距离矩阵中的R

Question

数据帧对我有下面的距离矩阵在R.

   s1   s2   s3   s4   s5   s6   s7   s8   s9 
s2 624667.8824                           
s3 618711.2948 526120.6529                       
s4 1023257.9362 1006497.8847 1025400.5256                    
s5 628679.9303 585435.1935 559319.5066 1031703.3141                 
s6 1023252.3053 1006489.4853 1025393.0225  156.4817 1031695.9148              
s7 1023263.1482 1006500.0433 1025404.4117  152.3829 1031707.4551  131.4696          
s8 619143.6849 557422.9677 513802.7576 1026714.3103 472012.4235 1026706.2563 1026718.1619       
s9 1023282.3175 1006518.0410 1025422.0552  196.2825 1031727.5514  158.4078  159.3760 1026737.7122    
s10 610640.0380 532209.9671 519005.5448 1019135.5176 561862.4551 1019128.4830 1019139.1512 516171.7835 1019158.3403 

如何可以转换上述矩阵成数据帧，如下（与行表头）以最有效的方式。

s1,s2 624667.8824 
s1,s3 618711.2948 
s1,s4 1023257.9362 
s1,s5 628679.9303 
. 
. 
. 
s9,s10 1019158.3403 

Answer 1

library(reshape2) 
mdat <- matrix(c(1,2,3, 11,12,13), nrow = 2, ncol = 3, byrow = TRUE, 
       dimnames = list(c("row1", "row2"), 
           c("C.1", "C.2", "C.3"))) 
melt(mdat) 

输出 -

> melt(mdat) 
    Var1 Var2 value 
1 row1 C.1  1 
2 row2 C.1 11 
3 row1 C.2  2 
4 row2 C.2 12 
5 row1 C.3  3 
6 row2 C.3 13 

Answer 2

呼唤你的距离矩阵d：

f <- function(i) { 
    c1 <- paste(colnames(d)[i],rownames(d)[i:nrow(d)],sep=",") 
    c2 <- d[i:nrow(d),i] 
    return(cbind(c1,c2)) 
} 
result <- do.call(rbind,lapply(1:9,f)) 
result <- data.frame(result) 
head(result) 
#  c1   c2 
# 1 s1,s2 624667.8824 
# 2 s1,s3 618711.2948 
# 3 s1,s4 1023257.9362 
# 4 s1,s5 628679.9303 
# 5 s1,s6 1023252.3053 
# 6 s1,s7 1023263.1482 

如果你愿意，你可以用匿名把所有这一切到一个语句功能：

result=data.frame(do.call(rbind, lapply(1:9,function(i){ 
     cbind(paste0(colnames(d)[i],",",rownames(d)[i:nrow(d)]), 
       d[i:nrow(d),i])})))