编写引用矩阵中元素的函数

Question

我有一个名为“关系”的数据框：

library(tidyverse)
library(igraph)


set.seed(123)

n=10
data = tibble(d = paste(1:n))

relations = tibble(
  from = sample(data$d),
  to = lead(from, default=from[1]),
)

relations = data.frame(relations)

“关系”数据框如下所示：

然后我有一个名为“m”的矩阵（这个矩阵的维度对应于“关系”数据框）：

m<-matrix(rnorm(100) , nrow = 10)

             [,1]       [,2]        [,3]       [,4]       [,5]       [,6]        [,7]       [,8]        [,9]      [,10]
 [1,] -1.66794194 -0.6407060  0.07796085  0.7399475  0.6879168 -0.3724388 -0.48378063  0.3104807  0.08473729 -1.2512714
 [2,] -0.38022652 -0.8497043 -0.96185663  1.9091036  2.1001089  0.9769734  0.51686204  0.4365235  0.75405379 -0.6111659
 [3,]  0.91899661 -1.0241288 -0.07130809 -1.4438932 -1.2870305 -0.3745809  0.36896453 -0.4583653 -0.49929202 -1.1854801
 [4,] -0.57534696  0.1176466  1.44455086  0.7017843  0.7877388  1.0527115 -0.21538051 -1.0633261  0.21444531  2.1988103
 [5,]  0.60796432 -0.9474746  0.45150405 -0.2621975  0.7690422 -1.0491770  0.06529303  1.2631852 -0.32468591  1.3124130
 [6,] -1.61788271 -0.4905574  0.04123292 -1.5721442  0.3322026 -1.2601552 -0.03406725 -0.3496504  0.09458353 -0.2651451
 [7,] -0.05556197 -0.2560922 -0.42249683 -1.5146677 -1.0083766  3.2410399  2.12845190 -0.8655129 -0.89536336  0.5431941
 [8,]  0.51940720  1.8438620 -2.05324722 -1.6015362 -0.1194526 -0.4168576 -0.74133610 -0.2362796 -1.31080153 -0.4143399
 [9,]  0.30115336 -0.6519499  1.13133721 -0.5309065 -0.2803953  0.2982276 -1.09599627 -0.1971759  1.99721338 -0.4762469
[10,]  0.10567619  0.2353866 -1.46064007 -1.4617556  0.5629895  0.6365697  0.03778840  1.1099203  0.60070882 -0.7886028

我想编写一个函数：

选择矩阵“m”中对应于“关系”数据框中每一行的“（从，到）数字”的元素
然后将所有这些元素加在一起

在我的例子中，这看起来像这样：

sums = m[3,10] + m[10,2] + m[2,8] + m[8,6] + m[6,9] + m[9,1] + m[1,7] + m[7,5] + m[5,4] + m[4,3]

 sums

[1] -0.8444946

我想也许我可以使用以下“技巧”“引用”矩阵中的正确元素：

sums = m[relations[1,1], relations[1,2]] + m[relations[2,1], relations[2,2]] + m[relations[3,1], relations[3,2]] + m[relations[4,1], relations[4,2]] + m[relations[5,1], relations[5,2]] + m[relations[6,1], relations[6,2]] + m[relations[7,1], relations[7,2]] + m[relations[8,1], relations[8,2]] + m[relations[9,1], relations[9,2]] + m[relations[10,1], relations[10,2]]

Error in m[relations[1, 1], relations[1, 2]] : subscript out of bounds

然后我尝试了这个：

sums = m[as.numeric(relations[1,1]), as.numeric(relations[1,2])] + 
m[as.numeric(relations[2,1]), as.numeric(relations[2,2])] 
+ m[as.numeric(relations[3,1]), as.numeric(relations[3,2])] 
+ m[as.numeric(relations[4,1]), as.numeric(relations[4,2])] 
+ m[as.numeric(relations[5,1]), as.numeric(relations[5,2])] +
 m[as.numeric(relations[6,1]), as.numeric(relations[6,2])] +
 m[as.numeric(relations[7,1]), as.numeric(relations[7,2])] + 
m[as.numeric(relations[8,1]), as.numeric(relations[8,2])] +
 m[as.numeric(relations[9,1]), as.numeric(relations[9,2])] +
 m[as.numeric(relations[10,1]), as.numeric(relations[10,2])]

        X9 
0.08593303

虽然我想错了，因为

-0.8444946 <> 0.08593303

如何编写一个函数，自动对任何“关系”和“m”执行以下操作？

sums = m[3,10] + m[10,2] + m[2,8] + m[8,6] + m[6,9] + m[9,1] + m[1,7] + m[7,5] + m[5,4] + m[4,3]

谢谢！

Answer 1

您可以使用 [ 为矩阵 m 编制索引，而另一个矩阵的列对应于 m 的行和列，就像您的 relations 对象一样。所以我们需要做的就是将 relations 隐藏到 matrix，用它来索引 m，然后 sum 全部：

my_sums = function(relations, m) {
  r = as.matrix(relations)
  if(mode(r) != "numeric") mode(r) = "numeric"
  sum(m[r])
}

my_sums(relations, m)

（未经测试，因为您没有共享 m，但应该可以正常工作。）

编写引用矩阵中元素的函数

Writing a Function That References Elements in a Matrix

r

sum

function

matrix