3
votes

Would it be possible to mutate DataFrame inplace with groupby statement?

import pandas as pd
dt = pd.DataFrame({
                   "LETTER": ["a", "b", "c", "a", "b"],
                   "VALUE" : [10 , 12 , 13,  0,  15]
                   })
def __add_new_col(dt_):
    dt_['NEW_COL'] = dt_['VALUE'] - dt_['VALUE'].mean()
    return dt_
pass


dt.groupby("LETTER").apply(__add_new_col)
  LETTER  VALUE  NEW_COL
0      a     10      5.0
1      b     12     -1.5
2      c     13      0.0
3      a      0     -5.0
4      b     15      1.5


dt
  LETTER  VALUE
0      a     10
1      b     12
2      c     13
3      a      0
4      b     15

In R data.table it is possible by using := operator e.g. dt[, col := ... , by ='LETTER']

2
Why not df['NEWCOL'] = dt.groupby('LETTER')['VALUE'].apply(lambda x: x - x.mean())? - Zero
@JohnGalt Is there an order guarantee? - Cron Merdek
Yes, you can validate it yourself too. - Zero

2 Answers

3
votes

I think you can use transform which return Series same length and same index as df with substracting:

print (dt.groupby("LETTER")['VALUE'].transform('mean'))
0     5.0
1    13.5
2    13.0
3     5.0
4    13.5
Name: VALUE, dtype: float64

dt['NEW_COL'] = dt['VALUE'] - dt.groupby("LETTER")['VALUE'].transform('mean')
print (dt)
  LETTER  VALUE  NEW_COL
0      a     10      5.0
1      b     12     -1.5
2      c     13      0.0
3      a      0     -5.0
4      b     15      1.5
1
votes

I'm quite sure you can't mutate the dataframe during a group by. You can do exactly the same operation mapping every lettering with it's mean and then perform the operation.

df['NEW_COL'] = df['VALUE'] - df['LETTER'].map(dt.groupby("LETTER")['VALUE'].mean()).values

This will deal with any possible ordering issue, which I wouldn't trust to be guarantee even if tested. Better safe than sorry :)

Also, I'm using .values accessor after the map because I'm not sure what the index of the "mapped" series will be the same of the 'VALUE' series, which sometime will result with NaN.