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
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
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.