使用str中的常量值在pandas df中添加日期列

时间:2016-11-29 06:08:41

标签: python python-2.7 python-3.x pandas

我在pandas df中有一张桌子

    product_id_x    product_id_y    count
0   2727846            7872456       1
1   29234              2932348       2
2   29346              9137500       1
3   29453              91365738      1
4   2933666            91323494      1

我想在str。

中添加一个新列'dates'
dateSelect = "'2016-11-06'"

所以我添加了一个新的常量列

df['dates'] = dateSelect 

但我得到了结果

   product_id_x   product_id_y    count   dates
0   2727846          7872456         1  '2016-11-06'
1   29234            2932348         2  '2016-11-06'
2   29346            9137500         1  '2016-11-06'
3   29453            91365738        1  '2016-11-06'
4   2933666          91323494        1  '2016-11-06'

日期中的值会出现在引号中。和

type(df['dates']) = str

但我希望它以日期格式,因为我将进一步将此表存储在我的mysql数据库中。我希望这个类型是约会的。

from sqlalchemy import create_engine
engine = create_engine('mysql+mysqldb://name:pwd@xxx.xx.xx.x/dbname', echo=False)
df.to_sql(name='tablename', con=engine, if_exists = 'append', index=False)

4 个答案:

答案 0 :(得分:3)

我认为您可以先使用空格replace ',然后使用to_datetime

dateSelect = pd.to_datetime("'2016-11-06'".replace("'",""))
print (dateSelect)
2016-11-06 00:00:00

print (type(dateSelect))
<class 'pandas.tslib.Timestamp'>
df['dates'] = pd.to_datetime("'2016-11-06'".replace("'",""))

print (df)
   product_id_x  product_id_y  count      dates
0       2727846       7872456      1 2016-11-06
1         29234       2932348      2 2016-11-06
2         29346       9137500      1 2016-11-06
3         29453      91365738      1 2016-11-06
4       2933666      91323494      1 2016-11-06

print (df.dtypes)
product_id_x             int64
product_id_y             int64
count                    int64
dates           datetime64[ns]
dtype: object

答案 1 :(得分:3)

最直接的路线

df['dates'] = pd.Timestamp('2016-11-06')
df

   product_id_x  product_id_y  count      dates
0       2727846       7872456      1 2016-11-06
1         29234       2932348      2 2016-11-06
2         29346       9137500      1 2016-11-06
3         29453      91365738      1 2016-11-06
4       2933666      91323494      1 2016-11-06

答案 2 :(得分:1)

啊! @jezrael首先到达那里......

 print timeit.timeit("""
import pandas as pd
import datetime as dt
import timeit
df = pd.read_csv('date_time_pandas.csv')
dateSelect_str = "2016-11-06"

# using standard datetime
dateSelect = dt.datetime.strptime(dateSelect_str,"%Y-%m-%d")
df['dates'] = dateSelect
#print(df['dates'])
""",number=100)


# Alternate method using pandas datetime
print timeit.timeit("""
import pandas as pd
import datetime as dt
import timeit
df = pd.read_csv('date_time_pandas.csv')
dateSelect_str = "2016-11-06"

dateSelect = pd.to_datetime(dateSelect_str, format='%Y-%m-%d', errors='ignore')
df['dates'] = dateSelect
#print df['dates']
""",number=100)

给出输出 -

0.228258825751
0.167258402887

平均而言。

结论在这种情况下使用pd_datetime效率更高

答案 3 :(得分:0)

在其中不要使用双引号避免将其定义为字符串。

dateSelect = '2016-11-06'  
df['dates'] = dateSelect