熊猫，GROUPBY和组发现的最大，返回值和计数

Question

我有日志数据的熊猫数据帧：

 host service 
0 this.com mail 
1 this.com mail 
2 this.com  web 
3 that.com mail 
4 other.net mail 
5 other.net  web 
6 other.net  web 

我要找到每一个主机给出了最错误的服务：

 host service no 
0 this.com mail 2 
1 that.com mail 1 
2 other.net  web 2 

我发现的唯一解决方案是按主机和服务分组，然后在索引的0级上迭代 。

任何人都可以推荐更好，更短的版本吗？没有迭代？

df = df_logfile.groupby(['host','service']).agg({'service':np.size}) 

df_count = pd.DataFrame() 
df_count['host'] = df_logfile['host'].unique() 
df_count['service'] = np.nan 
df_count['no'] = np.nan 

for h,data in df.groupby(level=0): 
    i = data.idxmax()[0] 
    service = i[1]    
    no = data.xs(i)[0] 
    df_count.loc[df_count['host'] == h, 'service'] = service 
    df_count.loc[(df_count['host'] == h) & (df_count['service'] == service), 'no'] = no 

全码https://gist.github.com/bjelline/d8066de66e305887b714

Answer 1

鉴于df，下一个步骤是组单独由host值和
骨料通过idxmax。这给你指定哪个 对应最大的服务值。然后，您可以使用df.loc[...]选择在df对应于最大贡献值的行：

import numpy as np 
import pandas as pd 

df_logfile = pd.DataFrame({ 
    'host' : ['this.com', 'this.com', 'this.com', 'that.com', 'other.net', 
       'other.net', 'other.net'], 
    'service' : ['mail', 'mail', 'web', 'mail', 'mail', 'web', 'web' ] }) 

df = df_logfile.groupby(['host','service'])['service'].agg({'no':'count'}) 
mask = df.groupby(level=0).agg('idxmax') 
df_count = df.loc[mask['no']] 
df_count = df_count.reset_index() 
print("\nOutput\n{}".format(df_count)) 

产生数据帧

 host service no 
0 other.net  web 2 
1 that.com mail 1 
2 this.com mail 2