8

假设我们有两个样本data1和data2它们各自的权重weight1，weight2并且我们想要计算两个加权样本之间的 Kolmogorov-Smirnov 统计量。

我们在 python 中这样做的方式如下：

import numpy as np

def ks_w(data1,data2,wei1,wei2):
    ix1=np.argsort(data1)
    ix2=np.argsort(data2)
    wei1=wei1[ix1]
    wei2=wei2[ix2]
    data1=data1[ix1]
    data2=data2[ix2]
    d=0.
    fn1=0.
    fn2=0.
    j1=0
    j2=0
    j1w=0.
    j2w=0.
    while(j1<len(data1))&(j2<len(data2)):
        d1=data1[j1]
        d2=data2[j2]
        w1=wei1[j1]
        w2=wei2[j2]
        if d1<=d2:
            j1+=1
            j1w+=w1
            fn1=(j1w)/sum(wei1)
        if d2<=d1:
            j2+=1
            j2w+=w2
            fn2=(j2w)/sum(wei2)
        if abs(fn2-fn1)>d:
            d=abs(fn2-fn1)
    return d

我们只是根据我们的目的修改在Press, Flannery, Teukolsky, Vetterling - Numerical Recipes in C - Cambridge University Press - 1992 - pag.626中实施的经典双样本 KS 统计量。

我们的问题是：

  • 有人知道其他方法吗？
  • python/R/* 中是否有任何库可以执行它？
  • 测试呢？它是否存在或者我们应该使用重新洗牌程序来评估统计数据？
4

3 回答 3

7

该解决方案基于以下代码scipy.stats.ks_2samp并以大约 1/10000 的时间运行（笔记本）：

import numpy as np

def ks_w2(data1, data2, wei1, wei2):
    ix1 = np.argsort(data1)
    ix2 = np.argsort(data2)
    data1 = data1[ix1]
    data2 = data2[ix2]
    wei1 = wei1[ix1]
    wei2 = wei2[ix2]
    data = np.concatenate([data1, data2])
    cwei1 = np.hstack([0, np.cumsum(wei1)/sum(wei1)])
    cwei2 = np.hstack([0, np.cumsum(wei2)/sum(wei2)])
    cdf1we = cwei1[[np.searchsorted(data1, data, side='right')]]
    cdf2we = cwei2[[np.searchsorted(data2, data, side='right')]]
    return np.max(np.abs(cdf1we - cdf2we))

以下是对其准确性和性能的测试：

ds1 = np.random.rand(10000)
ds2 = np.random.randn(40000) + .2
we1 = np.random.rand(10000) + 1.
we2 = np.random.rand(40000) + 1.

ks_w2(ds1, ds2, we1, we2)
# 0.4210415232236593
ks_w(ds1, ds2, we1, we2)
# 0.4210415232236593

%timeit ks_w2(ds1, ds2, we1, we2)
# 100 loops, best of 3: 17.1 ms per loop
%timeit ks_w(ds1, ds2, we1, we2)
# 1 loop, best of 3: 3min 44s per loop
于 2016-10-15T13:40:57.680 回答
2

为了补充 Luca Jokull 的答案，如果您还想返回一个 p 值（类似于未加权scipy.stats.ks_2samp函数），建议的ks_w2()函数可以修改如下：

from scipy.stats import distributions

def ks_weighted(data1, data2, wei1, wei2, alternative='two-sided'):
    ix1 = np.argsort(data1)
    ix2 = np.argsort(data2)
    data1 = data1[ix1]
    data2 = data2[ix2]
    wei1 = wei1[ix1]
    wei2 = wei2[ix2]
    data = np.concatenate([data1, data2])
    cwei1 = np.hstack([0, np.cumsum(wei1)/sum(wei1)])
    cwei2 = np.hstack([0, np.cumsum(wei2)/sum(wei2)])
    cdf1we = cwei1[np.searchsorted(data1, data, side='right')]
    cdf2we = cwei2[np.searchsorted(data2, data, side='right')]
    d = np.max(np.abs(cdf1we - cdf2we))
    # calculate p-value
    n1 = data1.shape[0]
    n2 = data2.shape[0]
    m, n = sorted([float(n1), float(n2)], reverse=True)
    en = m * n / (m + n)
    if alternative == 'two-sided':
        prob = distributions.kstwo.sf(d, np.round(en))
    else:
        z = np.sqrt(en) * d
        # Use Hodges' suggested approximation Eqn 5.3
        # Requires m to be the larger of (n1, n2)
        expt = -2 * z**2 - 2 * z * (m + 2*n)/np.sqrt(m*n*(m+n))/3.0
        prob = np.exp(expt)
    return d, prob

这是 scipy原始未加权函数使用的渐近方法。

于 2021-05-21T14:32:23.490 回答
2

这是一个 R 版本的双尾加权 KS 统计量，遵循 Monohan 的 Numerical Methods of Statistics，pg. 的建议。334 在 1E 和 pg。2E 中的 358 个。

ks_weighted <- function(vector_1,vector_2,weights_1,weights_2){
    F_vec_1 <- ewcdf(vector_1, weights = weights_1, normalise=FALSE)
    F_vec_2 <- ewcdf(vector_2, weights = weights_2, normalise=FALSE)
    xw <- c(vector_1,vector_2) 
    d <- max(abs(F_vec_1(xw) - F_vec_2(xw)))

    ## P-VALUE with NORMAL SAMPLE 
    # n_vector_1 <- length(vector_1)                                                           
    # n_vector_2<- length(vector_2)        
    # n <- n_vector_1 * n_vector_2/(n_vector_1 + n_vector_2)

    # P-VALUE EFFECTIVE SAMPLE SIZE as suggested by Monahan
    n_vector_1 <- sum(weights_1)^2/sum(weights_1^2)
    n_vector_2 <- sum(weights_2)^2/sum(weights_2^2)
    n <- n_vector_1 * n_vector_2/(n_vector_1 + n_vector_2)


    pkstwo <- function(x, tol = 1e-06) {
                if (is.numeric(x)) 
                    x <- as.double(x)
                else stop("argument 'x' must be numeric")
                p <- rep(0, length(x))
                p[is.na(x)] <- NA
                IND <- which(!is.na(x) & (x > 0))
                if (length(IND)) 
                    p[IND] <- .Call(stats:::C_pKS2, p = x[IND], tol)
                p
            }

    pval <- 1 - pkstwo(sqrt(n) * d)

    out <- c(KS_Stat=d, P_value=pval)
    return(out)
}
于 2019-04-13T09:58:25.260 回答