计算某些文本中多字串的出现次数

Question

所以对于某些文本中的单个单词子串计数，我可以使用some_text.split().count(single_word_substring)。对于某些文本中的多字子字符串计数，我该如何做到这一点？

例子：

text = 'he is going to school. abc is going to school. xyz is going to school.'
to_be_found = 'going to school'

计数应该是 3。

text = 'he is going to school. abc is going to school. xyz is going to school.'
to_be_found = 'going to'

计数应该是 3。

text = 'he is going to school. abc is going to school. xyz is going to school.'
to_be_found = 'go'

计数应为 0。

text = 'he is going to school. abc-xyz is going to school. xyz is going to school.'
to_be_found = 'school'

计数应该是 3。

text = 'he is going to school. abc-xyz is going to school. xyz is going to school.'
to_be_found = 'abc-xyz'

计数应该是 1。

假设 1： 一切都是小写的。 假设 2： 文本可以包含任何内容。 假设 3： to be found 也可以包含任何东西。比如car with 4 passengers、xyz & abc等。

注意：基于 REGEX 的解决方案是可以接受的。我只是好奇是否可以不使用正则表达式（很高兴拥有并且仅适用于将来可能对此感兴趣的其他人）。

Answer 1

这是一个使用正则表达式的有效解决方案：

import re

def occurrences(text,to_be_found):
    return len(re.findall(rf'\W{to_be_found}\W', text))

正则表达式中的大写 W 用于非单词字符，包括空格和其他标点符号。

Answer 2

设法让它与这段代码一起工作（但它根本不是 Pythonic 方式）：

text = 'he is going to school. abc is going to school. xyz is going to school.'
to_be_found = 'going to school'

def find_occurences(text, look_for):
    spec = [',','.','!','?']
    where = 0
    how_many = 0

    if not to_be_found in text:
        return how_many

    while True:
        i = text.find(look_for, where)

        if i != -1: #We have a match
            if (((text[i-1] == " ") and (text[i + len(look_for)] == " ")) #Check if the text is really alone
            or (((text[i-1] in spec) or ((text[i-1] == " "))) and (text[i + len(look_for)] in spec))): #Check if it is not surrounded by special characters such as ,.!?

                where = i + len(look_for)
                how_many += 1
            else:
                where = i + len(look_for)
        else:
            break
    
    return how_many

print("'{}' was in '{}' this many times: {}".format(to_be_found, text, find_occurences(text, to_be_found)))

1. 第一个条件：(text[i-1] == " ") and (text[i + len(look_for)] == " ") 检查子字符串是否没有被空格包围。
2. 第二个条件：((text[i-1] in spec) or ((text[i-1] == " "))) and (text[i + len(look_for)] in spec)) 检查子字符串是否没有被任何特殊字符和左边的空格包围。

---

示例 1：

to_be_found = 'going to school'
Output1: 3

示例 2：

to_be_found = 'going to'
Output2: 3

示例 3：

to_be_found = 'go'
Output3: 0

示例 4：

to_be_found = 'school'
Output4: 3

Answer 3

你试试这个：

text = 'he is going to school. abc is going to school. xyz is going to school.'
to_be_found = 'going to school'
i=0
r=0
while True :
  if text.find(to_be_found,i) <0 or i>len(text) :
    break
  elif text.find(to_be_found,i) >= 0 :
     r=r+1
     i=text.find(to_be_found,i)+len(to_be_found)


print(r)

Answer 4

1. 搜索子字符串的最佳本地方式仍然是计数。它可以根据需要与多字子字符串一起使用

   text = 'he is going to school. abc is going to school. xyz is going to school.'
   text.count('going to school') # 3
   text.count('going to') # 3
   text.count('school') # 3
   text.count('go') # 3
   

   对于 case 'go'，如果你需要 0，你可以搜索 'go'、'go' 或 'go' 来捕获单独的单词

2. 您也可以编写自己的方法来按字符搜索 https://stackoverflow.com/a/30863956/15080484