CODEWITHSUNDEEP
pythonstringdate
Rahul Rajaram
Rahul Rajaram

Reputation: 47

Converting non padded string to Date (in a given format) in python

I'm trying to convert non padded string to date in python and getting following error:

e_dataframe['Modified_Date'] =  pd.to_datetime(e_dataframe['Modified_Date'], format='%d/%m/%y %I:%M %p')

ValueError: time data '1/08/2018 4:41 PM' does not match format '%d/%m/%y %I:%M %p' (match)

The format is similar still getting an error.

Upvotes: 0

Views: 182

Answers (1)

Martijn Pieters
Martijn Pieters

Reputation: 1122002

%y matches a year with two digits, omitting the century. Your input has a 4 digit year, so you need to use %Y:

'%d/%m/%Y %I:%M %p'

When in doubt, use datetime.strptime() with a substring of your input so you can see what parts work and what parts fail:

>>> import datetime
>>> datetime.datetime.strptime('1', '%d')
datetime.datetime(1900, 1, 1, 0, 0)
>>> datetime.datetime.strptime('1/08', '%d/%m')
datetime.datetime(1900, 8, 1, 0, 0)
>>> datetime.datetime.strptime('1/08/2018', '%d/%m/%y')
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/Users/mj/Development/Library/buildout.python/parts/opt/lib/python3.7/_strptime.py", line 577, in _strptime_datetime
    tt, fraction, gmtoff_fraction = _strptime(data_string, format)
  File "/Users/mj/Development/Library/buildout.python/parts/opt/lib/python3.7/_strptime.py", line 362, in _strptime
    data_string[found.end():])
ValueError: unconverted data remains: 18

The error indicates that %y only processed the 20 part of the year, telling us that it was not the correct placeholder for the 2018 input.

Upvotes: 3

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