Question

我有一个从GCTA生成的关系矩阵，我可以使用以下函数导入到R中

var item = [
{name:'test1',variable:20},
{name:'test2',variable:20},
{name:'test3',variable:20},
{name:'test4',variable:20},
{name:'test5',variable:20},
{name:'test6',variable:20},
{name:'test7',variable:20},
{name:'test8',variable:20}]

//removing 1 element from number 2
var removeObj = item.splice(2,1);

console.log(item);

然后列出对角线和对角线。

ReadGRMBin=function(prefix, AllN=F, size=4){
  sum_i=function(i){
    return(sum(1:i))
  }
  BinFileName=paste(prefix,".grm.bin",sep="")
  NFileName=paste(prefix,".grm.N.bin",sep="")
  IDFileName=paste(prefix,".grm.id",sep="")
  id = read.table(IDFileName)
  n=dim(id)[1]
  BinFile=file(BinFileName, "rb");
  grm=readBin(BinFile, n=n*(n+1)/2, what=numeric(0), size=size)
  NFile=file(NFileName, "rb");
  if(AllN==T){
    N=readBin(NFile, n=n*(n+1)/2, what=numeric(0), size=size)
  }
  else N=readBin(NFile, n=1, what=numeric(0), size=size)
  i=sapply(1:n, sum_i)
  return(list(diag=grm[i], off=grm[-i], id=id, N=N))
}

我希望创建一个grm我可以从此反转，理想情况下在输出行，列，值

我已经尝试了以下代码，但它没有以正确的格式读取对角线

$ diag: num [1:850] 0.878 0.815 1.11 1.161 1.062 ...  
$ off : num [1:360825] 0.0181 -0.0304 -0.0663 -0.0211 -0.0583 ...  
$ n   : int 850

而不是将对角线值读为

2 1,3 1,3 2,4 1,2,4等

正在读取对角线的长度

2 1,3 1,4 1,5 1,6 1等

因此得到的矩阵（缩短）最终会像这样结束

  m <- matrix(NA, ncol = length(grm$diag), nrow = length(grm$diag))
    m[lower.tri(m)] <- grm$off
    m[upper.tri(m)] <- t(m)[upper.tri(t(m))]
    diag(m) <- grm$diag
    m
    want=cbind(which(!is.na(m),arr.ind = TRUE),na.omit(as.vector(m)))

当偏好是这个

时

      [,1]         [,2]        [,3]         [,4]        [,5]
[1,]  0.87798703  0.018129893 -0.03044302 -0.066282429 -0.02106927
[2,]  0.01812989  0.814602911  0.07577287 -0.004078172 -0.03182918
[3,] -0.03044302  0.075772874  1.10976517 -0.055698857 -0.03960679
[4,] -0.06628243 -0.004078172 -0.05569886  1.160611629 -0.01021352
[5,] -0.02106927 -0.031829182 -0.03960679 -0.010213521  1.06245303

如果您知道如何修改上述代码以提供所需的格式，那将非常感激。如果可能，最终期望输出将是长格式的矩阵的逆。感谢

      [,1]         [,2]        [,3]         [,4]        [,5]
[1,]  0.87798703   0.018129893  -0.03044302  -0.02106927  -0.04011643
[2,]  0.01812989   0.814602911  -0.06628243  -0.00582625  -0.06237402
[3,] -0.03044302  -0.06628243    1.10976517   0.1315616   -0.1601102
[4,] -0.02106927  -0.00582625    0.1315616    1.160611629 -0.1388046
[5,] -0.04011643  -0.06237402   -0.1601102   -0.1388046    1.06245303

Answer 1

一个简单的解决方案是调整代码以填充下面的上三角形（因为它是应按列顺序填充的上三角形）：

grm = list(
  diag = 1:5 * 11,
  off  = 0:9)

m <- diag(grm$diag)
m[upper.tri(m)] <- grm$off
m[lower.tri(m)] <- t(m)[lower.tri(t(m))]
#      [,1] [,2] [,3] [,4] [,5]
# [1,]   11    0    1    3    6
# [2,]    0   22    2    4    7
# [3,]    1    2   33    5    8
# [4,]    3    4    5   44    9
# [5,]    6    7    8    9   55

使用R从列表构造和反矩阵

1 个答案: