Question

汇总表

structure(list(X1 = c(1L, 2L, 3L, 4L, 2L, 5L), X2 = c(2L, 3L, 
4L, 5L, 3L, 6L), X3 = c(3L, 4L, 4L, 5L, 3L, 2L), X4 = c(2L, 4L, 
6L, 5L, 3L, 8L), X5 = c(1L, 3L, 2L, 4L, 6L, 4L)), .Names = c("X1", 
"X2", "X3", "X4", "X5"), class = "data.frame", row.names = c(NA, 
-6L))

我想创建一个 5 x 5 的距离矩阵, 显示匹配率和所有列之间的行数总数。例如, X4 和 X3 之间的距离应该是 0.5, 因为两列的匹配率是 6 次中的 3 次。

我尝试了使用包件“ 代理” 中的 dist( 测试, 方法=“ 简单匹配” ) < proxy', 但这种方法只对二进制数据有效。

Answer 1

使用 outer (再次:-)

my.dist <- function(x) {
 n <- nrow(x)
 d <- outer(seq.int(ncol(x)), seq.int(ncol(x)),
            Vectorize(function(i,j)sum(x[[i]] == x[[j]]) / n))
 rownames(d) <- names(x)
 colnames(d) <- names(x)
 return(d)
}

my.dist(x)
#           X1        X2  X3  X4        X5
# X1 1.0000000 0.0000000 0.0 0.0 0.3333333
# X2 0.0000000 1.0000000 0.5 0.5 0.1666667
# X3 0.0000000 0.5000000 1.0 0.5 0.0000000
# X4 0.0000000 0.5000000 0.5 1.0 0.0000000
# X5 0.3333333 0.1666667 0.0 0.0 1.0000000

Answer 2

在这里试一试(dt is your 矩阵) :

library(reshape)
df = expand.grid(names(dt),names(dt))
df$val=apply(df,1,function(x) mean(dt[x[1]]==dt[x[2]]))
cast(df,Var2~Var1)

Answer 3

这里 s 这里的解决方案比其他两个更快, 虽然略微丑一些。我想, 速度缓冲来自没有使用 mode () , 因为与 sum () 相比速度会慢一些, 并且只计算输出矩阵的一半, 然后手工填充下三角形。函数目前在对角线上留下 NA , 但是您可以很容易地将那些设置为一个, 以便将其它答案完全匹配为 diag(out) & lt; - 1 < /code > 。



FUN <- function(m) {
  #compute all the combinations of columns pairs
  combos <- t(combn(ncol(m),2))
  #compute the similarity index based on the criteria defined
  sim <- apply(combos, 1, function(x) sum(m[, x[1]] - m[, x[2]] == 0) / nrow(m))
  combos <- cbind(combos, sim)
  #dimensions of output matrix
  out <- matrix(NA, ncol = ncol(m), nrow = ncol(m))

  for (i in 1:nrow(combos)){
    #upper tri
    out[combos[i, 1], combos[i, 2]] <- combos[i,3]
    #lower tri
    out[combos[i, 2], combos[i, 1]] <- combos[i,3]
  }
  return(out)
}


我接受了另外两个答案, 把它们变成功能, 并做了一些基准:

library(rbenchmark)
benchmark(chase(m), flodel(m), blindJessie(m), 
          replications = 1000,
          order = "elapsed", 
          columns = c("test", "elapsed", "relative"))
#-----
       test elapsed relative
1  chase(m)   1.217 1.000000
2 flodel(m)   1.306 1.073131
3 blindJessie(m)  17.691 14.548520

Answer 4

I have got the answer as follows: 1st I have made some modifications on the row data as:

X1 = c(1L, 2L, 3L, 4L, 2L, 5L)
X2 = c(2L, 3L, 4L, 5L, 3L, 6L)
X3 = c(3L, 4L, 4L, 5L, 3L, 2L)
X4 = c(2L, 4L, 6L, 5L, 3L, 8L)
X5 = c(1L, 3L, 2L, 4L, 6L, 4L)
matrix_cor=rbind(x1,x2,x3,x4,x5)
matrix_cor

   [,1] [,2] [,3] [,4] [,5] [,6]
X1    1    2    3    4    2    5
X2    2    3    4    5    3    6
X3    3    4    4    5    3    2
X4    2    4    6    5    3    8
X5    1    3    2    4    6    4

然后:

dist(matrix_cor)

     X1       X2       X3       X4
X2 2.449490                           
X3 4.472136 4.242641                  
X4 5.000000 3.000000 6.403124         
X5 4.358899 4.358899 4.795832 6.633250

Answer 5

感谢各位的建议。基于您的回答,我制定了三行解决方案(“测试”是数据集的名称 ) 。

require(proxy)
ff <- function(x,y) sum(x == y) / NROW(x)
dist(t(test), ff, upper=TRUE)

以下是输出 :

          X1        X2        X3        X4        X5
X1           0.0000000 0.0000000 0.0000000 0.3333333
X2 0.0000000           0.5000000 0.5000000 0.1666667
X3 0.0000000 0.5000000           0.5000000 0.0000000
X4 0.0000000 0.5000000 0.5000000           0.0000000
X5 0.3333333 0.1666667 0.0000000 0.0000000

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