Вы можете использовать str.replace
:
df.columns = df.columns.str.replace('.','')
или list comprehension
с replace
:
df.columns = [x.replace('.','') for x in df.columns]
Образец :
df = pd.DataFrame({'traffic.seas1':list('abcdef'),
'traffic.seas2':[4,5,4,5,5,4],
'traffic.seas3':[7,8,9,4,2,3],
'D':[1,3,5,7,1,0],
'E':[5,3,6,9,2,4],
'F':list('aaabbb')})
print (df)
D E F traffic.seas1 traffic.seas2 traffic.seas3
0 1 5 a a 4 7
1 3 3 a b 5 8
2 5 6 a c 4 9
3 7 9 b d 5 4
4 1 2 b e 5 2
5 0 4 b f 4 3
df.columns = df.columns.str.replace('.','')
print (df)
D E F trafficseas1 trafficseas2 trafficseas3
0 1 5 a a 4 7
1 3 3 a b 5 8
2 5 6 a c 4 9
3 7 9 b d 5 4
4 1 2 b e 5 2
5 0 4 b f 4 3
Сроки :
N = 9000
df = pd.DataFrame(np.random.randint(10, size=(3, N))).add_prefix('traffic.seas')
print (df)
In [161]: %timeit df.columns = df.columns.str.replace('.','')
4.4 ms ± 267 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
In [162]: %timeit df.columns = [x.replace('.','') for x in df.columns]
2.53 ms ± 177 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)