我有一个excel doc,用户将日期和字符串放在同一列中。我想让每个字符串对象为null并保留所有日期。我怎么在熊猫里这样做?感谢。
答案 0 :(得分:4)
在@Feff中提到的在DataFrame中转换日期的简单方法是pandas.DataFrame.convert_objects,它还处理数字和时间数据。以下是使用它的示例:
# contents of Sheet1 of test.xlsx
x y date1 z date2 date3
1 fum 6/1/2016 7 9/1/2015 string3
2 fo 6/2/2016 alpha string0 10/1/2016
3 fi 6/3/2016 9 9/3/2015 10/2/2016
4 fee 6/4/2016 10 string1 string4
5 dumbledum 6/5/2016 beta string2 10/3/2015
6 dumbledee 6/6/2016 12 9/4/2015 string5
import pandas as pd
xl = pd.ExcelFile('test.xlsx')
df = xl.parse("Sheet1")
df1 = df.convert_objects(convert_dates='coerce')
# 'coerce' required for conversion to NaT on error
df1
Out[7]:
x y date1 z date2 date3
0 1 fum 2016-06-01 7 2015-09-01 NaT
1 2 fo 2016-06-02 alpha NaT 2016-10-01
2 3 fi 2016-06-03 9 2015-09-03 2016-10-02
3 4 fee 2016-06-04 10 NaT NaT
4 5 dumbledum 2016-06-05 beta NaT 2015-10-03
5 6 dumbledee 2016-06-06 12 2015-09-04 NaT
DataFrame中的各个列可以使用pandas.to_datetime进行转换,如@Jeff所指出的那样,以及使用pandas.Series.map,但是两者都没有完成。例如,使用pandas.to_datetime:
import pandas as pd
xl2 = pd.ExcelFile('test.xlsx')
df2 = xl2.parse("Sheet1")
for col in ['date1', 'date2', 'date3']:
df2[col] = pd.to_datetime(df2[col],coerce=True, infer_datetime_format=True)
df2
Out[8]:
x y date1 z date2 date3
0 1 fum 2016-06-01 7 2015-09-01 NaT
1 2 fo 2016-06-02 alpha NaT 2016-10-01
2 3 fi 2016-06-03 9 2015-09-03 2016-10-02
3 4 fee 2016-06-04 10 NaT NaT
4 5 dumbledum 2016-06-05 beta NaT 2015-10-03
5 6 dumbledee 2016-06-06 12 2015-09-04 NaT
使用pandas.Series.map:
import pandas as pd
import datetime
xl3 = pd.ExcelFile('test.xlsx')
df3 = xl3.parse("Sheet1")
for col in ['date1', 'date2', 'date3']:
df3[col] = df3[col].map(lambda x: x if isinstance(x,(datetime.datetime)) else None)
df3
Out[9]:
x y date1 z date2 date3
0 1 fum 2016-06-01 7 2015-09-01 NaT
1 2 fo 2016-06-02 alpha NaT 2016-10-01
2 3 fi 2016-06-03 9 2015-09-03 2016-10-02
3 4 fee 2016-06-04 10 NaT NaT
4 5 dumbledum 2016-06-05 beta NaT 2015-10-03
5 6 dumbledee 2016-06-06 12 2015-09-04 NaT
在excel文档中转换日期的一种前期方法是在解析其工作表时。这可以使用pandas.ExcelFile.parse的转换器选项来完成,其中函数派生自pandas.to_datetime,作为转换器中的函数dict并使用coerce = True启用它以强制错误到NaT。例如:
def converter(x):
return pd.to_datetime(x,coerce=True,infer_datetime_format=True)
# the following also works for this example
# return pd.to_datetime(x,format='%d/%m/%Y',coerce=True)
converters={'date1': converter,'date2': converter, 'date3': converter}
xl4 = pd.ExcelFile('test.xlsx')
df4 = xl4.parse("Sheet1",converters=converters)
df4
Out[10]:
x y date1 z date2 date3
0 1 fum 2016-06-01 7 2015-09-01 NaT
1 2 fo 2016-06-02 alpha NaT 2016-10-01
2 3 fi 2016-06-03 9 2015-09-03 2016-10-02
3 4 fee 2016-06-04 10 NaT NaT
4 5 dumbledum 2016-06-05 beta NaT 2015-10-03
5 6 dumbledee 2016-06-06 12 2015-09-04 NaT