pandas dataframe groupby 索引并将行值转换为列

Question

While I think I could do this naively and poorly, I'm interested to learn a more elegant and efficient approach.

Given the following dataframe:

In [42]: df = pd.DataFrame({'flavor':['goober','tronic','goober','tronic'], 'points':[42,55,31,101]}, index=['foo','foo','bar','bar'])

In [43]: df
Out[43]: 
     flavor  points
foo  goober      42
foo  tronic      55
bar  goober      31
bar  tronic     101

I would like to groupby the index, and convert values from flavor column into column headers themselves, completely throwing away the flavor and points . So the final result would look like:

In [44]: pd.DataFrame({'goober':[42,31], 'tronic':[55,101]}, index=['foo','bar'])
Out[44]: 
     goober  tronic
foo      42      55
bar      31     101

Thanks for any suggestions.

Answer 1

Use DataFrame.pivot with convert index to column first and then remove index and columns names by DataFrame.rename_axis :

df = df.reset_index().pivot('index', 'flavor','points').rename_axis(index=None,columns=None)
print (df)
     goober  tronic
bar      31     101
foo      42      55

Or use DataFrame.set_index with Series.unstack :

df = (df.set_index('flavor', append=True)['points']
        .unstack()
        .rename_axis(index=None, columns=None))
print (df)
     goober  tronic
bar      31     101
foo      42      55

Answer 2

You can use pandas.pivot_table :

print( pd.pivot_table(df, index=df.index, columns='flavor', values='points') )

Prints:

flavor  goober  tronic
bar         31     101
foo         42      55

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Note: If you don't need aggfunc , use .pivot() instead. As @jezrael pointed out in the comments.