Faster way to rank rows in subgroups in pandas dataframe

Question

I have a pandas data frame that has is composed of different subgroups.

df = pd.DataFrame({
'id':[1, 2, 3, 4, 5, 6, 7, 8], 
'group':['a', 'a', 'a', 'a', 'b', 'b', 'b', 'b'], 
'value':[.01, .4, .2, .3, .11, .21, .4, .01]
})

I want to find the rank of each id in its group with say, lower values being better. In the example above, in group A, Id 1 would have a rank of 1, Id 2 would have a rank of 4. In group B, Id 5 would have a rank of 2, Id 8 would have a rank of 1 and so on.

Right now I assess the ranks by:

Sorting by value.

df.sort('value', ascending = True, inplace=True)

Create a ranker function (it assumes variables already sorted)

def ranker(df):
    df['rank'] = np.arange(len(df)) + 1
    return df

Apply the ranker function on each group separately:

df = df.groupby(['group']).apply(ranker)

This process works but it is really slow when I run it on millions of rows of data. Does anyone have any ideas on how to make a faster ranker function.

Asked By: captain ahab

||

Source

Answer 1

rank is cythonized so should be very fast. And you can pass the same options as df.rank()
here are the docs for rank. As you can see, tie-breaks can be done in one of five different ways via the method argument.

Its also possible you simply want the .cumcount() of the group.

In [12]: df.groupby('group')['value'].rank(ascending=False)
Out[12]: 
0    4
1    1
2    3
3    2
4    3
5    2
6    1
7    4
dtype: float64

Answered By: Jeff

Answer 2

Working with a big DataFrame (13 million lines), the method rank with groupby maxed out my 8GB of RAM an it took a really long time. I found a workaround less greedy in memory , that I put here just in case:

df.sort_values('value')
tmp = df.groupby('group').size()
rank = tmp.map(range)
rank =[item for sublist in rank for item in sublist]
df['rank'] = rank

Answered By: Quentin Febvre

Faster way to rank rows in subgroups in pandas dataframe

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