问题
I've got a pandas Series containing datetime-like strings with 12h format, but without the am/pm abbreviations. It covers an entire month of data :
40 01/01/2017 11:51:00
41 01/01/2017 11:51:05
42 01/01/2017 11:55:05
43 01/01/2017 11:55:10
44 01/01/2017 11:59:30
45 01/01/2017 11:59:35
46 02/01/2017 12:00:05
47 02/01/2017 12:00:10
48 02/01/2017 12:13:20
49 02/01/2017 12:13:25
50 02/01/2017 12:24:50
51 02/01/2017 12:24:55
52 02/01/2017 12:33:30
Name: TS, dtype: object
(318621,) # shape
My goal is to convert it to datetime format, so as to obtain the appropriate unix timestamps values, and make comparisions/arithmetics with other datetime data with, this time, 24h format. So I already tried this :
pd.to_datetime(df.TS, format = '%d/%m/%Y %I:%M:%S') # %I for 12h format
Which outputs me :
64 2017-01-02 00:46:50
65 2017-01-02 00:46:55
66 2017-01-02 01:01:00
67 2017-01-02 01:01:05
68 2017-01-02 01:05:00
But the am/pm informations are not taken into account. I know that, as a rule, the am/pm first have to be specified in the strings, then one can use dt.dt.strptime() or pd.to_datetime() to parse them with the %p indicator.
So I wanted to know if there's an other way to deal with this issue through datetime or pandas datetime modules ? Or, do I have to manualy add the abbreviations 'am/pm' before the parsing ?
回答1:
You have data in 5 second intervals throughout multiple days. The desired end format is like this (with AM/PM column we need to add, because Pandas cannot possibly guess, since it looks at one value at a time):
31/12/2016 11:59:55 PM
01/01/2017 12:00:00 AM
01/01/2017 12:00:05 AM
01/01/2017 11:59:55 AM
01/01/2017 12:00:00 PM
01/01/2017 12:59:55 PM
01/01/2017 01:00:00 PM
01/01/2017 01:00:05 PM
01/01/2017 11:59:55 PM
02/01/2017 12:00:00 AM
First, we can parse the whole thing without AM/PM info, as you already showed:
ts = pd.to_datetime(df.TS, format = '%d/%m/%Y %I:%M:%S')
We have a small problem: 12:00:00 is parsed as noon, not midnight. Let's normalize that:
ts[ts.dt.hour == 12] -= pd.Timedelta(12, 'h')
Now we have times from 00:00:00 to 11:59:55, twice per day.
Next, note that the transitions are always at 00:00:00. We can easily detect these, as well as the first instance of each date:
twelve = ts.dt.time == datetime.time(0,0,0)
newdate = ts.dt.date.diff() > pd.Timedelta(0)
midnight = twelve & newdate
noon = twelve & ~newdate
Next, build an offset series, which should be easy to inspect for correctness:
offset = pd.Series(np.nan, ts.index, dtype='timedelta64[ns]')
offset[midnight] = pd.Timedelta(0)
offset[noon] = pd.Timedelta(12, 'h')
offset.fillna(method='ffill', inplace=True)
And finally:
ts += offset
来源：https://stackoverflow.com/questions/51018182/convert-incomplete-12h-datetime-like-strings-into-appropriate-datetime-type