如何正确计算字符串中给定单词的出现次数而不计算作为 Python 中不同单词的子字符串的单词？

Question

I want to calculate the occurrences of a given word in an article. I tried to use split method to cut the articles into n pieces and calculate the length like this.

def get_occur(str, word):
    lst = str.split(word)
    return len(lst) - 1

But the problem is, I will always count the word additionally if the word is a substring of another word. For example, I only want to count the number of "sad" in this sentence "I am very sad and she is a saddist" . It should be one, but because "sad" is part of "saddist", I will count it accidentally. If I use " sad " , I will omit words that are at the start and end of sentences. Plus, I am dealing with huge number of articles so it is most desirable that I don't have to compare each word. How can I address this? Much appreciated.

Answer 1

As mentioned by @ schwobaseggl in the comment this will miss the word before the comma and there may be other cases so I have updated the answer.

from nltk.tokenize import word_tokenize
text = word_tokenize(text)

This will give you a list of words. Now use the below code

count = 0
for word in text:
 if (word.lower() == 'sad'): # .lower to make it case-insensitive
   count += 1

Answer 2

You can use regular expressions:

import re

def count(text, pattern):
    return len(re.findall(rf"\b{pattern}\b", text, flags=re.IGNORECASE))

\\b marks word boundaries and the passed flag makes the matching case insensitive:

>>> count("Sadly, the SAD man is sad.", "sad")
2

If you want to only count lower-case occurrences, just omit the flag.

Answer 3

string = "I am very sad and she is sadder"
substring = " sad "

count = string.count(substring)