pandas get minimum of one column in group when groupby another

Question

I have a pandas dataframe that looks like this:

      c     y
0     9     0
1     8     0
2     3     1
3     6     2
4     1     3
5     2     3
6     5     3
7     4     4
8     0     4
9     7     4

I'd like to groupby y and get the min and max of c so that my new dataframe would look like this:

      c     y     min   max
0     9     0     8     9
1     8     0     8     9
2     3     1     3     3   
3     6     2     6     6 
4     1     3     1     5
5     2     3     1     5
6     5     3     1     5
7     4     4     0     7
8     0     4     0     7
9     7     4     0     7

I tried using df['min'] = df.groupby(['y'])['c'].min() but that gave me some weird results. The first 175 rows were populated in the min column but then it went to NaN for all the rest. is that not how you're supposed to use the groupby method?

Answer 1

Option 1 Use transform

In [13]: dfc = df.groupby('y')['c']

In [14]: df.assign(min=dfc.transform(min), max=dfc.transform(max))
Out[14]:
   c  y  max  min
0  9  0    9    8
1  8  0    9    8
2  3  1    3    3
3  6  2    6    6
4  1  3    5    1
5  2  3    5    1
6  5  3    5    1
7  4  4    7    0
8  0  4    7    0
9  7  4    7    0

Or

In [15]: df['min' ] = dfc.transform('min')

In [16]: df['max' ] = dfc.transform('max')

---

Option 2 Use join and agg

In [30]: df.join(df.groupby('y')['c'].agg(['min', 'max']), on='y')
Out[30]:
   c  y  min  max
0  9  0    8    9
1  8  0    8    9
2  3  1    3    3
3  6  2    6    6
4  1  3    1    5
5  2  3    1    5
6  5  3    1    5
7  4  4    0    7
8  0  4    0    7
9  7  4    0    7

---

Option 3 Use merge and agg

In [28]: df.merge(df.groupby('y')['c'].agg(['min', 'max']), right_index=True, left_on='y')
Out[28]:
   c  y  min  max
0  9  0    8    9
1  8  0    8    9
2  3  1    3    3
3  6  2    6    6
4  1  3    1    5
5  2  3    1    5
6  5  3    1    5
7  4  4    0    7
8  0  4    0    7
9  7  4    0    7