python list of strings parsing

Question

I have a list of strings representing dates:

>>> dates
['14.9.2016',
 '13.9.2016',
 '12.9.2016',
 '11.9.2016',
 '10.9.2016',
 '9.9.2016',
 '8.9.2016',
 '7.9.2016',
 '6.9.2016',
 '5.9.2016']

I need to zero-padd days & months and I cannot use standard calendar methods due to "artifical dates" I need to work with ("29.2.2015" for example)

following seems to be working:

>>> parsed_dates = []
>>> for l in [d.split('.') for d in dates]:
>>>    parsed_dates.append('.'.join([i.zfill(2) for i in l]))

>>> parsed_dates
['14.09.2016',
 '13.09.2016',
 '12.09.2016',
 '11.09.2016',
 '10.09.2016',
 '09.09.2016',
 '08.09.2016',
 '07.09.2016',
 '06.09.2016',
 '05.09.2016']

is it possible to achieve the same result using a single list comprehension? or some other, more elegant way? I have tried following, but cannot find a way to concatenate single list items to form date strings again...

>>> [i.zfill(2) for l in [d.split('.') for d in dates] for i in l]
['14',
 '09',
 '2016',
 '13',
 '09',
 '2016',
 '12',
 '09',
.
.
.

Martijn Pieters · Accepted Answer · 2016-08-19 10:08:39Z

7

Sure, just inline the expression you pass to the parsed_dates.append() call, with l substituted with d.split('.') from your for loop:

['.'.join([i.zfill(2) for i in d.split('.')]) for d in dates]

Demo:

>>> ['.'.join([i.zfill(2) for i in d.split('.')]) for d in dates]
['14.09.2016', '13.09.2016', '12.09.2016', '11.09.2016', '10.09.2016', '09.09.2016', '08.09.2016', '07.09.2016', '06.09.2016', '05.09.2016']

edited Aug 19, 2016 at 10:08

answered Aug 19, 2016 at 10:00

Martijn Pieters

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