How can I extract a number from a string in python without having to use regex? I have seen isinstance but the number could change to almost anything. Any ideas?
https://www.investopedia.com/articles/retirement/?page=6
5 Answers
It's a bit verbose, but I would use url parsing for this. The advantage overy regex is that you would get some input validation for free, and more readable code.
>>> from urllib.parse import urlparse, parse_qs
>>> url = 'https://www.investopedia.com/articles/retirement/?page=6'
>>> parsed = urlparse(url)
>>> query = parse_qs(parsed.query)
>>> [page] = query['page']
>>> int(page)
6
Comments
You can extract continuous groups of digits, anywhere on the string, using the following:
from itertools import groupby
url = 'https://www.investopedia.com/articles/retirement/?page=6&limit=10&offset=15'
print([int(''.join(group)) for key, group in groupby(iterable=url, key=lambda e: e.isdigit()) if key])
Output
[6, 10, 15]
Comments
This assumes that there isn't multiple blocks of integers (e.g. www.something212.com/page=?13)
You could try using list comprehensions and str.isdigit()
url = 'https://www.investopedia.com/articles/retirement/?page=6'
digits = [d for d in url if d.isdigit()]
digit = ''.join(digits)
digit
>>> 6
Edited: now works with digits above 9
15 Comments
vash_the_stampede
what happens if that 6 is 12?
yvesva
digits would produce
[6,12]. You could join the answer by using number = ''.join(map(int, digits))vash_the_stampede
I know, I'm saying why not address that in your answer?
vash_the_stampede
you could just
''.join(digits) since you already know whats in there
Kamikaze_goldfish
@vash_the_stampede I’m gonna have to agree with you. This is a pretty good bit of code.
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If the url always has that format with only digits at the end you could do this:
s = 'https://www.investopedia.com/articles/retirement/?page=25'
new = []
k = list(s)
[new.append(i) for i in k if i.isdigit()]
print(''.join(new))
(xenial)vash@localhost:~/python/stack_overflow$ python3.7 isdigit.py 25
Comments
I know you do not need re, but it is actually very powerful. Under the hood, most libraries make use of re. Here is my solution to handle this situation:
import re
url = "www.fake888.com/article/?article=123&page=9&group=8"
numbers = re.findall(r'(?<==)(\d+)', url)
print(f'Found: {" ".join(numbers)}')
varval = re.findall(r'(\w+)=(\d+)', url)
urldict = {}
for var in varval:
urldict[var[0]] = var[1]
print(urldict)
The output is
Found: 123 9 8
{'article': '123', 'page': '9', 'group': '8'}
1 Comment
Kamikaze_goldfish
I’m gonna have to check that out! Thanks for the link. I really have wanted to learn regex for a while because of its power. 💪
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