pandas groupby where you get the max of one column

2019-01-19 18:24发布

问题:

I have a dataframe as follows:

user    num1    num2
a       1       1
a       2       2
a       3       3
b       4       4
b       5       5

I want a dataframe which has the minimum from num1 for each user, and the maximum of num2 for each user.

The output should be like:

user    num1    num2
a       1       3
b       4       5

I know that if I wanted the max of both columns I could just do:

a.groupby('user')['num1', 'num2'].max()

Is there some equivalent without having to do something like:

series_1 = a.groupby('user')['num1'].min() 
series_2 = a.groupby('user')['num2'].max()

# converting from series to df so I can do a join on user
df_1 = pd.DataFrame(np.array([series_1]).transpose(), index=series_1.index, columns=['num1']) 
df_2 = pd.DataFrame(np.array([series_2]).transpose(), index=series_2.index, columns=['num2'])

df_1.join(df_2)

回答1:

Use groupby + agg by dict, so then is necessary order columns by subset or reindex_axis. Last add reset_index for convert index to column if necessary.

df = a.groupby('user').agg({'num1':'min', 'num2':'max'})[['num1','num2']].reset_index()
print (df)
  user  num1  num2
0    a     1     3
1    b     4     5

What is same as:

df = a.groupby('user').agg({'num1':'min', 'num2':'max'})
                      .reindex_axis(['num1','num2'], axis=1)
                      .reset_index()
print (df)
  user  num1  num2
0    a     1     3
1    b     4     5