如何使用正则表达式从句子中提取两位数？

Question

I am trying to make a function that only extracts the two digit interger out of a specific regex expression.

def extract_number(message_text):
    regex_expression = 'What are the top ([0-9]{2}) trends on facebook'
    regex= re.compile(regex_expression)
    matches = regex.finditer(message_text)
    for match in matches:
        return match.group()

    # if there were no matches, return None
    return None

So that when I print

message_text= 'What are the top 54 trends on facebook today'
print(extract_number(message_text))

I will get the number 54. If I write the following beneath, i get the whatever character that I put in (.+)...Why won't it work for numbers?

def extract_number(message_text):
    regex_expression = 'What are the top (.+) trends on facebook'
    regex= re.compile(regex_expression)
    matches = regex.finditer(message_text)
    for match in matches:
        return match.group()

message_text= 'What are the top fifty trends on facebook today'
print(extract_number(message_text))

Answer 1

The only problem with both your snippets is that you're not returning the capture-group result of interest, but the overall match:

return match.group()

is the same as return match.group(0) , ie, it'll return the overall match , which in your case is the entire input string .

By contrast, you want index 1 , ie, what the 1st capture group - the first subexpression enclosed in (...) , ([0-9]{2}) - matched:

return match.group(1)

To put it all together:

def extract_number(message_text):
    regex_expression = 'What are the top ([0-9]{2}) trends on facebook'
    regex= re.compile(regex_expression)
    matches = regex.finditer(message_text)
    # (See bottom of this answer for a loop-less alternative.)
    for match in matches:
        return match.group(1)  # index 1 returns what the 1st capture group matched

    # if there were no matches, return None
    return None

message_text= 'What are the top 54 trends on facebook today'
print(extract_number(message_text))

This yields the desired output:

54

---

Note: As @EvanL00 points out, given that only ever 1 match is needed, the use of regex.finditer() with a subsequent for loop that unconditionally returns in the first iteration is unnecessary and may obscure the intent of the code; the simpler and clearer approach is:

match = regex.search(message_text) # Get first match only.
if match:
    return match.group(1)

Answer 2

This should work for numeric/string:

def extract_number(message_text):
    regex_expression = 'What are the top ([a-zA-Z0-9]+) trends on facebook'
    regex= re.compile(regex_expression)
    matches = regex.findall(message_text)
    if matches:
        return matches[0]

message_text= 'What are the top fifty trends on facebook today'
print(extract_number(message_text))
message_text= 'What are the top 50 trends on facebook today'
print(extract_number(message_text))
message_text= 'What are the top -- trends on facebook today'
print(extract_number(message_text))

Output:

fifty
50
None