如何使用Python在ISO中格式化日期？

Question

I have some dates which format is d/m/yyyy (eg 2/28/1987). I would like to have it in the ISO format : 1987-02-28

I think we can do it that way, but it seems a little heavy:

str_date = '2/28/1987'
arr_str = re.split('/', str_date)
iso_date = arr_str[2]+'-'+arr_str[0][:2]+'-'+arr_str[1]

Is there another way to do it with Python?

Answer 1

You could use the datetime module :

datetime.datetime.strptime(str_date, '%m/%d/%Y').date().isoformat()

or, as running code:

>>> import datetime
>>> str_date = '2/28/1987'
>>> datetime.datetime.strptime(str_date, '%m/%d/%Y').date().isoformat()
'1987-02-28'

Answer 2

I would use the datetime module to parse it:

>>> from datetime import datetime
>>> date = datetime.strptime('2/28/1987', '%m/%d/%Y')
>>> date.strftime('%Y-%m-%d')
'1987-02-28'

Answer 3

What you have is very nearly how I would write it, the only major improvement I can suggest is using plain old split instead of re.split , since you do not need a regular expression:

arr_str = str_date.split('/')

If you needed to do anything more complicated like that I would recommend time.strftime , but that's significantly more expensive than string bashing.