captain ahab - 1 year ago 1163

Python 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.

Answer Source

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
```