CODEWITHSUNDEEP
pythonpython-3.x
Omegaspard
Omegaspard

Reputation: 1980

Extract number from a string using a pattern

I have strings like :

's3://bukcet_name/tables/name=moonlight/land/timestamp=2020-06-25 01:00:23.180745/year=2019/month=5'

And from it I would like to obtain a tuple contain the year value and the month value as first and second element of my tuple.

('2019', '5')

For now I did this :

([elem.split('=')[-1:][0] for elem in part[0].split('/')[-2:]][0], [elem.split('=')[-1:][0] for elem in part[0].split('/')[-2:]][1])

It isn't very elegant, how could I do better ?

Upvotes: 2

Views: 82

Answers (2)

Hachiko1337
Hachiko1337

Reputation: 23

Perhaps regular expressions could do it. I would use regular expressions to capture the strings 'year=2019' and 'month=5' then return the item at index [-1] by splitting these two with the character '='. Hold on, let me open up my Sublime and try to write actual code which suits your specific case.

import re  
search_string = 's3://bukcet_name/tables/name=moonlight/land/timestamp=2020-06-25 01:00:23.180745/year=2019/month=5'  
string1 = re.findall(r'year=\d+', search_string)  
string2 = re.findall(r'month=\d+', search_string)  
result = (string1[0].split('=')[-1], string2[0].split('=')[-1]) print(result)

Upvotes: -1

Shubham Sharma
Shubham Sharma

Reputation: 71707

Use, re.findall along with the given regex pattern:

import re
matches = re.findall(r'(?i)/year=(\d+)/month=(\d+)', string)

Result:

# print(matches)
[('2019', '5')]

Test the regex pattern here.

Upvotes: 3

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