Python Pandas: Assign Last Value of DataFrame Grou

Question

In Python Pandas, I have a DataFrame. I group this DataFrame by a column and want to assign the last value of a column to all rows of another column.

I know that I am able to select the last row of the group by this command:

import pandas as pd

df = pd.DataFrame({'a': (1,1,2,3,3), 'b':(20,21,30,40,41)})
print(df)
print("-")
result = df.groupby('a').nth(-1)
print(result)

Result:

   a   b
0  1  20
1  1  21
2  2  30
3  3  40
4  3  41
-
    b
a    
1  21
2  30
3  41

How would it be possible to assign the result of this operation back to the original dataframe so that I have something like:

   a   b b_new
0  1  20 21
1  1  21 21
2  2  30 30
3  3  40 41
4  3  41 41

Answer 1

Two possibilities, with groupby + nth + map or replace

df['b_new'] = df.a.map(df.groupby('a').b.nth(-1))

Or,

df['b_new'] = df.a.replace(df.groupby('a').b.nth(-1))

You can also replace nth(-1) with last() (in fact, doing so happens to make this a little faster), but nth gives you more flexibility over what item to pick from each group in b.

---

df

   a   b  b_new
0  1  20     21
1  1  21     21
2  2  30     30
3  3  40     41
4  3  41     41

Answer 2

I think this should be fast

df.merge(df.drop_duplicates('a',keep='last'),on='a',how='left')
Out[797]: 
   a  b_x  b_y
0  1   20   21
1  1   21   21
2  2   30   30
3  3   40   41
4  3   41   41

Answer 3

Use transform with last:

df['b_new'] = df.groupby('a')['b'].transform('last')

Alternative:

df['b_new'] = df.groupby('a')['b'].transform(lambda x: x.iat[-1])

print(df)
   a   b  b_new
0  1  20     21
1  1  21     21
2  2  30     30
3  3  40     41
4  3  41     41

Solution with nth and join:

df = df.join(df.groupby('a')['b'].nth(-1).rename('b_new'), 'a')
print(df)
   a   b  b_new
0  1  20     21
1  1  21     21
2  2  30     30
3  3  40     41
4  3  41     41

Timings:

N = 10000

df = pd.DataFrame({'a':np.random.randint(1000,size=N),
                   'b':np.random.randint(10000,size=N)})

#print (df)


def f(df):
    return df.join(df.groupby('a')['b'].nth(-1).rename('b_new'), 'a')

#cᴏʟᴅsᴘᴇᴇᴅ1
In [211]: %timeit df['b_new'] = df.a.map(df.groupby('a').b.nth(-1))
100 loops, best of 3: 3.57 ms per loop

#cᴏʟᴅsᴘᴇᴇᴅ2
In [212]: %timeit df['b_new'] = df.a.replace(df.groupby('a').b.nth(-1))
10 loops, best of 3: 71.3 ms per loop

#jezrael1
In [213]: %timeit df['b_new'] = df.groupby('a')['b'].transform('last')
1000 loops, best of 3: 1.82 ms per loop

#jezrael2
In [214]: %timeit df['b_new'] = df.groupby('a')['b'].transform(lambda x: x.iat[-1])
10 loops, best of 3: 178 ms per loop

#jezrael3
In [219]: %timeit f(df)
100 loops, best of 3: 3.63 ms per loop

Caveat

The results do not address performance given the number of groups, which will affect timings a lot for some of these solutions.