如何获得 pandas groupby 中值的最小出现次数

Question

          home_team_name  home_team_goal_count
0         Bayern München                     2
1         Bayern München                     2
2         Bayern München                     1
3                   Köln                     2
4                   Köln                     2

I groupby the data on the variable home_team_name.

df.groupby("home_team_name")

The values of home_team_goal_count can only be 2 or 1. I want to get the minimum number of occurrences of the values in each group. The result I would want is 1 for Bayern Munchen and 0 for Koln. To illustrate Bayern Munchen has 2 times 2 and 1 times 1, therefore the minimum is 1. Koln has 2 times 2 and 0 time 1 therefore the minimum is 0.

Answer 1

First count values by SeriesGroupBy.value_counts , reshape and add 0 for all combinations 1,2 and last get minimum by min :

s = (df.groupby("home_team_name")['home_team_goal_count']
       .value_counts()
       .unstack(fill_value=0)
       .min(axis=1))

print (s)
home_team_name
Bayern München    1
Köln              0
dtype: int64

Details :

print (df.groupby("home_team_name")['home_team_goal_count']
         .value_counts()
         .unstack(fill_value=0))
home_team_goal_count  1  2
home_team_name            
Bayern München        1  2
Köln                  0  2

If possible only 1 or only 2 values in input data is necessary reindex :

s = (df.groupby("home_team_name")['home_team_goal_count']
       .value_counts()
       .unstack(fill_value=0)
       .reindex([1, 2], axis=1, fill_value=0) 
       .min(axis=1))

Answer 2

Let's try using pd.crosstab :

pd.crosstab(df['home_team_name'], df['home_team_goal_count'])\
    .reindex([1, 2], axis=1, fill_value=0).min(1)

Result:

home_team_name
Bayern München    1
Köln              0
dtype: int64

Answer 3

import pandas as pd
import numpy as np
list1=['Bayern Munchen','Bayern Munchen','Bayern Munchen','FC Koln','FC Koln']
list2=[2,2,1,2,2]
d={'Home Team Name':list1,'Home Team Goal Count':list2}
data=pd.DataFrame(d)

data['Name']= data['Home Team Name'] +" "+ data['Home Team Goal Count'].astype(str)

data['Name']
Out[39]: 
0    Bayern Munchen 2
1    Bayern Munchen 2
2    Bayern Munchen 1
3           FC Koln 2
4           FC Koln 2

name,count=np.unique(data['Name'].tolist(),return_counts=True)

name=[' '.join(x.split(' ')[:-1]) for x in name]

name
Out[99]: ['Bayern Munchen', 'Bayern Munchen', 'FC Koln']

min_val=pd.DataFrame({"Name":name,"Count":count})

name=[]
min_val_count=[]
for x in min_val.Name.unique():
    name.append(min_val[min_val.Name!=x].min()[0])
if min_val[min_val.Name!=x].min()[1]==2:
    min_val_count.append(0)
else:
    min_val_count.append(min_val[min_val.Name!=x].min()[1])


minimum_val_dict=dict(zip(name,min_val_count))

minimum_val_dict
Out[104]: {'FC Koln': 0, 'Bayern Munchen': 1}

A slightly longer version as compared to the answers above.

Answer 4

Even another way to do this would be to use a cateorical variable, since there's a finite set of states. So:

(
    df
    .astype({"home_team_goal_count": "category"})
    .groupby("home_team_name")["home_team_goal_count"]
    .apply(lambda x: x.value_counts().min())
)

If you want to know which value occurred the least, you can call .idxmin() instead of .min() .