More efficient weighted Gini coefficient in Python

问题:

Per https://stackoverflow.com/a/48981834/1840471, this is an implementation of the weighted Gini coefficient in Python:

import numpy as np
def gini(x, weights=None):
    if weights is None:
        weights = np.ones_like(x)
    # Calculate mean absolute deviation in two steps, for weights.
    count = np.multiply.outer(weights, weights)
    mad = np.abs(np.subtract.outer(x, x) * count).sum() / count.sum()
    rmad = mad / np.average(x, weights=weights)
    # Gini equals half the relative mean absolute deviation.
    return 0.5 * rmad

This is clean and works well for medium-sized arrays, but as warned in its initial suggestion (https://stackoverflow.com/a/39513799/1840471) it's O(n²). On my computer that means it breaks after ~20k rows:

n = 20000  # Works, 30000 fails.
gini(np.random.rand(n), np.random.rand(n))

Can this be adjusted to work for larger datasets? Mine is ~150k rows.

回答1:

Here is a version which is much faster than the one you provided above, and also uses a simplified formula for the case without weight to get even faster results in that case.

def gini(x, w=None):
    # The rest of the code requires numpy arrays.
    x = np.asarray(x)
    if w is not None:
        w = np.asarray(w)
        sorted_indices = np.argsort(x)
        sorted_x = x[sorted_indices]
        sorted_w = w[sorted_indices]
        # Force float dtype to avoid overflows
        cumw = np.cumsum(sorted_w, dtype=float)
        cumxw = np.cumsum(sorted_x * sorted_w, dtype=float)
        return (np.sum(cumxw[1:] * cumw[:-1] - cumxw[:-1] * cumw[1:]) / 
                (cumxw[-1] * cumw[-1]))
    else:
        sorted_x = np.sort(x)
        n = len(x)
        cumx = np.cumsum(sorted_x, dtype=float)
        # The above formula, with all weights equal to 1 simplifies to:
        return (n + 1 - 2 * np.sum(cumx) / cumx[-1]) / n

Here is some test code to check we get (mostly) the same results:

>>> x = np.random.rand(1000000)
>>> w = np.random.rand(1000000)
>>> gini_max_ghenis(x, w)
0.33376310938610521
>>> gini(x, w)
0.33376310938610382

But the speed is very different:

%timeit gini(x, w)
203 ms ± 3.68 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

%timeit gini_max_ghenis(x, w)
55.6 s ± 3.35 s per loop (mean ± std. dev. of 7 runs, 1 loop each)

If you remove the pandas ops from the function, it is already much faster:

%timeit gini_max_ghenis_no_pandas_ops(x, w)
1.62 s ± 75 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

If you want to get the last drop of performance you could use numba or cython but that would only gain a few percent because most of the time is spent in sorting.

%timeit ind = np.argsort(x); sx = x[ind]; sw = w[ind]
180 ms ± 4.82 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

edit: gini_max_ghenis is the code used in Max Ghenis' answer

回答2:

Adapting the StatsGini R function from here:

import numpy as np
import pandas as pd

def gini(x, w=None):
    # Array indexing requires reset indexes.
    x = pd.Series(x).reset_index(drop=True)
    if w is None:
        w = np.ones_like(x)
    w = pd.Series(w).reset_index(drop=True)
    n = x.size
    wxsum = sum(w * x)
    wsum = sum(w)
    sxw = np.argsort(x)
    sx = x[sxw] * w[sxw]
    sw = w[sxw]
    pxi = np.cumsum(sx) / wxsum
    pci = np.cumsum(sw) / wsum
    g = 0.0
    for i in np.arange(1, n):
        g = g + pxi.iloc[i] * pci.iloc[i - 1] - pci.iloc[i] * pxi.iloc[i - 1]
    return g

This works for large vectors, at least up to 10M rows:

n = 1e7
gini(np.random.rand(n), np.random.rand(n))  # Takes ~15s.

It also produces the same result as the function provided in the question, for example giving 0.2553 for this example:

gini(np.array([3, 1, 6, 2, 1]), np.array([4, 2, 2, 10, 1]))