Pandas 系列：字符串日期到纪元 unix 秒

Question

I have a Pandas Dataframe where one column is in a string date format as below

0               time
1  September 20 2016  
2  September 20 2016     
3  September 19 2016     
4  September 16 2016

What would be a succinct way for replacing time to be in epoch unix seconds?

Answer 1

You can modify the values of a column using the Series' apply method by giving it a function containing the actions you want to perform on each of the values.

For handling datetimes you can use dateutil.parser.parse to parse arbitrary strings into datetime objects.

import datetime
import pandas as pd
from dateutil.parser import parse

s = pd.Series(['September 20 2016',
'September 20 2016',
'September 19 2016',
'September 16 2016'])
df = pd.DataFrame(s)

def dt2epoch(value):
    d = parse(value)
    return d.timestamp()
    
df[0].apply(dt2epoch)  # apples given function to each value of column

Result:

0    1474329600
1    1474329600
2    1474243200
3    1473984000
Name: 0, dtype: float64

Answer 2

You could try to_datetime .

import pandas as pd
your_df['time']=pd.to_datetime(your_df['time'])

Edit: To get the epoch from a datetime object, you can convert the series to an int64 object, which will give you the number of nanoseconds since the epoch, and divide by 10^9 (the number of nanoseconds in a second).

import numpy as np
your_df['time']  = (pd.to_datetime(your_df['time']).astype(np.int64)/10**9).astype(np.int64)

The last conversion is needed if you want to have it in integers (the division will give you floats instead)

Note: If you have NaT objects in your time series, they will show up as the integer value -9223372036, and you may want to either filter them out up-front, or have them being output as NaN (in which case, the resulting series must be of a float type instead of int).