Replace datetime characters in a string

Question

Consider the following example str of dates:

'20180101,20180102,20180103,20180104,20180105,20180106,20180107,20180108,20180109,20180110,20180111,20180112,20180113,20180114,20180115,20180116,20180117,20180118,20180119,20180120,20180121,20180122,20180123,20180124,20180125,20180126,20180127,20180128,20180129,20180130,20180131,20180201,20180202,20180203,20180204,20180205,20180206,20180207,20180208'.

Using .replace('2018','2018-'), or re.sub('2018', '2018-', foo_string) from the module re, a date changes to 2018-0101. However, I'd like to insert a hyphen for a range of dates. I have tried the following re.sub('{}'.format(range(2018,2019)), '2018-', foo_string) to no avail. The string is returned without any changes. For reference the years I'd like to range over for the hyphen are 1981 to 2018.

Additionally, I'm unsure of a way to insert a hyphen between the month and dates in a clean way. Help along these directions would be greatly appreciated.

So you want 20180101 to be 2018-01-01?

heemayl
– heemayl

2018-02-18 11:57:46 +00:00
Commented Feb 18, 2018 at 11:57 — heemayl
– heemayl, Commented Feb 18, 2018 at 11:57
@heemayl Yes, all dates are in this format.

sunspots
– sunspots

2018-02-18 11:59:39 +00:00
Commented Feb 18, 2018 at 11:59 — sunspots
– sunspots, Commented Feb 18, 2018 at 11:59

Matt. Stroh · Accepted Answer · 2018-02-18 12:27:30Z

4

I think that using datetime will be more pythonic,

from datetime import datetime

date_list = '20180101,20180102,20180103,20180104,20180105,20180106,20180107,20180108'

def new_date_format(old_format_str):
    date_format = datetime.strptime(old_format_str, '%Y%m%d')
    new_format = date_format.strftime("%Y-%m-%d")
    return new_format

for date_str in date_list.split(","):
    print(new_date_format(date_str))

This will produce the following output:

edited Feb 18, 2018 at 12:27

answered Feb 18, 2018 at 12:14

Matt. Stroh

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Comments

Superluminal · Accepted Answer · 2018-02-18 12:04:17Z

1

I would do

def insert_hyphen(stri):
    return ",".join([date[:4]+"-"+date[4:] for date in stri.split(',')])

if you want double hyphens then do

def insert_hyphen(stri):
    return ",".join([date[:4]+"-"+date[4:6]+"-"+date[6:] for date in stri.split(',')])

edited Feb 18, 2018 at 12:04

answered Feb 18, 2018 at 12:00

Superluminal

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1 Comment

sunspots Over a year ago

Works better if you combine KGSH's answer with yours, i.e.,

def insert_hyphen(stri):     return ",".join([date[:4:] + '-' + date[4:6:] + '-' + date[6::] for date in stri.split(',')])

jpp · Accepted Answer · 2018-02-18 12:32:32Z

If your list consists of dates, why not store your list as dates?

Python has a datetime module which facilitates just this:

from datetime import datetime

lst = '20180101,20180102'

# first convert to datetime
lst = [datetime.strptime(i, '%Y%M%d') for i in lst.split(',')]

# define year combiner
def year_range(lst, year_start, year_end):
    return [i.replace(year=j) for j in range(year_start, year_end+1) for i in lst]

# make list for year range from lst
lst = year_range(lst, 2015, 2018)

# [datetime.datetime(2015, 1, 1, 0, 1),
#  datetime.datetime(2016, 1, 1, 0, 1),
#  datetime.datetime(2017, 1, 1, 0, 1),
#  datetime.datetime(2018, 1, 1, 0, 1),
#  datetime.datetime(2015, 1, 2, 0, 1),
#  datetime.datetime(2016, 1, 2, 0, 1),
#  datetime.datetime(2017, 1, 2, 0, 1),
#  datetime.datetime(2018, 1, 2, 0, 1)]

Giovanni Valerio · Accepted Answer · 2018-02-18 14:03:41Z

1

if you DO need to use the re module, you just match the sequence of numbers and replace them with what you find intermixed with the hyphens:

import re

string = '20180101,20180102,20180103,20180104,20180105,20180106,20180107,20180108,20180109,20180110,20180111,20180112,20180113,20180114,20180115,20180116,20180117,20180118,20180119,20180120,20180121,20180122,20180123,20180124,20180125,20180126,20180127,20180128,20180129,20180130,20180131,20180201,20180202,20180203,20180204,20180205,20180206,20180207,20180208'

result = re.sub('([0-9]{4})([0-9]{2})([0-9]{2})','\\1-\\2-\\3',string,0)

print(result)

answered Feb 18, 2018 at 14:03

Giovanni Valerio

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Comments

Gareth Ma · Accepted Answer · 2018-02-18 11:56:22Z

0

Use this:

string = '20180203'
new_string = string[:4:] + '-' + string[4:6:] + '-' + string[6::]
print (new_string)

Result:

2018-02-03

Just make sure that the date is in the format YYYYMMDD.

answered Feb 18, 2018 at 11:56

Gareth Ma

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