topn 0.0.2

This package boosts a group-wise nlargest sort

topn

Utility function for string_grouper to use instead of pandas' nlargest() function (since pandas does it so slowly).

import pandas as pd
import numpy as np

r = np.array([0, 1, 2, 1, 2, 3, 2]) 
c = np.array([1, 1, 0, 3, 1, 2, 3]) 
d = np.array([0.0, 0.2, 0.1, 1.0, 0.9, 0.4, 0.6]) 
rcd = pd.DataFrame({'r': r, 'c': c, 'd': d})
rcd

	r	c	d
0	0	1	0.0
1	1	1	0.2
2	2	0	0.1
3	1	3	1.0
4	2	1	0.9
5	3	2	0.4
6	2	3	0.6

ntop = 2

rcd.set_index('c').groupby('r')['d'].nlargest(ntop).reset_index().sort_values(['r', 'd'], ascending = [True, False])

	r	c	d
0	0	1	0.0
1	1	3	1.0
2	1	1	0.2
3	2	1	0.9
4	2	3	0.6
5	3	2	0.4

Usage

from topn import awesome_topn

r, c, d = awesome_topn(r, c, d, ntop, n_jobs=7)
pd.DataFrame({'r': r, 'c': c, 'd': d})

	r	c	d
0	0	1	0.0
1	1	3	1.0
2	1	1	0.2
3	2	1	0.9
4	2	3	0.6
5	3	2	0.4

Short Description

def awesome_topn(r, c, d, ntop, use_threads=False, n_jobs=1):
    """
    r, c, and d are 1D numpy arrays all of the same length N. 
    This function will return arrays rn, cn, and dn of length n <= N such
    that the set of triples {(rn[i], cn[i], dn[i]) : 0 < i < n} is a subset of 
    {(r[j], c[j], d[j]) : 0 < j < N} and that for every distinct value 
    x = rn[i], dn[i] is among the first ntop existing largest d[j]'s whose 
    r[j] = x.

    Input:
        r and c: two 1D integer arrays of the same length
        d: 1D array of single or double precision floating point type of the
        same length as r or c
        ntop maximum number of maximum d's returned
        use_threads: use multi-thread or not
        n_jobs: number of threads, must be >= 1

    Output:
        (rn, cn, dn) where rn, cn, dn are all arrays as described above.
    """