如何让pandas.read_csv()从CSV文件列中推断出datetime和timedelt

In [59]:

import pandas as pd
import datetime as dt

def parse_dates(x):
    return dt.datetime.strptime(x,'%Y%m%d %H:%M:%S')
# dict for word lookup,conversion
word_to_int={'zero':0,'one':1,'two':2,'three':3,'four':4,'five':5,'six':6,'seven':7,'eight':8,'nine':9}


def str_to_time_delta(x):
    num = 0
    if 'hour' in x.lower():
        num = x[0:x.find(' ')].lower()
    return dt.timedelta( hours = word_to_int[num])
df = pd.read_csv(r'c:temp1.txt',parse_dates=[0],date_parser=parse_dates)
df.dtypes
Out[59]:
datetime     datetime64[ns]
timedelta            object
integer               int64
number              float64
boolean                bool
string               object
dtype: object
In [60]:

然后转换为timedeltas使用dict和函数来解析并转换为timedeltas

df['timedelta'] = df['timedelta'].map(str_to_time_delta)

In [61]:

df.dtypes
Out[61]:
datetime      datetime64[ns]
timedelta    timedelta64[ns]
integer                int64
number               float64
boolean                 bool
string                object
dtype: object
In [62]:

df
Out[62]:
             datetime  timedelta  integer  number boolean  string
0 2011-12-30 00:00:00   01:00:00       10     1.6    True  Foobar

[1 rows x 6 columns]

要回答您的主要问题,我不知道如何自动执行此操作.

编辑

而不是我复杂的映射功能,你可以这样做：

df['timedelta'] = pd.to_timedelta(df['timedelta'])

进一步编辑

正如@Jeff所述,你可以在阅读csv时使用strptime而不是使用strptime(虽然版本在0.13.1及更高版本中)：

df = pd.read_csv(r'c:temp1.txt',infer_datetime_format=True)