python - Find first value of column and assign to separate coulmn pandas -


assume have following dataset

table = [[datetime.datetime(2015, 1, 1), 1, 0.5],          [datetime.datetime(2015, 1, 27), 1, 0.5],          [datetime.datetime(2015, 1, 31), 1, 0.5],          [datetime.datetime(2015, 2, 1), 1, 2],          [datetime.datetime(2015, 2, 3), 1, 2],          [datetime.datetime(2015, 2, 15), 1, 2],           [datetime.datetime(2015, 2, 28), 1, 2],          [datetime.datetime(2015, 3, 1), 1, 3],          [datetime.datetime(2015, 3, 17), 1, 3],          [datetime.datetime(2015, 3, 31), 1, 3]]  df = pd.dataframe(table, columns=['date', 'id', 'value'])

how find first element of column value , assign separate column in df grouped id? such df like

        date  id  value  first 0 2015-01-01   1    0.5   0.5 1 2015-01-27   1    0.5   0.5 2 2015-01-31   1    0.5   0.5 3 2015-02-01   1    2.0   0.5 4 2015-02-03   1    2.0   0.5 5 2015-02-15   1    2.0   0.5 6 2015-02-28   1    2.0   0.5 7 2015-03-01   1    3.0   0.5 8 2015-03-17   1    3.0   0.5 9 2015-03-31   1    3.0   0.5

this seems easy problem, can't figure out solution. grateful suggestions, thanks.

you can group on 'id' column, call first return first value group, returns series 'id' index, can call map on orig df 'id' column perform lookup , assign corresponding value each 'id':

in [127]: df['first'] = df['id'].map(df.groupby('id')['value'].first()) df  out[127]:         date  id  value  first 0 2015-01-01   1    0.5    0.5 1 2015-01-27   1    0.5    0.5 2 2015-01-31   1    0.5    0.5 3 2015-02-01   1    2.0    0.5 4 2015-02-03   1    2.0    0.5 5 2015-02-15   1    2.0    0.5 6 2015-02-28   1    2.0    0.5 7 2015-03-01   1    3.0    0.5 8 2015-03-17   1    3.0    0.5 9 2015-03-31   1    3.0    0.5

output first:

in [128]: df.groupby('id')['value'].first()  out[128]: id 1    0.5 name: value, dtype: float64

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