Pandas add new second level column to column multiindex based on other columns

Question

I have a DataFrame with column multi-index:

System   A                B
Trial    Exp1    Exp2     Exp1    Exp2
1        NaN     1        2       3
2        4       5        NaN     NaN
3        6       NaN      7       8

Turns out for each system (A, B) and each measurement (1, 2, 3 in index), results from Exp1 is always superior to Exp2. So I want to generate a 3rd column for each system, call it Final, that should take Exp1 whenever available, and default to Exp2 otherwise. The desired result is

System   A                       B
Trial    Exp1    Exp2    Final   Exp1    Exp2    Final
1        NaN     1       1       2       3       2
2        4       5       4       NaN     NaN     NaN
3        6       NaN     6       7       8       7

What is the best way to do this?

I've tried to use groupby on the columns:

grp = df.groupby(level=0, axis=1)

And was thinking of using either transform or apply combined by assign to achieve it. But am not able to find either a working or an efficient way of doing it. Specifically I am avoiding native python for loops for efficiency reasons (else the problem is trivial).

Answer 1

Use stack for reshape, add column with fillna and then reshape back by unstack with swaplevel + sort_index:

df = df.stack(level=0)
df['Final'] = df['Exp1'].fillna(df['Exp1'])
df = df.unstack().swaplevel(0,1,axis=1).sort_index(axis=1)
print (df)
System    A               B           
Trial  Exp1 Exp2 Final Exp1 Exp2 Final
1       NaN  1.0   NaN  2.0  3.0   2.0
2       4.0  5.0   4.0  NaN  NaN   NaN
3       6.0  NaN   6.0  7.0  8.0   7.0

Another solution with xs for select DataFrames, create new DataFrame by combine_first, but there is missing second level - was added by MultiIndex.from_product and last concat both DataFrames together:

a = df.xs('Exp1', axis=1, level=1)
b = df.xs('Exp2', axis=1, level=1)
df1 =  a.combine_first(b)
df1.columns = pd.MultiIndex.from_product([df1.columns, ['Final']])
df = pd.concat([df, df1], axis=1).sort_index(axis=1)
print (df)
System    A               B           
Trial  Exp1 Exp2 Final Exp1 Exp2 Final
1       NaN  1.0   1.0  2.0  3.0   2.0
2       4.0  5.0   4.0  NaN  NaN   NaN
3       6.0  NaN   6.0  7.0  8.0   7.0

Similar solution with rename:

a = df.xs('Exp1', axis=1, level=1, drop_level=False)
b = df.xs('Exp2', axis=1, level=1, drop_level=False)
df1 = a.rename(columns={'Exp1':'Final'}).combine_first(b.rename(columns={'Exp2':'Final'}))
df = pd.concat([df, df1], axis=1).sort_index(axis=1)
print (df)
System    A               B           
Trial  Exp1 Exp2 Final Exp1 Exp2 Final
1       NaN  1.0   1.0  2.0  3.0   2.0
2       4.0  5.0   4.0  NaN  NaN   NaN
3       6.0  NaN   6.0  7.0  8.0   7.0

Answer 2

• stack with your first level of the column index stack(0) leaving ['Exp1', 'Exp2'] in the column index
• Use a lambda function that gets applied to the whole dataframe within an assign call.
• Finally, unstack, swaplevel, sort_index to clean it up and put everything where it belongs.

---

f = lambda x: x.Exp1.fillna(x.Exp2)
df.stack(0).assign(Final=f).unstack() \
    .swaplevel(0, 1, 1).sort_index(1)

     A               B           
  Exp1 Exp2 Final Exp1 Exp2 Final
1  NaN  1.0   1.0  2.0  3.0   2.0
2  4.0  5.0   4.0  NaN  NaN   NaN
3  6.0  NaN   6.0  7.0  8.0   7.0

---

Another concept using xs

d1 = df.xs('Exp1', 1, 1).fillna(df.xs('Exp2', 1, 1))
d1.columns = [d1.columns, ['Final'] * len(d1.columns)]
pd.concat([df, d1], axis=1).sort_index(1)


     A               B           
  Exp1 Exp2 Final Exp1 Exp2 Final
1  NaN  1.0   1.0  2.0  3.0   2.0
2  4.0  5.0   4.0  NaN  NaN   NaN
3  6.0  NaN   6.0  7.0  8.0   7.0