写一个正则表达式来提取'/'之前的数字

Question

I don't want to use string split because I have numbers 1-99, and a column of string that contain '#/#' somewhere in the text.

How can I write a regex to extract the number 10 in the following example:

He got 10/19 questions right.

Answer 1

Use a lookahead to match on the / , like this:

\d+(?=/)

You may need to escape the / if your implementation uses it as its delimiter.

Live example: https://regex101.com/r/xdT4vq/1

Answer 2

You can still use str.split() if you carefully construct logic around it:

t = "He got 10/19 questions right."
t2 = "He/she got 10/19 questions right"


for q in [t,t2]:


    # split whole string at spaces
    # split each part at / 
    # only keep parts that contain / but not at 1st position and only consists
    # out of numbers elsewise
    numbers = [x.split("/") for x in q.split() 
               if "/" in x and all(c in "0123456789/" for c in x)
              and not x.startswith("/")]
    if numbers:
        print(numbers[0][0])

Output:

10
10

Answer 3

res = re.search('(\d+)/\d+', r'He got 10/19 questions right.')
res.groups()
('10',)

Answer 4

import re
myString = "He got 10/19 questions right."
oldnumber = re.findall('[0-9]+/', myString)  #find one or more digits followed by a slash.
newNumber = oldnumber[0].replace("/","")  #get rid of the slash.

print(newNumber)
>>>10

Answer 5

Find all numbers before the forward-slash and exclude the forward-slash by using start-stop parentheses.

>>> import re
>>> myString = 'He got 10/19 questions right.'
>>> stringNumber = re.findall('([0-9]+)/', myString)
>>> stringNumber
['10']

This returns all numbers ended with a forward-slash, but in a list of strings. if you want integers, you should map your list with int , then make a list again.

>>> intNumber = list(map(int, stringNumber))
>>> intNumber
[10]