pandas-列中每个唯一字符串/组的新计算行

Question

I have a dataframe df like:

GROUP  TYPE  COUNT
A       1     5
A       2     10
B       1     3
B       2     9
C       1     20
C       2     100

I would like to add a row for each group such that the new row calculates the quotient of COUNT where TYPE equals 2 and COUNT where TYPE equals 1 for each GROUP ala:

GROUP  TYPE  COUNT
A       1     5
A       2     10
A             .5
B       1     3
B       2     9
B             .33
C       1     20
C       2     100
C             .2

Thanks in advance.

Answer 1

df2 = df.pivot(index='GROUP', columns='TYPE', values='COUNT')
df2['div'] = df2[1]/df2[2]
df2.reset_index().melt('GROUP').sort_values('GROUP')

Output:

  GROUP TYPE       value
0     A    1    5.000000
3     A    2   10.000000
6     A  div    0.500000
1     B    1    3.000000
4     B    2    9.000000
7     B  div    0.333333
2     C    1   20.000000
5     C    2  100.000000
8     C  div    0.200000

My approach would be to reshape the dataframe by pivoting, so every type has its own column. Then the division is very easy, and then by melting you reshape it back to the original shape. In my opinion this is also a very readable solution.

Of course, if you prefer np.nan to div as a type, you can replace it very easily, but I'm not sure if that's what you want.

Answer 2

s=df[df.TYPE.isin([1,2])].sort_values(['GROUP','TYPE']).groupby('GROUP').COUNT.apply(lambda x : x.iloc[0]/x.iloc[1])
# I am sort and filter your original df ,to make they are ordered and only have type 1 and 2 
pd.concat([df,s.reset_index()]).sort_values('GROUP') 
# cancat your result back 

Out[77]: 
        COUNT GROUP  TYPE
0    5.000000     A   1.0
1   10.000000     A   2.0
0    0.500000     A   NaN
2    3.000000     B   1.0
3    9.000000     B   2.0
1    0.333333     B   NaN
4   20.000000     C   1.0
5  100.000000     C   2.0
2    0.200000     C   NaN

Answer 3

You can do:

import numpy as np
import pandas as pd

def add_quotient(x):
    last_row = x.iloc[-1]
    last_row['COUNT'] = x[x.TYPE == 1].COUNT.min() / x[x.TYPE == 2].COUNT.max()
    last_row['TYPE'] = np.nan
    return x.append(last_row)


print(df.groupby('GROUP').apply(add_quotient))

Output

        GROUP  TYPE       COUNT
GROUP                          
A     0     A   1.0    5.000000
      1     A   2.0   10.000000
      1     A   NaN    0.500000
B     2     B   1.0    3.000000
      3     B   2.0    9.000000
      3     B   NaN    0.333333
C     4     C   1.0   20.000000
      5     C   2.0  100.000000
      5     C   NaN    0.200000

Note that the function select the min of the TYPE == 1 and the max of the TYPE == 2 , in case there is more than one value per group. And the TYPE is set to np.nan , but that can be easily changed.

Answer 4

Here's a way first using sort_values' by '['GROUP', 'TYPE'] so ensuring that TYPE 2 comes before 1 and then GroupBy GROUP .

Then use first and last to compute the quocient and outer merging with df :

g = df.sort_values(['GROUP', 'TYPE']).groupby('GROUP')
s = (g.first()/ g.nth(1)).COUNT.reset_index()
df.merge(s, on = ['GROUP','COUNT'], how='outer').fillna(' ').sort_values('GROUP')

   GROUP TYPE       COUNT
0     A    1    5.000000
1     A    2   10.000000
6     A         0.500000
2     B    1    3.000000
3     B    2    9.000000
7     B         0.333333
4     C    1   20.000000
5     C    2  100.000000
8     C         0.200000