您可以使用groupby / transform来准备一个布尔掩码,对于您想要的行,该掩码为True,对于您不想要的行,则为False.一旦有了这样的布尔掩码,就可以使用df.loc [mask]选择子DataFrame：

import numpy as np

import pandas as pd

df = pd.DataFrame(

{'ID': [10001, 10001, 10001, 10002, 10002, 10002, 10003, 10003, 10003],

'PRICE': [14.5, 14.5, 14.5, 15.125, 14.5, 14.5, 14.5, 14.5, 15.0],

'date': [19920103, 19920106, 19920107, 19920108, 19920109, 19920110,

19920113, 19920114, 19920115]},

index = range(1,10))

def mask_first(x):

result = np.ones_like(x)

result[0] = 0

return result

mask = df.groupby(['ID'])['ID'].transform(mask_first).astype(bool)

print(df.loc[mask])

产量

ID PRICE date

2 10001 14.5 19920106

3 10001 14.5 19920107

5 10002 14.5 19920109

6 10002 14.5 19920110

8 10003 14.5 19920114

9 10003 15.0 19920115

既然你对效率感兴趣,这里有一个基准：

import timeit

import operator

import numpy as np

import pandas as pd

N = 10000

df = pd.DataFrame(

{'ID': np.random.randint(100, size=(N,)),

'PRICE': np.random.random(N),

'date': np.random.random(N)})

def using_mask(df):

def mask_first(x):

result = np.ones_like(x)

result[0] = 0

return result

mask = df.groupby(['ID'])['ID'].transform(mask_first).astype(bool)

return df.loc[mask]

def using_apply(df):

return df.groupby('ID').apply(lambda group: group.iloc[1:, 1:])

def using_apply_alt(df):

return df.groupby('ID', group_keys=False).apply(lambda x: x[1:])

timing = dict()

for func in (using_mask, using_apply, using_apply_alt):

timing[func] = timeit.timeit(

'{}(df)'.format(func.__name__),

'from __main__ import df, {}'.format(func.__name__), number=100)

for func, t in sorted(timing.items(), key=operator.itemgetter(1)):

print('{:16}: {:.2f}'.format(func.__name__, t))

报告

using_mask : 0.85

using_apply_alt : 2.04

using_apply : 3.70