Pandas groupby为每个值添加新列

Question

I hope the title speaks for itself; I'd just like to add that it can be assumed that each key has the same amount of values. Online searching the title yielded the following solution:

Split pandas dataframe based on groupby

Which supposed to be solving my problem, although it does not. I'll give an example:

Input:

pd.DataFrame(data={'a':['foo','foo','foo','bar','bar','bar'],'b':[1,2,3,4,5,6]})

Output:

pd.DataFrame(data={'a':['foo','bar'],'b':[1,4],'c':[2,5],'d':[3,6]})

Intuitively, it would be a groupby function without an aggregation function, or an aggregation function that makes a list out of the keys.

Obviously, it can be done 'manually' using for loops etc., but using for loops with large data sets is very expensive computationally.

Answer 1

Use GroupBy.cumcount for Series or column g , then reshape by DataFrame.set_index + Series.unstack or DataFrame.pivot , last data cleaning by DataFrame.add_prefix , DataFrame.rename_axis with DataFrame.reset_index :

g = df1.groupby('a').cumcount()
df = (df1.set_index(['a', g])['b']
         .unstack()
         .add_prefix('new_')
         .reset_index()
         .rename_axis(None, axis=1))
print (df)
     a  new_0  new_1  new_2
0  bar      4      5      6
1  foo      1      2      3

Or:

df1['g'] = df1.groupby('a').cumcount()
df = df1.pivot('a','g','b').add_prefix('new_').reset_index().rename_axis(None, axis=1)
print (df)
     a  new_0  new_1  new_2
0  bar      4      5      6
1  foo      1      2      3

Answer 2

Here is an alternative approach, using groupby.apply and string.ascii_lowercase if column names are important:

from string import ascii_lowercase

df = pd.DataFrame(data={'a':['foo','foo','foo','bar','bar','bar'],'b':[1,2,3,4,5,6]})

# Groupby 'a'
g = df.groupby('a')['b'].apply(list)

# Construct new DataFrame from g
new_df = pd.DataFrame(g.values.tolist(), index=g.index).reset_index()

# Fix column names
new_df.columns = [x for x in ascii_lowercase[:new_df.shape[1]]]

print(new_df)

     a  b  c  d
0  bar  4  5  6
1  foo  1  2  3