In python 3.6.0 and pandas 0.20.0
there is a date column yyyy-mm-dd
date
2017-08-16
2017-08-17
2017-08-18
There is the same question here
Convert a column of datetimes to epoch in Python
But sadly none of the solutions in the post works
df['date']=df['date'].astype('int64')//1e9
ValueError: invalid literal for int() with base 10: '2017-08-16'
df['date']=(df['date'] - dt.datetime(1970,1,1)).dt.total_seconds()
NameError: name 'dt' is not defined
Any thoughts? Thank you.
Instead of doing it manually, you can just convert it to datetime and use timestamp()
Something like this.
from datetime import datetime
s = '2017-08-16'
epoch = datetime.strptime(s, "%Y-%m-%d").timestamp()
print(epoch)
# Output -- 1502821800.0
try with pd.DateTimeIndex and multiply of 1000 is to make it milliseconds ( you can ignore multiplication if you want in seconds)
df['Date'] = (pd.DatetimeIndex(df['Date']).astype(np.int64) // 10**9) * 1000
print(df)
Date
1502841600000
1502928000000
1503014400000
Try this:
from datetime import datetime as dt
# String for the date
s = '2017-08-16'
#Convert string to datetime
s_dt = dt.strptime(s, '%Y-%m-%d')
#Initialize epoch
epoch = dt.utcfromtimestamp(0)
#Get your difference from epoch in ms
linux_time = (s_dt - dt.utcfromtimestamp(0)).total_seconds()*1000
#Your output
linux_time
1502841600000.0
This is a code for a single date object. You could create a function and use lambda to apply it to the pandas column.
Hope this helps.
