how to get each output of iteration - r

Question

i have to do 1000 iteration for this SIMPLS function to get the value of the coefficient. my problem is how to get the value of the coefficient for each iteration? can I print the output for iteration?
n = 10
k = 20
a = 2
coef = matrix(0,nrow=20, ncol=10)
for (i in 1:1000) {
t[,i] = matrix(rnorm(n%*%a,0,1), ncol=a) # n x a
p[,i] = matrix(rnorm(k%*%a,0,1), ncol=a) # k x a
B[,i] = matrix(rnorm(k,0,0.001), nrow=k, ncol=1) # k x 1
e[,i] = matrix(rcauchy(n,location=0,scale=1), nrow=n, ncol=1)##standard cauchy
x[,i] = t%*%t(p) ## explanatary variable xi
y[,i] = (t%*%(t(p)%*%B)) + e ## response variable yi
simpls <- function(y, x, a) {
n <- nrow(x)
k <- ncol(x)
m <- NCOL(y)
y <- matrix(y)
Ps <- matrix(0, k, a)
Cs <- matrix(0, m, a)
Rs <- matrix(0, k, a)
Ts <- matrix(0, n, a)
mx <- apply(x, 2, mean)
sdx <- apply(x, 2, sd)
x <- sapply(1:k, function(i) (x[,i]-mx[i]))
my <- apply(y, 2, mean)
sdy <- apply(y, 2, sd)
y <- sapply(1:m, function(i) (y[,i]-my[i]))
S <- t(x)%*%y
Snew <- S
for (i in 1:a) {
rs <- svd(Snew)$u[,1,drop=FALSE]
rs <- rs/norm(rs,type="F")
ts <- x%*%rs
ts <- ts/norm(ts,type="F")
ps <- t(x)%*%ts
cs <- t(y)%*%ts
Rs[,i] <- rs
Ts[,i] <- ts
Ps[,i] <- ps
Cs[,i] <- cs
Snew <- Snew-Ps[,1:i]%*%solve(t(Ps[,1:i])%*%Ps[,1:i])%*%t(Ps[,1:i])%*%Snew
}
coef[,i] <- matrix(drop(Rs%*%(solve(t(Ps)%*%Rs)%*%t(Cs))))
yfit <- x%*%coef
orgyfit <- yfit+my
res <- y-yfit
SSE <- sum((y-yfit)^2)
scale <- sqrt(SSE/(n-a))
stdres <- sapply(1:m, function(i) (res[,i]-mean(res[,i]))/sqrt(var(res[,i])))
hatt <- diag(Ts%*%solve(t(Ts)%*%Ts)%*%t(Ts))
result <- list(coef=coef, fit=orgyfit, res=res, SSE=SSE,scale=scale, stdres=stdres, leverage=hatt,Ts=Ts,Rs=Rs,Ps=Ps,Cs=Cs)
}
}
print(coef)

Answer 1

You can just add your coef to a vector for every iteration. I've created an example here:
coef_vector <- NULL
for (i in 1:10) {
loop_coef <- i*2
coef_vector <- c(coef_vector, loop_coef)
}
Result:
> coef_vector
[1] 2 4 6 8 10 12 14 16 18 20
>
Of course, if your coef is more complex than a variable, you can add it to a list instead of a vector.