在python 3中查找字符串中出现的单词

问题描述

我试图找到一个字在字符串中出现的次数。

  word =dog
 str1 =狗吠
  
 
 
 
我使用以下方法来计算出现次数：
  count = str1.count（word）
  
问题是我想要一个完全匹配。所以这句话的计数为0. 
是可能的吗？
解决方案
 ：
  import re 
 count = sum（1 for _ in re.finditer（r'\％ \ b'％re.escape（word），input_string））
  
创建任何中间列表（不像 split（）），因此对于大的 input_string  
 
 
它也有正确工作标点符号的好处 - 它会正确返回 1 作为短语Mike看到一只狗。（而无参数 split（）不会）。它使用 \b  regex标志，它匹配字边界（ \w  aka  [a-zA-Z0-9 _] 和任何其他）。
 
 
 
如果您需要担心ASCII字符以外的语言设置，您可能需要调整正则表达式以正确匹配这些语言中的非字符字符，但对于许多应用程序，这将是一个过于复杂，在许多其他情况下，设置正则表达式的Unicode和/或区域设置标志就足够了。 / p> 
I'm trying to find the number of occurrences of a word in a string. 
word = "dog"
str1 = "the dogs barked"
I used the following to count the occurrences:
count = str1.count(word)
The issue is I want an exact match. So the count for this sentence would be 0.Is that possible?
 解决方案 
If you're going for efficiency:
import re
count = sum(1 for _ in re.finditer(r'\b%s\b' % re.escape(word), input_string))
This doesn't need to create any intermediate lists (unlike split()) and thus will work efficiently for large input_string values.
It also has the benefit of working correctly with punctuation - it will properly return 1 as the count for the phrase "Mike saw a dog." (whereas an argumentless split() would not). It uses the \b regex flag, which matches on word boundaries (transitions between \w a.k.a [a-zA-Z0-9_] and anything else).
If you need to worry about languages beyond the ASCII character set, you may need to adjust the regex to properly match non-word characters in those languages, but for many applications this would be an overcomplication, and in many other cases setting the unicode and/or locale flags for the regex would suffice.
                        
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