R-如何取得列&距离矩阵中匹配元素的列下标 [英] R - How to get row & column subscripts of matched elements from a distance matrix

问题描述

我有一个整数矢量vec1，并且正在使用dist函数生成一个远距矩阵.我想在距离矩阵中获取某个值的元素的坐标(行和列).从本质上讲，我想将这对元素分开d距离.例如:

I have an integer vector vec1 and I am generating a distant matrix using dist function. I want to get the coordinates (row and column) of element of certain value in the distance matrix. Essentially I would like to get the pair of elements that are d-distant apart. For example:

vec1 <- c(2,3,6,12,17)
distMatrix <- dist(vec1)

#   1  2  3  4
#2  1         
#3  4  3      
#4 10  9  6   
#5 15 14 11  5

说，我对向量中相距5个单位的一对元素感兴趣.我想获得分别是距离矩阵的行和坐标2的列的coordinate1.在这个玩具示例中，我希望

Say, I am interested in pair of elements in the vector that are 5 unit apart. I wanted to get the coordinate1 which are the rows and coordinate2 which are the columns of the distance matrix. In this toy example, I would expect

coord1  
# [1] 5
coord2
# [1] 4

我想知道是否有一种有效的方法来获取这些值，而不涉及将dist对象转换为矩阵或遍历矩阵?

I am wondering if there is an efficient way to get these values that doesn't involve converting the dist object to a matrix or looping through the matrix?

推荐答案

距离矩阵是压缩格式的下三角矩阵，其中下三角按列存储为一维矢量.您可以通过

A distance matrix is a lower triangular matrix in packed format, where the lower triangular is stored as a 1D vector by column. You can check this via

str(distMatrix)
# Class 'dist'  atomic [1:10] 1 4 10 15 3 9 14 6 11 5
# ...

即使我们调用dist(vec1, diag = TRUE, upper = TRUE)，向量也仍然相同；仅打印样式会更改.也就是说，无论您如何调用dist，都始终会得到一个向量.

Even if we call dist(vec1, diag = TRUE, upper = TRUE), the vector is still the same; only the printing styles changes. That is, no matter how you call dist, you always get a vector.

此答案集中于如何在1D和2D索引之间进行转换，因此您可以使用"dist"对象，而无需先使用as.matrix使其成为完整的矩阵.如果确实要使其成为矩阵，则对距离对象使用 as.matrix中定义的dist2mat函数非常慢；如何使其更快?.

This answer focus on how to transform between 1D and 2D index, so that you can work with a "dist" object without first making it a complete matrix using as.matrix. If you do want to make it a matrix, use the dist2mat function defined in as.matrix on a distance object is extremely slow; how to make it faster?.

为这些索引转换编写向量化的R函数很容易.我们只需要处理越界"索引，应该为其返回NA.

It is easy to write vectorized R functions for those index transforms. We only need some care dealing with "out-of-bound" index, for which NA should be returned.

## 2D index to 1D index
f <- function (i, j, dist_obj) {
  if (!inherits(dist_obj, "dist")) stop("please provide a 'dist' object")
  n <- attr(dist_obj, "Size")
  valid <- (i >= 1) & (j >= 1) & (i > j) & (i <= n) & (j <= n)
  k <- (2 * n - j) * (j - 1) / 2 + (i - j)
  k[!valid] <- NA_real_
  k
  }

## 1D index to 2D index
finv <- function (k, dist_obj) {
  if (!inherits(dist_obj, "dist")) stop("please provide a 'dist' object")
  n <- attr(dist_obj, "Size")
  valid <- (k >= 1) & (k <= n * (n - 1) / 2)
  k_valid <- k[valid]
  j <- rep.int(NA_real_, length(k))
  j[valid] <- floor(((2 * n + 1) - sqrt((2 * n - 1) ^ 2 - 8 * (k_valid - 1))) / 2)
  i <- j + k - (2 * n - j) * (j - 1) / 2
  cbind(i, j)
  }

这些函数在内存使用上非常便宜，因为它们使用索引而不是矩阵.

These functions are extremely cheap in memory usage, as they work with index instead of matrices.

您可以使用

vec1 <- c(2,3,6,12,17)
distMatrix <- dist(vec1)

finv(which(distMatrix == 5), distMatrix)
#     i j
#[1,] 5 4

通常，距离矩阵包含浮点数.使用==判断两个浮点数是否相等是有风险的.阅读为什么这些数字不相等?以了解更多可行的策略.

Generally speaking, a distance matrix contains floating point numbers. It is risky to use == to judge whether two floating point numbers are equal. Read Why are these numbers not equal? for more and possible strategies.

在距离对象上使用 as.matrix中提供的dist2mat函数非常慢；如何使其更快?，我们可能会使用which(, arr.ind = TRUE).

Using the dist2mat function given in as.matrix on a distance object is extremely slow; how to make it faster?, we may use which(, arr.ind = TRUE).

library(Rcpp)
sourceCpp("dist2mat.cpp")
mat <- dist2mat(distMatrix, 128)
which(mat == 5, arr.ind = TRUE)
#  row col
#5   5   4
#4   4   5

---

附录:图片的Markdown(需要MathJax支持)

## 2D index to 1D index

The lower triangular looks like this: $$\begin{pmatrix} 0 & 0 & \cdots & 0\\ \times & 0 & \cdots & 0\\ \times & \times & \cdots & 0\\ \vdots & \vdots & \ddots & 0\\ \times & \times & \cdots & 0\end{pmatrix}$$ If the matrix is $n \times n$, then there are $(n - 1)$ elements ("$\times$") in the 1st column, and $(n - j)$ elements in the j<sup>th</sup> column. Thus, for element $(i,\  j)$ (with $i > j$, $j < n$) in the lower triangular, there are $$(n - 1) + \cdots (n - (j - 1)) = \frac{(2n - j)(j - 1)}{2}$$ "$\times$" in the previous $(j - 1)$ columns, and it is the $(i - j)$<sup>th</sup> "$\times$" in the $j$<sup>th</sup> column. So it is the $$\left\{\frac{(2n - j)(j - 1)}{2} + (i - j)\right\}^{\textit{th}}$$ "$\times$" in the lower triangular.

----

## 1D index to 2D index

Now for the $k$<sup>th</sup> "$\times$" in the lower triangular, how can we find its matrix index $(i,\ j)$? We take two steps: 1> find $j$;  2> obtain $i$ from $k$ and $j$.

The first "$\times$" of the $j$<sup>th</sup> column, i.e., $(j + 1,\ j)$, is the $\left\{\frac{(2n - j)(j - 1)}{2} + 1\right\}^{\textit{th}}$ "$\times$" of the lower triangular, thus $j$ is the maximum value such that $\frac{(2n - j)(j - 1)}{2} + 1 \leq k$. This is equivalent to finding the max $j$ so that $$j^2 - (2n + 1)j + 2(k + n - 1) \geq 0.$$ The LHS is a quadratic polynomial, and it is easy to see that the solution is the integer no larger than its first root (i.e., the root on the left side): $$j = \left\lfloor\frac{(2n + 1) - \sqrt{(2n-1)^2 - 8(k-1)}}{2}\right\rfloor.$$ Then $i$ can be obtained from $$i = j + k - \left\{\frac{(2n - j)(j - 1)}{2}\right\}.$$

这篇关于R-如何取得列&距离矩阵中匹配元素的列下标的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！