Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
470 views
in Technique[技术] by (71.8m points)

python - Pandas: How to create a datetime object from Week and Year?

I have a dataframe that provides two integer columns with the Year and Week of the year:

import pandas as pd
import numpy as np
L1 = [43,44,51,2,5,12]
L2 = [2016,2016,2016,2017,2017,2017]
df = pd.DataFrame({"Week":L1,"Year":L2})

df
Out[72]: 
   Week  Year
0    43  2016
1    44  2016
2    51  2016
3     2  2017
4     5  2017
5    12  2017

I need to create a datetime-object from these two numbers.

I tried this, but it throws an error:

df["DT"] = df.apply(lambda x: np.datetime64(x.Year,'Y') + np.timedelta64(x.Week,'W'),axis=1)

Then I tried this, it works but gives the wrong result, that is it ignores the week completely:

df["S"] = df.Week.astype(str)+'-'+df.Year.astype(str)
df["DT"] = df["S"].apply(lambda x: pd.to_datetime(x,format='%W-%Y'))

df
Out[74]: 
   Week  Year        S         DT
0    43  2016  43-2016 2016-01-01
1    44  2016  44-2016 2016-01-01
2    51  2016  51-2016 2016-01-01
3     2  2017   2-2017 2017-01-01
4     5  2017   5-2017 2017-01-01
5    12  2017  12-2017 2017-01-01

I'm really getting lost between Python's datetime, Numpy's datetime64, and pandas Timestamp, can you tell me how it's done correctly?

I'm using Python 3, if that is relevant in any way.

EDIT:

Starting with Python 3.8 the problem is easily solved with a newly introduced method on datetime.date objects: https://docs.python.org/3/library/datetime.html#datetime.date.fromisocalendar

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

Try this:

In [19]: pd.to_datetime(df.Year.astype(str), format='%Y') + 
             pd.to_timedelta(df.Week.mul(7).astype(str) + ' days')
Out[19]:
0   2016-10-28
1   2016-11-04
2   2016-12-23
3   2017-01-15
4   2017-02-05
5   2017-03-26
dtype: datetime64[ns]

Initially I have timestamps in s

It's much easier to parse it from UNIX epoch timestamp:

df['Date'] = pd.to_datetime(df['UNIX_Time'], unit='s')

Timing for 10M rows DF:

Setup:

In [26]: df = pd.DataFrame(pd.date_range('1970-01-01', freq='1T', periods=10**7), columns=['date'])

In [27]: df.shape
Out[27]: (10000000, 1)

In [28]: df['unix_ts'] = df['date'].astype(np.int64)//10**9

In [30]: df
Out[30]:
                       date    unix_ts
0       1970-01-01 00:00:00          0
1       1970-01-01 00:01:00         60
2       1970-01-01 00:02:00        120
3       1970-01-01 00:03:00        180
4       1970-01-01 00:04:00        240
5       1970-01-01 00:05:00        300
6       1970-01-01 00:06:00        360
7       1970-01-01 00:07:00        420
8       1970-01-01 00:08:00        480
9       1970-01-01 00:09:00        540
...                     ...        ...
9999990 1989-01-05 10:30:00  599999400
9999991 1989-01-05 10:31:00  599999460
9999992 1989-01-05 10:32:00  599999520
9999993 1989-01-05 10:33:00  599999580
9999994 1989-01-05 10:34:00  599999640
9999995 1989-01-05 10:35:00  599999700
9999996 1989-01-05 10:36:00  599999760
9999997 1989-01-05 10:37:00  599999820
9999998 1989-01-05 10:38:00  599999880
9999999 1989-01-05 10:39:00  599999940

[10000000 rows x 2 columns]

Check:

In [31]: pd.to_datetime(df.unix_ts, unit='s')
Out[31]:
0         1970-01-01 00:00:00
1         1970-01-01 00:01:00
2         1970-01-01 00:02:00
3         1970-01-01 00:03:00
4         1970-01-01 00:04:00
5         1970-01-01 00:05:00
6         1970-01-01 00:06:00
7         1970-01-01 00:07:00
8         1970-01-01 00:08:00
9         1970-01-01 00:09:00
                  ...
9999990   1989-01-05 10:30:00
9999991   1989-01-05 10:31:00
9999992   1989-01-05 10:32:00
9999993   1989-01-05 10:33:00
9999994   1989-01-05 10:34:00
9999995   1989-01-05 10:35:00
9999996   1989-01-05 10:36:00
9999997   1989-01-05 10:37:00
9999998   1989-01-05 10:38:00
9999999   1989-01-05 10:39:00
Name: unix_ts, Length: 10000000, dtype: datetime64[ns]

Timing:

In [32]: %timeit pd.to_datetime(df.unix_ts, unit='s')
10 loops, best of 3: 156 ms per loop

Conclusion: I think 156 milliseconds for converting 10.000.000 rows is not that slow


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

1.4m articles

1.4m replys

5 comments

57.0k users

...