使用groupby和idxmax方法:

将col date转换为datetime: df(“日期”)= pd.to_datetime (df(“日期”)) 在groupyby ad_id之后，获取列日期的Max索引: idx = df.groupby(=“ad_id”)(“日期”).idxmax () 获取所需数据: df_max = df.loc (idx,)

[54]:

ad_id  price       date
7     22      2 2018-06-11
6     23      2 2018-06-22
2     24      2 2018-06-30
3     28      5 2018-06-22

In [1]: df
Out[1]:
    Sp  Mt Value  count
0  MM1  S1     a      3
1  MM1  S1     n      2
2  MM1  S3    cb      5
3  MM2  S3    mk      8
4  MM2  S4    bg     10
5  MM2  S4   dgd      1
6  MM4  S2    rd      2
7  MM4  S2    cb      2
8  MM4  S2   uyi      7

In [2]: df.groupby(['Mt'], sort=False)['count'].max()
Out[2]:
Mt
S1     3
S3     8
S4    10
S2     7
Name: count

要获得原始DF的指数，您可以这样做:

In [3]: idx = df.groupby(['Mt'])['count'].transform(max) == df['count']

In [4]: df[idx]
Out[4]:
    Sp  Mt Value  count
0  MM1  S1     a      3
3  MM2  S3    mk      8
4  MM2  S4    bg     10
8  MM4  S2   uyi      7

注意，如果每个组有多个最大值，则将返回所有最大值。

更新

碰碰运气，这就是OP要求的:

In [5]: df['count_max'] = df.groupby(['Mt'])['count'].transform(max)

In [6]: df
Out[6]:
    Sp  Mt Value  count  count_max
0  MM1  S1     a      3          3
1  MM1  S1     n      2          3
2  MM1  S3    cb      5          8
3  MM2  S3    mk      8          8
4  MM2  S4    bg     10         10
5  MM2  S4   dgd      1         10
6  MM4  S2    rd      2          7
7  MM4  S2    cb      2          7
8  MM4  S2   uyi      7          7

df = pd.DataFrame({
'sp' : ['MM1', 'MM1', 'MM1', 'MM2', 'MM2', 'MM2', 'MM4', 'MM4','MM4'],
'mt' : ['S1', 'S1', 'S3', 'S3', 'S4', 'S4', 'S2', 'S2', 'S2'],
'val' : ['a', 'n', 'cb', 'mk', 'bg', 'dgb', 'rd', 'cb', 'uyi'],
'count' : [3,2,5,8,10,1,2,2,7]
})

df.groupby(['sp', 'mt']).apply(lambda grp: grp.nlargest(1, 'count'))

您可以根据计数对dataFrame进行排序，然后删除重复项。我认为这样更简单:

df.sort_values('count', ascending=False).drop_duplicates(['Sp','Mt'])

综上所述，有很多方法，但哪一种更快呢?

import pandas as pd
import numpy as np
import time

df = pd.DataFrame(np.random.randint(1,10,size=(1000000, 2)), columns=list('AB'))

start_time = time.time()
df1idx = df.groupby(['A'])['B'].transform(max) == df['B']
df1 = df[df1idx]
print("---1 ) %s seconds ---" % (time.time() - start_time))

start_time = time.time()
df2 = df.sort_values('B').groupby(['A']).tail(1)
print("---2 ) %s seconds ---" % (time.time() - start_time))

start_time = time.time()
df3 = df.sort_values('B').drop_duplicates(['A'],keep='last')
print("---3 ) %s seconds ---" % (time.time() - start_time))

start_time = time.time()
df3b = df.sort_values('B', ascending=False).drop_duplicates(['A'])
print("---3b) %s seconds ---" % (time.time() - start_time))

start_time = time.time()
df4 = df[df['B'] == df.groupby(['A'])['B'].transform(max)]
print("---4 ) %s seconds ---" % (time.time() - start_time))

start_time = time.time()
d = df.groupby('A')['B'].nlargest(1)
df5 = df.iloc[[i[1] for i in d.index], :]
print("---5 ) %s seconds ---" % (time.time() - start_time))

获胜者是……

——1)0.03337574005126953秒—— ——2)0.1346898078918457秒—— ——3)0.10243558883666992秒—— ——3b) 0.1004343032836914秒—— ——4)0.028397560119628906秒—— ——5)0.07552886009216309秒——

意识到“应用”“nmaximum”到groupby对象同样有效:

额外的优势-也可以获取前n个值，如果需要:

In [85]: import pandas as pd

In [86]: df = pd.DataFrame({
    ...: 'sp' : ['MM1', 'MM1', 'MM1', 'MM2', 'MM2', 'MM2', 'MM4', 'MM4','MM4'],
    ...: 'mt' : ['S1', 'S1', 'S3', 'S3', 'S4', 'S4', 'S2', 'S2', 'S2'],
    ...: 'val' : ['a', 'n', 'cb', 'mk', 'bg', 'dgb', 'rd', 'cb', 'uyi'],
    ...: 'count' : [3,2,5,8,10,1,2,2,7]
    ...: })

## Apply nlargest(1) to find the max val df, and nlargest(n) gives top n values for df:
In [87]: df.groupby(["sp", "mt"]).apply(lambda x: x.nlargest(1, "count")).reset_index(drop=True)
Out[87]:
   count  mt   sp  val
0      3  S1  MM1    a
1      5  S3  MM1   cb
2      8  S3  MM2   mk
3     10  S4  MM2   bg
4      7  S2  MM4  uyi