我有一个数据框,它提供两个整数列,分别是年份和年份:
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
我需要从这两个数字创建一个datetime-object。
我试过了,但是抛出一个错误:
df["DT"] = df.apply(lambda x: np.datetime64(x.Year,'Y') + np.timedelta64(x.Week,'W'),axis=1)
然后我尝试了一下,它可以工作,但是给出了错误的结果,即它完全忽略了一周:
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
我真的迷失在Python datetime
,Numpy datetime64
和Pandas之间Timestamp
,您能告诉我它是如何正确完成的吗?
我正在使用Python 3,如果这在任何方面都有意义的话。
编辑:
从Python 3.8开始,可以使用针对datetime.date对象的新引入的方法轻松解决此问题:https : //docs.python.org/3/library/datetime.html#datetime.date.fromisocalendar
尝试这个:
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]
最初我有时间戳
s
从UNIX纪元时间戳解析它要容易得多:
df['Date'] = pd.to_datetime(df['UNIX_Time'], unit='s')
10M行DF的计时:
设定:
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]
校验:
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]
定时:
In [32]: %timeit pd.to_datetime(df.unix_ts, unit='s')
10 loops, best of 3: 156 ms per loop
结论:我认为156毫秒转换10.000.000行并不那么慢
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