構建元素

Question

之間關係的對稱矩陣有一個矩陣：

infor <- cbind(c("1st","2nd","3rd","4th","5th","6th"), c("a;b;c","c;d;e;f","a;c;d","b;g;h","b;d;e","e;h")) 
infor 
    [,1] [,2]  
[1,] "1st" "a;b;c" 
[2,] "2nd" "c;d;e;f" 
[3,] "3rd" "a;c;d" 
[4,] "4th" "b;g;h" 
[5,] "5th" "b;d;e" 
[6,] "6th" "e;h" 

我想在infor[, 1]來算，每兩個元素之間的重疊和構建這樣一個對稱矩陣：

> result 
    1st 2nd 3rd 4th 5th 6th 
1st 3 1 2 1 1 0 
2nd 1 4 2 1 2 1 
3rd 2 2 3 1 1 0 
4th 1 1 1 3 1 1 
5th 1 2 1 1 3 1 
6th 0 1 0 1 1 2 

然後我bulid兩個矩陣aa和bb：

> aa <- matrix(rep(infor[, 2], dim(infor)[1]), nrow=dim(infor)[1]) 
> aa 
    [,1]  [,2]  [,3]  [,4]  [,5]  [,6]  
[1,] "a;b;c" "a;b;c" "a;b;c" "a;b;c" "a;b;c" "a;b;c" 
[2,] "c;d;e;f" "c;d;e;f" "c;d;e;f" "c;d;e;f" "c;d;e;f" "c;d;e;f" 
[3,] "a;c;d" "a;c;d" "a;c;d" "a;c;d" "a;c;d" "a;c;d" 
[4,] "b;g;h" "b;g;h" "b;g;h" "b;g;h" "b;g;h" "b;g;h" 
[5,] "b;d;e" "b;d;e" "b;d;e" "b;d;e" "b;d;e" "b;d;e" 
[6,] "e;h"  "e;h"  "e;h"  "e;h"  "e;h"  "e;h"  
> bb <- t(aa) 
> bb 
    [,1] [,2]  [,3] [,4] [,5] [,6] 
[1,] "a;b;c" "c;d;e;f" "a;c;d" "b;g;h" "b;d;e" "e;h" 
[2,] "a;b;c" "c;d;e;f" "a;c;d" "b;g;h" "b;d;e" "e;h" 
[3,] "a;b;c" "c;d;e;f" "a;c;d" "b;g;h" "b;d;e" "e;h" 
[4,] "a;b;c" "c;d;e;f" "a;c;d" "b;g;h" "b;d;e" "e;h" 
[5,] "a;b;c" "c;d;e;f" "a;c;d" "b;g;h" "b;d;e" "e;h" 
[6,] "a;b;c" "c;d;e;f" "a;c;d" "b;g;h" "b;d;e" "e;h" 

> Overlaps <- function(a, b){ 
    spliteA <- strsplit(a, ";") 
    spliteB <- strsplit(b, ";") 
    score <- length(intersect(spliteA, spliteB)) 
    return(score) 
    } 

我不知道是否e是一個函數，我可以用來實現我的目標，就像

function(aa, bb, Overlaps) 

順便說一下，我不喜歡循環。 ^^

Answer 1

你可以試試：

x<-strsplit(infor[,2],";") 
y<-expand.grid(x,x) 
matrix(mapply(function(.x,.y) 
    length(intersect(.x,.y)),y[[1]],y[[2]]), 
    nrow=nrow(infor),dimnames=list(infor[,1],infor[,1])) 
# 1st 2nd 3rd 4th 5th 6th 
#1st 3 1 2 1 1 0 
#2nd 1 4 2 0 2 1 
#3rd 2 2 3 0 1 0 
#4th 1 0 0 3 1 1 
#5th 1 2 1 1 3 1 
#6th 0 1 0 1 1 2