Pandas, groupby and finding maximum in groups, returning value and count
I have a pandas DataFrame with log data:
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
And I want to find the service on every host that gives the most errors:
host service no
0 this.com mail 2
1 that.com mail 1
2 other.net web 2
The only solution I found was grouping by host and service, and then iterating over the level 0 of the index.
Can anyone suggest a better, shorter version? without the Iteration?
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
full code https://gist.github.com/bjelline/d8066de66e305887b714
Given df , the next step is to group by the host value alone and
aggregate by idxmax . This gives you the index which corresponds the the greatest service value. You can then use df.loc[...] to select the rows in df which correspond to the greatest service values:
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("nOutputn{}".format(df_count))
yields the DataFrame
host service no
0 other.net web 2
1 that.com mail 1
2 this.com mail 2