将多种日期格式转换为一种格式python

Question

The date variable in my data is in multiple formats like DD/MM/YYYY D/MM/YY DD/M/YYYY 12/8/2017 27/08/17 8/9/2017 10/9/2017 15/09/17..

I need to change these multiple formats into one single format like DD/MM/YYYY

Tried to create a parsing function

def parse_date(date):
if date == '':
    return None
else:
    return dt.strptime(date, '%d/%m/%y').date()

and when I apply this function to my dataset, it throws me the following error..

"ValueError Traceback (most recent call last) in () ----> 1 data.Date = data.Date.apply(parse_date)

Unconverted Data Remains Error ValueError: unconverted data remains: 17"

How can I solve the unconverted data remains error?

Answer 1

You can use the dateutil module to do this

import dateutil.parser as dparser
a = ["12/8/2017", "27/08/17", "8/9/2017", "10/9/2017", "15/09/17"]

for i in a:
    print dparser.parse(i,fuzzy=True).date()

Result:

2017-12-08
2017-08-27
2017-08-09
2017-10-09
2017-09-15

Answer 2

This is because %y expects a 4 digit number.

In order to cover multiple date formats, you can have a look at the dateparser library. ( Docs )

Otherwise you will have to manually go through possible types or extend the dates yourself. If you are sure you only need to extend the year part you can do something like this before feeding the string to the parser:

date_parts = date.split('/')
if len(date_parts[2]) == 2:
    date_parts[2] = "20" + date_parts[2]
date = '/'.join(date_parts)

I think using the dateparser library is the way to go, as it is more extendible.

Answer 3

A basic approach is to split the strings on the slashes, and then re-join them with the correct number of digits. A simple approach:

date = "12/8/2017"

parts = date.split("/")

print(parts) # ['12', '8', '2017']

if len(parts[0]) == 1:
    parts[0] = "0" + parts[0]
if len(parts[1]) == 1:
    parts[1] = "0" + parts[1]
if len(parts[2]) == 2:
    parts[2] = "20" + parts[2]
newDate = "/".join(parts)
# or 
newDate = parts[0] + "/" + parts[1] + "/" + parts[2]

print(newDate) # 12/08/2017

Then you have a consistent date format throughout. (An additional check is required if your dates extend into the last century.)

I would test this first, and consider the other answers' approaches if this is not performant.