Help me to edit my code in the link

Question

When I enter date in 00/00/0000 format, it should output sensible date. For example,
11/20/2000 should output 20/11/2000.
https://code.sololearn.com/cCrw0rsFpEA4/?ref=app

Answer 1

Good try pal but there are more efficient solutions...
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SOLUTION 1 (Best)

from datetime import datetime

df = ['01/31/2014', '02/06/2013']
f = [datetime.strptime(d, "%d/%m/%Y").strftime("%m/%d/%Y") for d in df]
print(f)
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The way you tried, just a smaller ver.
SOLUTION 2

df = ['01/31/2014', '02/06/2013']

for a in df:
s = list(a)
d = [s[3],s[4],s[2],s[0],s[1],s[5],s[6],s[7],s[8],s[9]]
f.append("".join(d))

print (f)

Answer 2

Arun Bhattacharya Thanks for your honest help. But may you please explain your code!

Answer 3

My Solution 1 Explanation:

from datetime import datetime
df = ['01/31/2014', '02/06/2013']
f = [datetime.strptime(d, "%d/%m/%Y").strftime("%m/%d/%Y") for d in df]
print(f)

Line 1: Importing Datetime Module

Line 2: Considering any two dates in a form of a tuple (stored in "df")

Line 3: datetime.strptime(d, "%d/%m/%Y")
-> strptime is a method in datetime
-> The first parameter "d" -> is the dates in "df"
-> strptime parses the date
-> The Second parameter parses the "%d"->The Date, "%m"-> The Month, "%y"-> The Year

.strftime("%m/%d/%Y")
-> The "." brings the earlier statement together
-> strftime is a method in datetime
-> Here the parameter is the format in which we parsed earlier, so, now "%m"-> The Month, "%d"-> The Date, "%y"-> The Year

for d in df
-> This, in simpler manner, the "d" anywhere in the statement is the "df[0]" and then "df[1]"

Line4: You understand whats "print(f)"

strptime -> strptime means string parser, this will convert a string format to datetime.
strftime -> strftime means string formatter, this will format a datetime object to string format.

Answer 4

Thanks once again