CODEWITHSUNDEEP
pythonpandasdataframe
jdoe
jdoe

Reputation: 79

unsupported operand type(s) for -: 'datetime.date' and 'Timestamp'

I have a csv file which is a list of dates as strings e.g. 2018-05-04

I'm trying to make a new column in a dataframe that is the difference in days. I'm having a problem with different formats of date as apparently you can't subtract a pandas to_datetime from a datetime.date

This is what i have

import pandas as pd
import datetime as dt

today=dt.date.today() #<class 'datetime.date'>

df=pd.read_csv('test.csv')
df.columns=['ATH']
df['ATH']=pd.to_datetime(df['ATH'])  #<class 'pandas._libs.tslibs.timestamps.Timesta...
df['diff']=df['ATH'].apply(lambda x : abs(today-x).days)

and this is what happens: TypeError: unsupported operand type(s) for -: 'datetime.date' and 'Timestamp'

I can see in the pandas to_datetime documentation that it can return "scalar: Timestamp (or datetime.datetime)" and whilst this must be what's happening, I can't figure out what I should be doing instead.

If I print df after pd.to_datetime it's still just a list of dates

I found d.dtypes and ATH is datetime64[ns]

Upvotes: 1

Views: 13833

Answers (1)

user2246849
user2246849

Reputation: 4407

You are subtracting a pd.Timestamp object from a datetime.date instance. The latter has no notion of time. You should convert today to a datetime.datetime object instead:

# Initialises a datetime.datetime instance at midnight of the given date (today).
today = dt.datetime.combine(dt.date.today(), dt.datetime.min.time())

Or get a Timestamp with Pandas:

today = pd.to_datetime('today').normalize()

Upvotes: 2

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