Getting the corresponding value of columnC while g

Question

Building on this answer, and given that

>>> df
  columnA  columnB  columnC
0    cat1        3      400
1    cat1        2       20
2    cat1        5     3029
3    cat2        1      492
4    cat2        4       30
5    cat3        2      203
6    cat3        6      402
7    cat3        4      391

>>> df.groupby(['columnA']).agg({'columnA':'size','columnB':'min'}).rename(columns={'columnA':'size'})

         size  min
columnA           
cat1        3    2
cat2        2    1
cat3        3    2

I want to obtain a DataFrame containing also the value of columnC corresponding to (on the same row of) the displayed minimum value of columnB, that is:

         size  min  columnC
columnA           
cat1        3    2       20
cat2        2    1      492
cat3        3    2      203

Of course this is possible only for those aggregating functions (like min or max) which 'pick' a value from the group rather than 'aggregate' (like sum or average).

Any clue?

Thanks in advance.

Answer 1

You can use idxmin to pull out the row indices of those rows:

In [11]: g = df.groupby(['columnA'])

In [12]: res = g.agg({'columnA': 'size', 'columnB': 'min'})

In [13]: g['columnB'].idxmin()
Out[13]:
columnA
cat1    1
cat2    3
cat3    5
Name: columnB, dtype: int64

In [14]: df["columnC"].iloc[g['columnB'].idxmin()]
Out[14]:
1     20
3    492
5    203
Name: columnC, dtype: int64

You can append this as a column to res:

In [15]: res["columnC"] = df["columnC"].iloc[g['columnB'].idxmin()].values

In [16]: res
Out[16]:
         columnA  columnB  columnC
columnA
cat1           3        2       20
cat2           2        1      492
cat3           3        2      203

Answer 2

Since the result you are looking for is essentially a join on ['columnA', 'columnB'], you can obtain the desired DataFrame using

result = pd.merge(result, df, on=['columnA', 'columnB'], how='left')

provided we setup result with the right column names:

import pandas as pd

df = pd.DataFrame(
    {'columnA': ['cat1', 'cat1', 'cat1', 'cat2', 'cat2', 'cat3', 'cat3', 'cat3'],
     'columnB': [3, 2, 5, 1, 4, 2, 6, 4],
     'columnC': [400, 20, 3029, 492, 30, 203, 402, 391]})

result = df.groupby('columnA').agg({'columnA':'size', 'columnB':'min'})
result = result.rename(columns={'columnA':'size'})
result = result.reset_index()
result = pd.merge(result, df, on=['columnA', 'columnB'], how='left')
result = result.set_index('columnA')
result = result.rename(columns={'columnB':'min'})
print(result)

yields

         min  size  columnC
columnA                    
cat1       2     3       20
cat2       1     2      492
cat3       2     3      203

---

On reason why you might want to use pd.merge instead of groupby/apply is because groupby/apply calls a function for each group. If there are a lot of groups, this can be slow.

For example, if you had a 10000-row DataFrame with 1000 groups,

import numpy as np
import pandas as pd

N = 10000
df = pd.DataFrame(
    {'columnA': np.random.choice(['cat{}'.format(i) for i in range(N//10)], 
                                 size=N),
     'columnB': np.random.randint(10, size=N),
     'columnC': np.random.randint(100, size=N)})

then using_merge (below) is ~ 250x faster than using_apply:

def using_merge(df):
    result = df.groupby('columnA').agg({'columnA':'size', 'columnB':'min'})
    result = result.rename(columns={'columnA':'size'})
    result = result.reset_index()
    result = pd.merge(result, df, on=['columnA', 'columnB'], how='left')
    result = result.set_index('columnA')
    result = result.rename(columns={'columnB':'min'})
    return result

def using_apply(df):
    return (df.groupby("columnA")
            .apply(lambda g: (g[g.columnB == g.columnB.min()]
                   .assign(size = g.columnA.size)
                   .rename(columns={'columnB': 'min'})
                   .drop('columnA', 1)))
            .reset_index(level=1, drop=True))

---

In [80]: %timeit using_merge(df)
100 loops, best of 3: 7.99 ms per loop

In [81]: %timeit using_apply(df)
1 loop, best of 3: 2.06 s per loop

In [82]: 2060/7.99
Out[82]: 257.8222778473091