Python中更有效的加权基尼系数

Question

根据/sf/answers/3428728411/,这是Python中加权基尼系数的实现:

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

这很干净,适用于中型阵列,但正如其初步建议(/sf/answers/2765965961/)所述,它是O(n ²).在我的计算机上,这意味着它在大约20k行之后中断:

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

可以调整它以适用于更大的数据集吗？我的行是~150k行.

Answer 1

这是一个比上面提供的版本快得多的版本,并且在没有重量的情况下使用简化的公式来获得更快的结果.

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

这里有一些测试代码来检查我们得到(大多数)相同的结果:

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

但速度差异很大:

%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)

如果从函数中删除pandas ops,它已经快得多:

%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)

如果你想获得最后一滴性能,你可以使用numba或cython,但这只会获得几个百分点,因为大部分时间都花在排序上.

%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)

编辑:gini_max_ghenis是Max Ghenis回答中使用的代码

Answer 2

StatsGini从这里调整R 函数：

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

这适用于大型向量，至少高达 10M 行：

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

它还产生与问题中提供的函数相同的结果，例如，本例给出 0.2553：

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