根据
https://stackoverflow.com/a/48981834/1840471,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,但正如其最初的建议(https://stackoverflow.com/a/39513799/1840471)所警告的那样是O(n2).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.
解决方法
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_slow(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_slow(x,w)
- 55.6 s ± 3.35 s per loop (mean ± std. dev. of 7 runs,1 loop each)
- %timeit gini_slow2(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)