计算加权成对马氏距离

Question

我知道HDMD 包有一个名为pairwise.mahalanobis 的函数，它应该计算成对的马氏距离。但是，我也想在这个距离上引入权重，这个函数是不可行的。因此，我开发了自己的代码。为了测试它是否运行良好，我首先保持简单，即没有权重，并将其结果与pairwise.mahalanobis 函数的结果进行比较。但是结果不匹配...下面是我使用的功能：

dist.maha <- function (X) {
  diff = pair.diff(X) # pairwise difference of rows 
  V <- cov(X)  ## empirical covariance; positive definite
  L <- t(chol(V))  ## lower triangular factor
  stdX <- t(forwardsolve(L, t(diff)))  # solving the system of linear equations
  return(stdX)
}

这是它在玩具数据上的两种替代方案的实现：

data = as.matrix(c(100, 54, 56, 79, 12))
dist_manuel = dist.maha(data)

# This is to convert dist_manuel from a vector to a distance matrix 

ind_1 = vector(length = choose(nrow(data),2))
ind_2 = vector(length = choose(nrow(data),2))
k =1  
for (j in 1:(nrow(data)-1)){
  for(i in (j+1):nrow(data)){
    ind_1[k] = i
    ind_2[k] = j    
    k = k + 1
  }
}
dist_manuel = cbind(ind_1,ind_2,dist_manuel)
dist_mat = matrix(data = NA, nrow = nrow(data), ncol = nrow(data))

for (j in 1:(nrow(data)-1)){
  for(i in (j+1):nrow(data)){
    dist_mat[i,j] = dist_manuel[which(dist_manuel[,1] == i & dist_manuel[,2] == j),3]
  }
}

# This is the HDMD alternative 

id = c(1,2,3,4,5)
data = cbind(id,data) 
HDMD = pairwise.mahalanobis(as.data.frame(data[,2]), grouping = data[,1])
dist_HDMD = HDMD$distance

# The outputs

dist_HDMD

#     [,1] [,2] [,3] [,4] [,5]
#[1,]    0    1    4    9   16
#[2,]    1    0    1    4    9
#[3,]    4    1    0    1    4
#[4,]    9    4    1    0    1
#[5,]   16    9    4    1    0

dist_mat

#          [,1]        [,2]       [,3]     [,4] [,5]
#[1,]        NA          NA         NA       NA   NA
#[2,] 1.4002541          NA         NA       NA   NA
#[3,] 1.3393735 -0.06088061         NA       NA   NA
#[4,] 0.6392465 -0.76100768 -0.7001271       NA   NA
#[5,] 2.6787470  1.27849290  1.3393735 2.039501   NA

pairwise.mahalanobis 函数的结果对我来说似乎完全荒谬。对于初学者，它为data[2] 和data[3] 和data[2] 和data[1] 分配距离1，当人们查看它们的值时这没有任何意义。另一方面，我的函数给出了一致的结果。比如我们比较一下data[1]&data[2]和data[1]&data[3]之间的距离比。

(100 - 54) / (100 - 56) = 46/44 = 1.045455

现在，这个比率也应该适用于我的函数产生的距离。

dist_mat[2,1]/dist_mat[3,1]
#[1] 1.045455

确实如此！这是否意味着我的功能运行良好，而 pairwise.mahalanobis 是错误的？ （或者我是否以某种方式错误地使用它？）我对R不是很有经验，所以我不敢自己得出这个结论。如果比我更有经验的人能证实我的逻辑，那就太好了。

Answer 1

您的dist.maha 函数中存在错误。这很明显，因为它计算的一些距离是负数——所以它们不可能是实际距离！幸运的是，这很容易解决，因为您只需将 stdX 向量平方即可。

library("HDMD")
library("tidyverse")

# Convert a vector of pairwise distances from to a distance matrix
# (Simplified approach which doesn't use for-loops)
pairwise_dist_to_dist_matrix <- function(dist, n) {
  stopifnot(length(dist) == n*(n-1)/2)
  mat <- matrix(NA_real_, n, n)
  diag(mat) <- 0
  mat[lower.tri(mat)] <- dist
  mat
}

dist.maha <- function (X) {
  diff <- pair.diff(X)                 # pairwise difference of rows
  V <- cov(X)                          # empirical covariance; positive definite
  L <- t(chol(V))                      # lower triangular factor
  stdX <- t(forwardsolve(L, t(diff)))  # solving the system of linear equations
  dist <- stdX * stdX                  # Don't forget to square!
  dist <- rowSums(dist)                # And add up the differences in each dimension.
  pairwise_dist_to_dist_matrix(dist, nrow(X))
}

# An alternative computation, for an additional check
dist.maha2 <- function(X) {
  diff <- pair.diff(X)
  V <- cov(X)
  Vinv <- solve(V)
  dist <- rowSums(diff %*% Vinv * diff)
  pairwise_dist_to_dist_matrix(dist, nrow(X))
}


# Slightly more complex data matrix to check if
# functions work in higher dimensions
data <- matrix(c(100, 54, 56, 79, 12, 1, 2, 3, 4, 5), ncol = 2)
dist.maha(data)
#>          [,1]      [,2]     [,3]     [,4] [,5]
#> [1,] 0.000000        NA       NA       NA   NA
#> [2,] 2.275210 0.0000000       NA       NA   NA
#> [3,] 1.974017 0.9742819 0.000000       NA   NA
#> [4,] 4.759700 7.5842687 3.250906 0.000000   NA
#> [5,] 7.896067 3.6213875 1.974017 5.690146    0
dist.maha2(data)
#>          [,1]      [,2]     [,3]     [,4] [,5]
#> [1,] 0.000000        NA       NA       NA   NA
#> [2,] 2.275210 0.0000000       NA       NA   NA
#> [3,] 1.974017 0.9742819 0.000000       NA   NA
#> [4,] 4.759700 7.5842687 3.250906 0.000000   NA
#> [5,] 7.896067 3.6213875 1.974017 5.690146    0

您似乎也没有正确使用pairwise.mahalanobis。您必须计算并传递协方差矩阵（cov 参数）。

# This is the HDMD alternative
id = c(1,2,3,4,5)

# Incorrect:

# You have to specify the `cov` argument.
# Otherwise `pairwise.mahalanobis` doesn't compute it correctly
# as each sample is assumed to be in its own group.

pairwise.mahalanobis(data, grouping = id)$distance
#>           [,1]      [,2]      [,3]     [,4]      [,5]
#> [1,]     0.000 4345.2805 3840.7349  759.689 15362.940
#> [2,]  4345.280    0.0000   16.7591 1487.209  3369.708
#> [3,]  3840.735   16.7591    0.0000 1194.197  3840.735
#> [4,]   759.689 1487.2090 1194.1967    0.000  9310.174
#> [5,] 15362.940 3369.7076 3840.7349 9310.174     0.000

# Correct:

# NOTE: Ignore the warning message; there seems to be a small bug in `pairwise.mahalanobis`.
pairwise.mahalanobis(data, grouping = id, cov = cov(data))$distance
#> Warning in if (dim(cov) != c(p, p)) stop("cov matrix not of dim = (p,p)
#> \n"): the condition has length > 1 and only the first element will be used
#>          [,1]      [,2]      [,3]     [,4]     [,5]
#> [1,] 0.000000 2.2752099 1.9740168 4.759700 7.896067
#> [2,] 2.275210 0.0000000 0.9742819 7.584269 3.621388
#> [3,] 1.974017 0.9742819 0.0000000 3.250906 1.974017
#> [4,] 4.759700 7.5842687 3.2509059 0.000000 5.690146
#> [5,] 7.896067 3.6213875 1.9740168 5.690146 0.000000

^{由reprex package (v0.2.1) 于 2019 年 3 月 24 日创建}