Python 根据标签列表中某个单词的索引位置，查找该单词在字符串中的起始位置和结束位置

我有一个句子

str = 'cold weather gives me cold' 

和一份清单

 tag = ['O','O','O','O','disease'] 

这表明句子中的第五个单词是一种疾病类型。现在我需要得到第五个单词的开始和结束位置

---

如果我只使用'cold'进行字符串搜索，它将给出首先出现的'cold'的起始位置

以下内容将输出给定单词的起始和结束位置，假设单词之间用空格分隔：

str = 'cold weather gives me cold'
word_idx = 4 # index of the word we are looking for

split_str = str.split(' ')
print(split_str[word_idx]) # outputs 'cold'

start_pos = 0
for i in range(word_idx):
    start_pos += len(split_str[i]) + 1 # add one because of the spaces between words
end_pos = start_pos + len(split_str[word_idx]) - 1

print(start_pos) # prints 22
print(end_pos) # prints 25

---

您可以简单地拆分字符串，然后再次连接它，但这有点尴尬

string_list = string.split(" ")
word_start = len(" ".join(string_list[:4])) + 1
word_end = word_start + len(string_list[4])

---

使用

itertools

和

re

：

import re
from itertools import accumulate

def find_index(string, n):
    words = string.split()
    len_word = len(words[n])
    end_index = list(accumulate(map(len, re.split('(\s)' , string))))[::2][n]
    return end_index - len_word, end_index - 1

使用它：

find_index('cold weather gives me cold', 4) #5th word means 4 in indexing

输出：

(22, 25)

(22, 25)

[[22, 26]]

这应该可以做到

def get(str,target_index):
  start = len(" ".join(str.split(" ")[:target_index])) + 1
  end = start + len(str.replace('.','').split(' ')[target_index])
  return (start,end)

str = 'cold weather gives me cold.' 
tag = ['O','O','O','O','disease']
start,end = get(str,tag.index('disease'))
print(start,end,str[start:end]) # outputs 22 26 cold

str = 'cold weather gives me cold'
tag = ['O','O','O','O','disease']
start,end = get(str,tag.index('disease'))
print(start,end,str[start:end]) # outputs 22 26 cold

str = 'cold weather gives me cold and cough' 
tag = ['O','O','O','O','disease']
start,end = get(str,tag.index('disease'))
print(start,end,str[start:end]) # outputs 22 26 cold

看到它在行动

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希望有帮助

首先从标签中查找疾病索引，然后从数据中查找疾病名称，然后查找开始和结束索引：

strData = 'cold weather gives me cold' 
tag = ['O','O','O','O','disease']
diseaseIndex = tag.index('disease')
diseaseName = strData.split()[diseaseIndex]
print(diseaseName)
diseaseNameStartIndex = sum(len(word) for (index, word) in enumerate(strData.split()) if index< diseaseIndex ) + diseaseIndex
diseaseNameEndIndex = diseaseNameStartIndex + len(diseaseName) -1
print("diseaseNameStartIndex = ",diseaseNameStartIndex)
print("diseaseNameEndIndex = ",diseaseNameEndIndex)

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尝试使用此功能：

def find_index(s, n):
    length = len(s.split()[n])
    index = [(0, len(s.split()[0]) - 1)]
    for i in s.split():
        index.append((index[-1][0] + len(i), index[-1][1] + len(i)))
    return index[n + 1]
print(find_index('cold weather gives me cold', 4))

输出：

(22, 25)

(22, 25)

[[22, 26]]

---

如果您必须对一个长行执行此操作，最好使用

迭代器

，该迭代器将使用

re.finditer

方法生成单词的起始和结束位置，然后使用

islice

>>> str = 'cold weather gives me cold' 
>>> word_pos = iter((match.group(), match.span(1)) for match in re.finditer(r'(\S+)\S', string))
>>>
>>> n=4
>>> next(islice(word_pos, n, n+1))
('cold', (22, 25))

---

您可以将

re

与列表一起使用：

import re
s = 'cold weather gives me cold' 
new_s = re.findall('\w+|\s+', s)
l = [(a, sum(map(len, new_s[:i]))) for i, a in enumerate(new_s) if a != ' ']

---

输出：

(22, 25)

(22, 25)

[[22, 26]]

---

所以你基本上需要字符串的第五个字？或者你需要它的索引吗？你想要COLD的第一个和最后一个字吗？我只想要最后一个COLD，字符串索引的第五个字。看起来你只需要

str.split（）[4]

btw，好吧，这里只是一个例子，但是'str'变量名将隐藏相应的内置项。