound', 'for', 'St', 'Petersburg', ',', 'crashed', 'in', 'Egypt', "'s", 'Sinai', 'desert', 'just', '23', 'minutes', 'after', 'take-off', 'from', 'Sharm', 'el-Sheikh', 'on', 'Saturday', '.']
>>> find_offsets(tokens, s)
[(0, 3), (4, 9), (9, 10), (11, 16), (17, 20), (21, 23), (24, 34),
        (34, 35), (36, 43), (44, 46), (47, 52), (52, 54), (55, 60), (61, 67),
        (68, 72), (73, 75), (76, 83), (84, 89), (90, 98), (99, 103), (104, 109),
        (110, 119), (120, 122), (123, 131), (131, 132)]

输出说明，第一个子字符串“The”可以通过使用字符串

(start, end)

使用

s

索引找到。所以从所需的输出。

因此，如果我们遍历所需输出的所有整数元组，我们将返回子字符串列表，即

>>> [s[start:end] for start, end in out]
['The', 'plane', ',', 'bound', 'for', 'St', 'Petersburg', ',', 'crashed', 'in', 'Egypt', "'s", 'Sinai', 'desert', 'just', '23', 'minutes', 'after', 'take-off', 'from', 'Sharm', 'el-Sheikh', 'on', 'Saturday', '.']

我试过：

def find_offset(tokens, s):
    index = 0
    offsets = []
    for token in tokens:
        start = s[index:].index(token) + index
        index = start + len(token)
        offsets.append((start, index))
    return offsets

还有其他方法可以从字符串中找到子串列表的位置吗？

Answer 1

第一个解决方案：

#use list comprehension and list.index function.
[tuple((s.index(e),s.index(e)+len(e))) for e in t]

第二个解决方案纠正第一个解决方案中的问题：

def find_offsets(tokens, s):
    tid = [list(e) for e in tokens]
    i = 0
    for id_token,token in enumerate(tid):
        while (token[0]!=s[i]):            
            i+=1
        tid[id_token] = tuple((i,i+len(token)))
        i+=len(token)

    return tid


find_offsets(tokens, s)
Out[201]: 
[(0, 3),
 (4, 9),
 (9, 10),
 (11, 16),
 (17, 20),
 (21, 23),
 (24, 34),
 (34, 35),
 (36, 43),
 (44, 46),
 (47, 52),
 (52, 54),
 (55, 60),
 (61, 67),
 (68, 72),
 (73, 75),
 (76, 83),
 (84, 89),
 (90, 98),
 (99, 103),
 (104, 109),
 (110, 119),
 (120, 122),
 (123, 131),
 (131, 132)]   

#another test
s = 'The plane, plane'
t = ['The', 'plane', ',', 'plane']
find_offsets(t,s)
Out[212]: [(0, 3), (4, 9), (9, 10), (11, 16)]

Answer 2

import re

s = "The plane, bound for St Petersburg, crashed in Egypt's Sinai desert just 23 minutes after take-off from Sharm el-Sheikh on Saturday."
tokens = ['The', 'plane', ',', 'bound', 'for', 'St', 'Petersburg', ',', 'crashed', 'in', 'Egypt', "'s", 'Sinai', 'desert', 'just', '23', 'minutes', 'after', 'take-off', 'from', 'Sharm', 'el-Sheikh', 'on', 'Saturday', '.']


for token in tokens:
  pattern = re.compile(re.escape(token))
  print(pattern.search(s).span())

结果

(0, 3)
(4, 9)
(9, 10)
(11, 16)
(17, 20)
(21, 23)
(24, 34)
(9, 10)
(36, 43)
(44, 46)
(47, 52)
(52, 54)
(55, 60)
(61, 67)
(68, 72)
(73, 75)
(76, 83)
(84, 89)
(90, 98)
(99, 103)
(104, 109)
(110, 119)
(120, 122)
(123, 131)
(131, 132)

Answer 3

如果我们不知道子字符串，除了为每个子字符串重新扫描整个文本之外别无他法。

如果从数据看来，我们知道这些是文本的连续片段，按文本顺序给出，那么很容易在每次匹配后只扫描文本的rest。不过，每次都剪切文本是没有意义的。

def spans(text, fragments):
    result = []
    point = 0  # Where we're in the text.
    for fragment in fragments:
        found_start = text.index(fragment, point)
        found_end = found_start + len(fragment)
        result.append((found_start, found_end))
        point = found_end
    return result

测试：

>>> spans('foo in bar', ['foo', 'in', 'bar'])
[(0, 3), (4, 6), (7, 10)]

这假设每个片段都出现在文本中正确的位置。您的输出格式没有提供不匹配报告的示例。使用

.find

而不是

.index

可能会有所帮助，尽管只是部分帮助。

Answer 4

def spans2(text, fragments):
    result = []
    for fragment in fragments:
        found_start = text.index(fragment)
        found_end = found_start + len(fragment)
        result.append((found_start, found_end))
    return(result)

测试：

txt = "foo man bit"
frag = ["bit","man"]

spans2(txt,frag)

[(8, 11), (4, 7)]