我有一个自然语言处理课程的数据矩阵,我需要Euclidean规范化矩阵,我对R
很新,所以我无法弄清楚我的代码出错了什么,继承人代码:
A=matrix(
c(1,1,1,0,0,0,0,0,0,0,0,0,1,1,1,1,0,0,0,0,0,0,0,0,1,0,0,1,1,0,0,0,0,1,0,0,1,2,0,0,0,0,0,0,1,0,0,0,1,1,0,0,0,0,0,0,0,0,0,0,0,0,1,1,0,0,0,0,0,0,0,0,0,0,1,1,0,0,0,0,0,0,0,0,0,1,1,1,0,0,0,0,0,0,0,0,1,0,1,0,0,0,0,0,0,1,0,0,0,1),
nrow=11,
ncol=10)
norm_vec <- function(x) sqrt(sum(x^2))
B=matrix(
nrow=11,
ncol=10)
for(n in 1:10)
length <- norm_vec(A[,n])
for(j in 1:11)
B[j,n]<- A[j,n]/length
为此,我得到的输出只有最后一列是正确的,其他列是NA
,我不明白为什么会这样。我做错了什么?
答案 0 :(得分:2)
@ Sathish的解决方案可以完全矢量化
A / matrix(sqrt(colSums(A*A)), nrow=11, ncol=10, byrow=TRUE)
# [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
#[1,] 0.5773503 0.0 0.0000000 0.4082483 0.5773503 0.0000000 0.0000000 0.0000000 0.0000000 0.0000000
#[2,] 0.5773503 0.5 0.0000000 0.0000000 0.0000000 0.0000000 0.0000000 0.0000000 0.0000000 0.0000000
#[3,] 0.5773503 0.5 0.5773503 0.0000000 0.0000000 0.0000000 0.0000000 0.0000000 0.0000000 0.0000000
#[4,] 0.0000000 0.5 0.0000000 0.4082483 0.0000000 0.0000000 0.0000000 0.0000000 0.0000000 0.0000000
#[5,] 0.0000000 0.5 0.0000000 0.8164966 0.5773503 0.0000000 0.0000000 0.0000000 0.0000000 0.0000000
#[6,] 0.0000000 0.0 0.5773503 0.0000000 0.5773503 0.0000000 0.0000000 0.0000000 0.0000000 0.0000000
#[7,] 0.0000000 0.0 0.5773503 0.0000000 0.0000000 0.0000000 0.0000000 0.0000000 0.0000000 0.7071068
#[8,] 0.0000000 0.0 0.0000000 0.0000000 0.0000000 0.7071068 0.0000000 0.0000000 0.0000000 0.0000000
#[9,] 0.0000000 0.0 0.0000000 0.0000000 0.0000000 0.7071068 0.7071068 0.5773503 0.7071068 0.0000000
#[10,] 0.0000000 0.0 0.0000000 0.0000000 0.0000000 0.0000000 0.7071068 0.5773503 0.0000000 0.0000000
#[11,] 0.0000000 0.0 0.0000000 0.0000000 0.0000000 0.0000000 0.0000000 0.5773503 0.7071068 0.7071068
答案 1 :(得分:0)
A=matrix(c(1,1,1,0,0,0,0,0,0,0,0,0,1,1,1,1,0,0,0,0,0,0,0,0,1,0,0,1,1,0,0,0,0,1,0,0,1,2,0,0,0,0,0,0,1,0,0,0,1,1,0,0,0,0,0,0,0,0,0,0,0,0,1,1,0,0,0,0,0,0,0,0,0,0,1,1,0,0,0,0,0,0,0,0,0,1,1,1,0,0,0,0,0,0,0,0,1,0,1,0,0,0,0,0,0,1,0,0,0,1), nrow=11, ncol=10)
norm_vec <- function(x){
sqrt(sum(x^2))
}
B=matrix(nrow=11, ncol=10)
Euc.lenA <- c()
for(n in 1:10){
Euc.lenA <- c(Euc.lenA, norm_vec(A[,n]))
}
for(j in 1:10){
B[,j]<- A[,j]/Euc.lenA[j]
}
Euc.lenA
[1] 1.732051 2.000000 1.732051 2.449490 1.732051 1.414214 1.414214 1.732051 1.414214 1.414214