How can I get the years from a string and put it into a separate list?

Question

Code:

str1 = "2016-02-30T14:44:22.477Z, 2012-08-03T06:27:14.873Z, 2017-03-30T14:44:21.383Z"

I want only the years put into a separate list with output: ("2016", "2012", "2017")

Answer 1

You can use regular expressions. You will first need to import the regular expressions module import re and then

years = re.findall(r'\d\d\d\d', str1)

The findall method will return every substring of str1 in which the regular expression is true. The \d character describes a digit.

years will be a list ['2016', '2012', '2017']

Answer 2

Method 1 - Using split function

str1 = "2016-02-30T14:44:22.477Z, 2012-08-03T06:27:14.873Z, 2017-03-30T14:44:21.383Z"

years=[]
s = str1.split(",")

for i in s:
    years.append(i.split("-")[0])
    
print(years)

Method 2 - Using regex

str1 = "2016-02-30T14:44:22.477Z, 2012-08-03T06:27:14.873Z, 2017-03-30T14:44:21.383Z"

years = re.findall(r'\d{4}', str1)