Regex to identify all occurrences of years

Question

I made this Regex to capture the occurrences of years in a sequence of numbers (between 2010-2029).

text = '0412020982012'
rg = r'20[1-2][0-9]'
years = re.compile("(%s)" % (rg)).findall(text)

Works perfectly.

But if the string is 0420202198, it doesn’t work because it only picks up the occurrence of 2020, but not of 2021, because it’s sharing the digits.

In my case, I always want to take the years identified more to the right. How can I solve keeping the use of Regex?

Answer 1

Use:

.*(20[1-2][0-9])

Group 1 will always be the last sequence 20XX (where XX is in accordance with the regex criteria):

3420201298      -> grupo 1: 2012
    ¯¯¯¯
34352020982012  -> grupo 1: 2012
          ¯¯¯¯

The .* will fetch 0 or more characters to the left of the group, i.e., will consider the last sequence (2012) and not the first (2020).

REGEXR.COM

Answer 2

The problem is that findall traverse the string from left to right, and each time you find a match, the next search starts after the last found stretch.

In this case, in the string 0420202198, after finding the 2020, the search starts from the third 2 (what begins the stretch 2198), and therefore the 2021 is not found.

To do what you need, an option is to use the method match, passing the position where the search will begin:

import re

text = '0420202198'
r = re.compile('20[1-2][0-9]')
years = []
# fazer buscas começando em todas as posições da string
for pos in range(0, len(text)):
    m = r.match(text, pos)
    if m:
        years.append(m.group())

print(years)

In this case you do not need to include the parentheses in regex (they serve to create capture groups, who are returned by findall), for the method group, when called without parameters, already returns all the chunk that was found.

Besides, I didn’t have to raw string (the r before quotation marks). This syntax is useful when regex has characters such as \ (So I don’t have to write it as \\), but in this case it is not necessary.

And I also put the results on a list, so that the return is the same as findall (that returns a list of the results).

The exit is:

['2020', 'in 2021']

---

How your regex necessarily needs 4 digits to find a match, I could optimize a little bit the loop and use range(0, len(text) - 3) in the for (so I avoid iterating in the last 3 positions, because I know that from then on there are not enough characters to satisfy the regex).

---

You can also use the syntax of comprehensilist on, well over pythonic (and usually more succinct, but in this particular case, I do not know if it is so much):

import re

text = '0420202198'
r = re.compile('20[1-2][0-9]')    
years = [m.group() for pos in range(0, len(text) - 3) for m in [r.match(text, pos)] if m]
print(years)

Or else:

years = [m.group() for m in (r.match(text, pos) for pos in range(0, len(text) - 3)) if m]

Both options above removed from here.

Answer 3

Try changing the variable rg to obtain more than one occurrence of a group of numbers

rg = r'20[1-2]\d+'

For in this way its regex searches for one or more occurrence of digits. The part \d is equivalent to "[0-9]"

Answer 4

Follow Pattern to catch only the year 2012 which is the rightmost year as you said

(?:.*?\K20[1-2]{1}[2-9]{1}){2}

Click here to see the Example