how to create new column in pandas dataframe by indexing desirable parts of other columns





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this is what i have:



fertility_rate =[3.97,3.68,3.14,2.73,2.27,2.02,1.86,1.70,1.68,1.32,1.31, 1.31,1.39,1.45,1.50,1.55,1.59]
year = ['1950-1955','1955-1960','1960-1965','1965-1970','1970-1975','1975-1980','1980-1985','1985-1990','1990-1995','1995-2000','2000-2005','2005-2010','2010-2015','2015-2020','2020-2025','2025-2030','2030-2035']
fertility = pd.DataFrame({'year': year, 'fertility': fertility_rate})
fertility.set_index('year', inplace=True)


fertility['fertility_high']=fertility['fertility'] +0.5


This code makes new column by adding 0.5 to every row in the column.
My goal is to for first 12 rows ( from 1955 to 2015 year) to copy values from a first column (fertility), and for 12+ rows to add 0.5.










share|improve this question























  • Use .iloc to index into both the rows and columns.

    – Alexander Reynolds
    Nov 26 '18 at 19:36


















1















this is what i have:



fertility_rate =[3.97,3.68,3.14,2.73,2.27,2.02,1.86,1.70,1.68,1.32,1.31, 1.31,1.39,1.45,1.50,1.55,1.59]
year = ['1950-1955','1955-1960','1960-1965','1965-1970','1970-1975','1975-1980','1980-1985','1985-1990','1990-1995','1995-2000','2000-2005','2005-2010','2010-2015','2015-2020','2020-2025','2025-2030','2030-2035']
fertility = pd.DataFrame({'year': year, 'fertility': fertility_rate})
fertility.set_index('year', inplace=True)


fertility['fertility_high']=fertility['fertility'] +0.5


This code makes new column by adding 0.5 to every row in the column.
My goal is to for first 12 rows ( from 1955 to 2015 year) to copy values from a first column (fertility), and for 12+ rows to add 0.5.










share|improve this question























  • Use .iloc to index into both the rows and columns.

    – Alexander Reynolds
    Nov 26 '18 at 19:36














1












1








1








this is what i have:



fertility_rate =[3.97,3.68,3.14,2.73,2.27,2.02,1.86,1.70,1.68,1.32,1.31, 1.31,1.39,1.45,1.50,1.55,1.59]
year = ['1950-1955','1955-1960','1960-1965','1965-1970','1970-1975','1975-1980','1980-1985','1985-1990','1990-1995','1995-2000','2000-2005','2005-2010','2010-2015','2015-2020','2020-2025','2025-2030','2030-2035']
fertility = pd.DataFrame({'year': year, 'fertility': fertility_rate})
fertility.set_index('year', inplace=True)


fertility['fertility_high']=fertility['fertility'] +0.5


This code makes new column by adding 0.5 to every row in the column.
My goal is to for first 12 rows ( from 1955 to 2015 year) to copy values from a first column (fertility), and for 12+ rows to add 0.5.










share|improve this question














this is what i have:



fertility_rate =[3.97,3.68,3.14,2.73,2.27,2.02,1.86,1.70,1.68,1.32,1.31, 1.31,1.39,1.45,1.50,1.55,1.59]
year = ['1950-1955','1955-1960','1960-1965','1965-1970','1970-1975','1975-1980','1980-1985','1985-1990','1990-1995','1995-2000','2000-2005','2005-2010','2010-2015','2015-2020','2020-2025','2025-2030','2030-2035']
fertility = pd.DataFrame({'year': year, 'fertility': fertility_rate})
fertility.set_index('year', inplace=True)


fertility['fertility_high']=fertility['fertility'] +0.5


This code makes new column by adding 0.5 to every row in the column.
My goal is to for first 12 rows ( from 1955 to 2015 year) to copy values from a first column (fertility), and for 12+ rows to add 0.5.







python pandas






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asked Nov 26 '18 at 19:34









waltexwqwaltexwq

103




103













  • Use .iloc to index into both the rows and columns.

    – Alexander Reynolds
    Nov 26 '18 at 19:36



















  • Use .iloc to index into both the rows and columns.

    – Alexander Reynolds
    Nov 26 '18 at 19:36

















Use .iloc to index into both the rows and columns.

– Alexander Reynolds
Nov 26 '18 at 19:36





Use .iloc to index into both the rows and columns.

– Alexander Reynolds
Nov 26 '18 at 19:36












1 Answer
1






active

oldest

votes


















0














IIUC, you need same value as fertility in fertility high for the first 12 rows and increment by 0.5 for the rest



fertility['fertility_high']=fertility['fertility'].copy() 

fertility.iloc[12:, 1]+=0.5

fertility fertility_high
year
1950-1955 3.97 3.97
1955-1960 3.68 3.68
1960-1965 3.14 3.14
1965-1970 2.73 2.73
1970-1975 2.27 2.27
1975-1980 2.02 2.02
1980-1985 1.86 1.86
1985-1990 1.70 1.70
1990-1995 1.68 1.68
1995-2000 1.32 1.32
2000-2005 1.31 1.31
2005-2010 1.31 1.31
2010-2015 1.39 1.89
2015-2020 1.45 1.95
2020-2025 1.50 2.00
2025-2030 1.55 2.05
2030-2035 1.59 2.09





share|improve this answer
























  • thank you. that is it.

    – waltexwq
    Nov 26 '18 at 19:52











  • @waltexwq, great that it works. Do remember to upvote/accept the answer if it answers your query fully

    – Vaishali
    Nov 26 '18 at 19:54














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1 Answer
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IIUC, you need same value as fertility in fertility high for the first 12 rows and increment by 0.5 for the rest



fertility['fertility_high']=fertility['fertility'].copy() 

fertility.iloc[12:, 1]+=0.5

fertility fertility_high
year
1950-1955 3.97 3.97
1955-1960 3.68 3.68
1960-1965 3.14 3.14
1965-1970 2.73 2.73
1970-1975 2.27 2.27
1975-1980 2.02 2.02
1980-1985 1.86 1.86
1985-1990 1.70 1.70
1990-1995 1.68 1.68
1995-2000 1.32 1.32
2000-2005 1.31 1.31
2005-2010 1.31 1.31
2010-2015 1.39 1.89
2015-2020 1.45 1.95
2020-2025 1.50 2.00
2025-2030 1.55 2.05
2030-2035 1.59 2.09





share|improve this answer
























  • thank you. that is it.

    – waltexwq
    Nov 26 '18 at 19:52











  • @waltexwq, great that it works. Do remember to upvote/accept the answer if it answers your query fully

    – Vaishali
    Nov 26 '18 at 19:54


















0














IIUC, you need same value as fertility in fertility high for the first 12 rows and increment by 0.5 for the rest



fertility['fertility_high']=fertility['fertility'].copy() 

fertility.iloc[12:, 1]+=0.5

fertility fertility_high
year
1950-1955 3.97 3.97
1955-1960 3.68 3.68
1960-1965 3.14 3.14
1965-1970 2.73 2.73
1970-1975 2.27 2.27
1975-1980 2.02 2.02
1980-1985 1.86 1.86
1985-1990 1.70 1.70
1990-1995 1.68 1.68
1995-2000 1.32 1.32
2000-2005 1.31 1.31
2005-2010 1.31 1.31
2010-2015 1.39 1.89
2015-2020 1.45 1.95
2020-2025 1.50 2.00
2025-2030 1.55 2.05
2030-2035 1.59 2.09





share|improve this answer
























  • thank you. that is it.

    – waltexwq
    Nov 26 '18 at 19:52











  • @waltexwq, great that it works. Do remember to upvote/accept the answer if it answers your query fully

    – Vaishali
    Nov 26 '18 at 19:54
















0












0








0







IIUC, you need same value as fertility in fertility high for the first 12 rows and increment by 0.5 for the rest



fertility['fertility_high']=fertility['fertility'].copy() 

fertility.iloc[12:, 1]+=0.5

fertility fertility_high
year
1950-1955 3.97 3.97
1955-1960 3.68 3.68
1960-1965 3.14 3.14
1965-1970 2.73 2.73
1970-1975 2.27 2.27
1975-1980 2.02 2.02
1980-1985 1.86 1.86
1985-1990 1.70 1.70
1990-1995 1.68 1.68
1995-2000 1.32 1.32
2000-2005 1.31 1.31
2005-2010 1.31 1.31
2010-2015 1.39 1.89
2015-2020 1.45 1.95
2020-2025 1.50 2.00
2025-2030 1.55 2.05
2030-2035 1.59 2.09





share|improve this answer













IIUC, you need same value as fertility in fertility high for the first 12 rows and increment by 0.5 for the rest



fertility['fertility_high']=fertility['fertility'].copy() 

fertility.iloc[12:, 1]+=0.5

fertility fertility_high
year
1950-1955 3.97 3.97
1955-1960 3.68 3.68
1960-1965 3.14 3.14
1965-1970 2.73 2.73
1970-1975 2.27 2.27
1975-1980 2.02 2.02
1980-1985 1.86 1.86
1985-1990 1.70 1.70
1990-1995 1.68 1.68
1995-2000 1.32 1.32
2000-2005 1.31 1.31
2005-2010 1.31 1.31
2010-2015 1.39 1.89
2015-2020 1.45 1.95
2020-2025 1.50 2.00
2025-2030 1.55 2.05
2030-2035 1.59 2.09






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answered Nov 26 '18 at 19:43









VaishaliVaishali

22.6k41438




22.6k41438













  • thank you. that is it.

    – waltexwq
    Nov 26 '18 at 19:52











  • @waltexwq, great that it works. Do remember to upvote/accept the answer if it answers your query fully

    – Vaishali
    Nov 26 '18 at 19:54





















  • thank you. that is it.

    – waltexwq
    Nov 26 '18 at 19:52











  • @waltexwq, great that it works. Do remember to upvote/accept the answer if it answers your query fully

    – Vaishali
    Nov 26 '18 at 19:54



















thank you. that is it.

– waltexwq
Nov 26 '18 at 19:52





thank you. that is it.

– waltexwq
Nov 26 '18 at 19:52













@waltexwq, great that it works. Do remember to upvote/accept the answer if it answers your query fully

– Vaishali
Nov 26 '18 at 19:54







@waltexwq, great that it works. Do remember to upvote/accept the answer if it answers your query fully

– Vaishali
Nov 26 '18 at 19:54






















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