您可以使用groupby/transform
来准备一个布尔掩码,该掩码对于您想要的行为True,对于您不想要的行为False。一旦您拥有这样的布尔掩码,您就可以使用df.loc[mask]
选择子数据框:
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
既然您关心效率,这里有一个基准测试:
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
[1:, 1:]
中我们需要两个1:
? - Jason Goal