Create multiple columns based on date





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I started to work with pandas recently and during testing with 'date' I have found this challenge. Given this dataframe:



df = pd.DataFrame({'id': [123, 431, 652, 763, 234],
'time': ['8/1/2017', '6/1/2015', '7/1/2016', '9/1/2014', '12/1/2018']})



Create the new dataframe with backdate columns look like this:



    id        time       time1       time2       time3       time4      time5

0   123 2017-08-01  2017-07-01  2017-06-01  2017-05-01  2017-04-01  2017-03-01

1   431 2015-06-01  2015-05-01  2015-04-01  2015-03-01  2015-02-01  2015-01-01

2   652 2016-07-01  2016-06-01  2016-05-01  2016-04-01  2016-03-01  2016-02-01

3   763 2014-09-01  2014-08-01  2014-07-01  2014-06-01  2014-05-01  2014-04-01

4   234 2018-12-01  2018-11-01  2018-10-01  2018-09-01  2018-08-01  2018-07-01


I tries with these codes:



df['time'] = pd.to_datetime(df['time'], errors='coerce') #Object to Date 

df['time1'] = df['time'] - pd.DateOffset(months=1)

df['time2'] = df['time'] - pd.DateOffset(months=2)

df['time3'] = df['time'] - pd.DateOffset(months=3)

df['time4'] = df['time'] - pd.DateOffset(months=4)

df['time5'] = df['time'] - pd.DateOffset(months=5)


Are there anyway to solve this problem faster and more efficient? I've already tested several methods to create the backdate. However I don't know how to do it with multiple columns. Because if the data requires to backdate 24 months, I have to copy and paste a lot (manually).






python python-3.x pandas date dataframe







share|improve this question




















  • 1





    Possible duplicate of Offset date for a Pandas DataFrame date index

    – Stephen Rauch
    Jan 4 at 2:56


















2















I started to work with pandas recently and during testing with 'date' I have found this challenge. Given this dataframe:



df = pd.DataFrame({'id': [123, 431, 652, 763, 234],
'time': ['8/1/2017', '6/1/2015', '7/1/2016', '9/1/2014', '12/1/2018']})



Create the new dataframe with backdate columns look like this:



    id        time       time1       time2       time3       time4      time5

0   123 2017-08-01  2017-07-01  2017-06-01  2017-05-01  2017-04-01  2017-03-01

1   431 2015-06-01  2015-05-01  2015-04-01  2015-03-01  2015-02-01  2015-01-01

2   652 2016-07-01  2016-06-01  2016-05-01  2016-04-01  2016-03-01  2016-02-01

3   763 2014-09-01  2014-08-01  2014-07-01  2014-06-01  2014-05-01  2014-04-01

4   234 2018-12-01  2018-11-01  2018-10-01  2018-09-01  2018-08-01  2018-07-01


I tries with these codes:



df['time'] = pd.to_datetime(df['time'], errors='coerce') #Object to Date 

df['time1'] = df['time'] - pd.DateOffset(months=1)

df['time2'] = df['time'] - pd.DateOffset(months=2)

df['time3'] = df['time'] - pd.DateOffset(months=3)

df['time4'] = df['time'] - pd.DateOffset(months=4)

df['time5'] = df['time'] - pd.DateOffset(months=5)


Are there anyway to solve this problem faster and more efficient? I've already tested several methods to create the backdate. However I don't know how to do it with multiple columns. Because if the data requires to backdate 24 months, I have to copy and paste a lot (manually).






python python-3.x pandas date dataframe







share|improve this question




















  • 1





    Possible duplicate of Offset date for a Pandas DataFrame date index

    – Stephen Rauch
    Jan 4 at 2:56














2












2








2








I started to work with pandas recently and during testing with 'date' I have found this challenge. Given this dataframe:



df = pd.DataFrame({'id': [123, 431, 652, 763, 234],
'time': ['8/1/2017', '6/1/2015', '7/1/2016', '9/1/2014', '12/1/2018']})



Create the new dataframe with backdate columns look like this:



    id        time       time1       time2       time3       time4      time5

0   123 2017-08-01  2017-07-01  2017-06-01  2017-05-01  2017-04-01  2017-03-01

1   431 2015-06-01  2015-05-01  2015-04-01  2015-03-01  2015-02-01  2015-01-01

2   652 2016-07-01  2016-06-01  2016-05-01  2016-04-01  2016-03-01  2016-02-01

3   763 2014-09-01  2014-08-01  2014-07-01  2014-06-01  2014-05-01  2014-04-01

4   234 2018-12-01  2018-11-01  2018-10-01  2018-09-01  2018-08-01  2018-07-01


I tries with these codes:



df['time'] = pd.to_datetime(df['time'], errors='coerce') #Object to Date 

df['time1'] = df['time'] - pd.DateOffset(months=1)

df['time2'] = df['time'] - pd.DateOffset(months=2)

df['time3'] = df['time'] - pd.DateOffset(months=3)

df['time4'] = df['time'] - pd.DateOffset(months=4)

df['time5'] = df['time'] - pd.DateOffset(months=5)


Are there anyway to solve this problem faster and more efficient? I've already tested several methods to create the backdate. However I don't know how to do it with multiple columns. Because if the data requires to backdate 24 months, I have to copy and paste a lot (manually).






python python-3.x pandas date dataframe







share|improve this question
















I started to work with pandas recently and during testing with 'date' I have found this challenge. Given this dataframe:



df = pd.DataFrame({'id': [123, 431, 652, 763, 234],
'time': ['8/1/2017', '6/1/2015', '7/1/2016', '9/1/2014', '12/1/2018']})



Create the new dataframe with backdate columns look like this:



    id        time       time1       time2       time3       time4      time5

0   123 2017-08-01  2017-07-01  2017-06-01  2017-05-01  2017-04-01  2017-03-01

1   431 2015-06-01  2015-05-01  2015-04-01  2015-03-01  2015-02-01  2015-01-01

2   652 2016-07-01  2016-06-01  2016-05-01  2016-04-01  2016-03-01  2016-02-01

3   763 2014-09-01  2014-08-01  2014-07-01  2014-06-01  2014-05-01  2014-04-01

4   234 2018-12-01  2018-11-01  2018-10-01  2018-09-01  2018-08-01  2018-07-01


I tries with these codes:



df['time'] = pd.to_datetime(df['time'], errors='coerce') #Object to Date 

df['time1'] = df['time'] - pd.DateOffset(months=1)

df['time2'] = df['time'] - pd.DateOffset(months=2)

df['time3'] = df['time'] - pd.DateOffset(months=3)

df['time4'] = df['time'] - pd.DateOffset(months=4)

df['time5'] = df['time'] - pd.DateOffset(months=5)


Are there anyway to solve this problem faster and more efficient? I've already tested several methods to create the backdate. However I don't know how to do it with multiple columns. Because if the data requires to backdate 24 months, I have to copy and paste a lot (manually).






python python-3.x pandas date dataframe




python python-3.x pandas date dataframe






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edited Jan 4 at 3:59







Long_NgV

















asked Jan 4 at 2:51









Long_NgVLong_NgV

426




426








  • 1





    Possible duplicate of Offset date for a Pandas DataFrame date index

    – Stephen Rauch
    Jan 4 at 2:56














  • 1





    Possible duplicate of Offset date for a Pandas DataFrame date index

    – Stephen Rauch
    Jan 4 at 2:56








1




1





Possible duplicate of Offset date for a Pandas DataFrame date index

– Stephen Rauch
Jan 4 at 2:56





Possible duplicate of Offset date for a Pandas DataFrame date index

– Stephen Rauch
Jan 4 at 2:56












1 Answer
1






active

oldest

votes


















2














Here is one way using date_range with concat



s=df.time.apply(lambda x : pd.date_range(end=x,periods =6,freq='MS')[::-1].tolist())

df=pd.concat([df,pd.DataFrame(s.tolist(),index=df.index).add_prefix('Time').iloc[:,1:]],axis=1)

df

    id       time      Time1      Time2      Time3      Time4      Time5

0  123 2017-08-01 2017-07-01 2017-06-01 2017-05-01 2017-04-01 2017-03-01

1  431 2015-06-01 2015-05-01 2015-04-01 2015-03-01 2015-02-01 2015-01-01

2  652 2016-07-01 2016-06-01 2016-05-01 2016-04-01 2016-03-01 2016-02-01

3  763 2014-09-01 2014-08-01 2014-07-01 2014-06-01 2014-05-01 2014-04-01

4  234 2018-12-01 2018-11-01 2018-10-01 2018-09-01 2018-08-01 2018-07-01





share|improve this answer
























  • Thank you sir, could you please explain a little bit about how to calculate 's'?

    – Long_NgV
    Jan 4 at 10:28












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1 Answer
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oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2














Here is one way using date_range with concat



s=df.time.apply(lambda x : pd.date_range(end=x,periods =6,freq='MS')[::-1].tolist())

df=pd.concat([df,pd.DataFrame(s.tolist(),index=df.index).add_prefix('Time').iloc[:,1:]],axis=1)

df

    id       time      Time1      Time2      Time3      Time4      Time5

0  123 2017-08-01 2017-07-01 2017-06-01 2017-05-01 2017-04-01 2017-03-01

1  431 2015-06-01 2015-05-01 2015-04-01 2015-03-01 2015-02-01 2015-01-01

2  652 2016-07-01 2016-06-01 2016-05-01 2016-04-01 2016-03-01 2016-02-01

3  763 2014-09-01 2014-08-01 2014-07-01 2014-06-01 2014-05-01 2014-04-01

4  234 2018-12-01 2018-11-01 2018-10-01 2018-09-01 2018-08-01 2018-07-01





share|improve this answer
























  • Thank you sir, could you please explain a little bit about how to calculate 's'?

    – Long_NgV
    Jan 4 at 10:28
















2














Here is one way using date_range with concat



s=df.time.apply(lambda x : pd.date_range(end=x,periods =6,freq='MS')[::-1].tolist())

df=pd.concat([df,pd.DataFrame(s.tolist(),index=df.index).add_prefix('Time').iloc[:,1:]],axis=1)

df

    id       time      Time1      Time2      Time3      Time4      Time5

0  123 2017-08-01 2017-07-01 2017-06-01 2017-05-01 2017-04-01 2017-03-01

1  431 2015-06-01 2015-05-01 2015-04-01 2015-03-01 2015-02-01 2015-01-01

2  652 2016-07-01 2016-06-01 2016-05-01 2016-04-01 2016-03-01 2016-02-01

3  763 2014-09-01 2014-08-01 2014-07-01 2014-06-01 2014-05-01 2014-04-01

4  234 2018-12-01 2018-11-01 2018-10-01 2018-09-01 2018-08-01 2018-07-01





share|improve this answer
























  • Thank you sir, could you please explain a little bit about how to calculate 's'?

    – Long_NgV
    Jan 4 at 10:28














2












2








2







Here is one way using date_range with concat



s=df.time.apply(lambda x : pd.date_range(end=x,periods =6,freq='MS')[::-1].tolist())

df=pd.concat([df,pd.DataFrame(s.tolist(),index=df.index).add_prefix('Time').iloc[:,1:]],axis=1)

df

    id       time      Time1      Time2      Time3      Time4      Time5

0  123 2017-08-01 2017-07-01 2017-06-01 2017-05-01 2017-04-01 2017-03-01

1  431 2015-06-01 2015-05-01 2015-04-01 2015-03-01 2015-02-01 2015-01-01

2  652 2016-07-01 2016-06-01 2016-05-01 2016-04-01 2016-03-01 2016-02-01

3  763 2014-09-01 2014-08-01 2014-07-01 2014-06-01 2014-05-01 2014-04-01

4  234 2018-12-01 2018-11-01 2018-10-01 2018-09-01 2018-08-01 2018-07-01





share|improve this answer













Here is one way using date_range with concat



s=df.time.apply(lambda x : pd.date_range(end=x,periods =6,freq='MS')[::-1].tolist())

df=pd.concat([df,pd.DataFrame(s.tolist(),index=df.index).add_prefix('Time').iloc[:,1:]],axis=1)

df

    id       time      Time1      Time2      Time3      Time4      Time5

0  123 2017-08-01 2017-07-01 2017-06-01 2017-05-01 2017-04-01 2017-03-01

1  431 2015-06-01 2015-05-01 2015-04-01 2015-03-01 2015-02-01 2015-01-01

2  652 2016-07-01 2016-06-01 2016-05-01 2016-04-01 2016-03-01 2016-02-01

3  763 2014-09-01 2014-08-01 2014-07-01 2014-06-01 2014-05-01 2014-04-01

4  234 2018-12-01 2018-11-01 2018-10-01 2018-09-01 2018-08-01 2018-07-01






share|improve this answer












share|improve this answer



share|improve this answer










answered Jan 4 at 3:05









Wen-BenWen-Ben

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  • Thank you sir, could you please explain a little bit about how to calculate 's'?

    – Long_NgV
    Jan 4 at 10:28



















  • Thank you sir, could you please explain a little bit about how to calculate 's'?

    – Long_NgV
    Jan 4 at 10:28

















Thank you sir, could you please explain a little bit about how to calculate 's'?

– Long_NgV
Jan 4 at 10:28





Thank you sir, could you please explain a little bit about how to calculate 's'?

– Long_NgV
Jan 4 at 10:28




















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