Question

我有一个带有日志数据的pandas DataFrame：

        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

我想在每个主机上找到提供最多错误的服务：

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

我找到的唯一解决方案是按主机和服务进行分组，然后进行迭代超过索引的0级。

有人能建议更好，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

完整代码https://gist.github.com/bjelline/d8066de66e305887b714

Answer 1

鉴于df，下一步是单独按host值分组按idxmax汇总。这给你索引对应最大的服务价值。然后，您可以使用df.loc[...]选择df中与最大服务值对应的行：

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DataFrame

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

Pandas，groupby并在组中找到最大值，返回值和计数

1 个答案: