如何使用python正则表达式计算单词后跟特殊字符的单词的出现次数

Question

I want to count the number of occurrences of the word 'people' in a text using python. For that I use Counter and Python's regular expression:

    for j in range(len(paragraphs)):
        text = paragraphs[j].text
        count[j] = Counter(re.findall(r'\bpeople\b' ,text))

Yet, here my code does not take into account of the occurrences of people. people! people? How can I modify it to also count the cases when the word is followed by a specific character?

Thank you for you help,

Answer 1

You can use an optional character-group in your regex:

r'\bpeople[.,!?]?\b'

The ? specifies it can occure 0 or 1 times - the [] specifies what characters are allowed. There is no need to escape the . (or fe ()*+? ) inside [] although they have special meaning for regex. If you wanted to use a - inside [] you would need to escape it as it is used to denote ranges in sets [1-5] == 12345 .

See: https://docs.python.org/3/library/re.html#regular-expression-syntax

[] Used to indicate a set of characters. In a set:

Characters can be listed individually, eg [amk] will match 'a', 'm', or 'k'. Ranges of characters can be indicated by giving two characters and separating them by a '-', for example [az] will match any lowercase ASCII letter, [0-5][0-9] will match all the two-digits numbers from 00 to 59, and [0-9A-Fa-f] will match any hexadecimal digit. [...]

Answer 2

people[?.!]

This will allow you to only match with people? people. and/or people!

So if you add a few more Counter(re.finall( you will be able to do something like this

#This will only match people
count[j] = Counter(re.findall(r'people\s' ,text))

#This will only match people?
count[j] = Counter(re.findall(r'people\?' ,text))

#This will only match people.
count[j] = Counter(re.findall(r'people\.' ,text))

#This will only match people!
count[j] = Counter(re.findall(r'people\!' ,text))

You need to use the \\ to escape the special characters

Also this is a good resource when you are experimenting with python regular expressions: https://pythex.org/ The site also has a regular expression cheat sheet

Answer 3

You can use a modifier statement at the end of the 'people' part of your Regex pattern. Try the following:

for j in range(len(paragraphs)):
    text = paragraphs[j].text
    count[j] = Counter(re.findall('r\bpeople[.?!]?\b', text)

The ? is for zero or more quantifier. The above pattern seems to work on regex101.com but I haven't tried in out in a Python shell yet.

Answer 4

Does it have to use regex? Why not just:

len(text.split("people"))-1