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Python:如何创建一个固定日期+另一列中的天的列

date pandas python

Pythonista anonymous · 技术社区 · 6 年前

import numpy as np
import pandas as pd
df=pd.DataFrame()
df['a']=np.arange(10,20)
df['date from index']=df.apply( lambda x: pd.to_datetime('15-2-2019') + pd.DateOffset(days=x.index), axis=1 )

TypeError:('must be str,not int','occurred at index 0')

我承认我不明白。我尝试创建一个显式列来代替索引:

df=pd.DataFrame()
df['a']=np.arange(10,20)
df['counter']=np.arange(0,df.shape[0])
df['date from counter']=df.apply( lambda x: pd.to_datetime('15-2-2019') + pd.DateOffset(days=x['counter']), axis=1 )

但这给了我:

TypeError:('timedelta days组件不支持的类型: numpy.int32','发生在索引0')

我做错什么了?

2 回复 | 直到 5 年前

jezrael 6 年前

使用 to_timedelta 用于将值转换为日时间增量或使用 origin 使用参数指定开始日期 unit 在里面 to_datetime

df['date from index']= pd.to_datetime('15-2-2019') + pd.to_timedelta(df.index, 'd')
df['date from counter']= pd.to_datetime('15-2-2019') + pd.to_timedelta(df['counter'], 'd')

df['date from index1']= pd.to_datetime(df.index, origin='15-02-2019', unit='d')
df['date from counter1']= pd.to_datetime(df['counter'], origin='15-02-2019', unit='d')
print(df.head())
    a  counter date from index date from counter date from index1  \
0  10        0      2019-02-15        2019-02-15       2019-02-15   
1  11        1      2019-02-16        2019-02-16       2019-02-16   
2  12        2      2019-02-17        2019-02-17       2019-02-17   
3  13        3      2019-02-18        2019-02-18       2019-02-18   
4  14        4      2019-02-19        2019-02-19       2019-02-19   

  date from counter1  
0         2019-02-15  
1         2019-02-16  
2         2019-02-17  
3         2019-02-18  
4         2019-02-19

cs95 abhishek58g 6 年前

你可以用 pd.to_timedelta

# pd.to_timedelta(df.index, unit='d') + pd.to_datetime('15-2-2019') # whichever
pd.to_timedelta(df.a, unit='d') + pd.to_datetime('15-2-2019')

0   2019-02-25
1   2019-02-26
2   2019-02-27
3   2019-02-28
4   2019-03-01
5   2019-03-02
6   2019-03-03
7   2019-03-04
8   2019-03-05
9   2019-03-06
Name: a, dtype: datetime64[ns]

df['date_from_counter'] = (
    pd.to_timedelta(df.a, unit='d') + pd.to_datetime('15-2-2019'))
df

    a  counter date_from_counter
0  10        0        2019-02-25
1  11        1        2019-02-26
2  12        2        2019-02-27
3  13        3        2019-02-28
4  14        4        2019-03-01
5  15        5        2019-03-02
6  16        6        2019-03-03
7  17        7        2019-03-04
8  18        8        2019-03-05
9  19        9        2019-03-06

如你所料,你可以打电话 pd.to\ U时间增量 Timedelta 日期时间算术列。

要使代码正常工作,似乎需要通过 int np.int

dt = pd.to_datetime('15-2-2019')
df['date from counter'] = df.apply(
    lambda x: dt + pd.DateOffset(days=x['counter'].item()), axis=1)
df

    a  counter date from counter
0  10        0        2019-02-15
1  11        1        2019-02-16
2  12        2        2019-02-17
3  13        3        2019-02-18
4  14        4        2019-02-19
5  15        5        2019-02-20
6  16        6        2019-02-21
7  17        7        2019-02-22
8  18        8        2019-02-23
9  19        9        2019-02-24