Pandas GroupBy and select rows with the minimum value in a specific column

Question

I am grouping my dataset by column A and then would like to take the minimum value in column B and the corresponding value in column C.

data = pd.DataFrame({'A': [1, 2], 'B':[ 2, 4], 'C':[10, 4]})
data  
    A   B   C
0   1   4   3
1   1   5   4
2   1   2   10
3   2   7   2
4   2   4   4
5   2   6   6  

and I would like to get :

    A   B   C
0   1   2   10
1   2   4   4

For the moment I am grouping by A, and creating a value that indicates me the rows I will keep in my dataset:

a = data.groupby('A').min()
a['A'] = a.index
to_keep = [str(x[0]) + str(x[1]) for x in a[['A', 'B']].values]
data['id'] = data['A'].astype(str) + data['B'].astype('str')
data[data['id'].isin(to_keep)]

I am sure that there is a much more straight forward way to do this. I have seen many answers here that use multi-indexing but I would like to do this without adding multi-index to my dataframe. Thank you for your help.

Answer 1

I feel like you're overthinking this. Just use groupby and idxmin:

df.loc[df.groupby('A').B.idxmin()]

   A  B   C
2  1  2  10
4  2  4   4

---

df.loc[df.groupby('A').B.idxmin()].reset_index(drop=True)

   A  B   C
0  1  2  10
1  2  4   4

Answer 2

Had a similar situation but with a more complex column heading (e.g. "B val") in which case this is needed:

df.loc[df.groupby('A')['B val'].idxmin()]

Answer 3

The accepted answer (suggesting idxmin) cannot be used with the pipe pattern. A pipe-friendly alternative is to first sort values and then use groupby with DataFrame.head:

data.sort_values('B').groupby('A').apply(DataFrame.head, n=1)

This is possible because by default groupby preserves the order of rows within each group, which is stable and documented behaviour (see pandas.DataFrame.groupby).

This approach has additional benefits:

• it can be easily expanded to select n rows with smallest values in specific column
• it can break ties by providing another column (as a list) to .sort_values(), e.g.:

  data.sort_values(['final_score', 'midterm_score']).groupby('year').apply(DataFrame.head, n=1)
  

As with other answers, to exactly match the result desired in the question .reset_index(drop=True) is needed, making the final snippet:

df.sort_values('B').groupby('A').apply(DataFrame.head, n=1).reset_index(drop=True)

Answer 4

I found an answer a little bit more wordy, but a lot more efficient:

This is the example dataset:

data = pd.DataFrame({'A': [1,1,1,2,2,2], 'B':[4,5,2,7,4,6], 'C':[3,4,10,2,4,6]})
data

Out:
   A  B   C
0  1  4   3
1  1  5   4
2  1  2  10
3  2  7   2
4  2  4   4
5  2  6   6 

First we will get the min values on a Series from a groupby operation:

min_value = data.groupby('A').B.min()
min_value

Out:
A
1    2
2    4
Name: B, dtype: int64

Then, we merge this series result on the original data frame

data = data.merge(min_value, on='A',suffixes=('', '_min'))
data

Out:
   A  B   C  B_min
0  1  4   3      2
1  1  5   4      2
2  1  2  10      2
3  2  7   2      4
4  2  4   4      4
5  2  6   6      4

Finally, we get only the lines where B is equal to B_min and drop B_min since we don't need it anymore.

data = data[data.B==data.B_min].drop('B_min', axis=1)
data

Out:
   A  B   C
2  1  2  10
4  2  4   4

I have tested it on very large datasets and this was the only way I could make it work in a reasonable time.