How to obtain different maturities from a yield curve

Question

I have a code for the construction of the yield curve, but I kinda want to extract the yields for every 0.25 months. I don't know, where to change it in the code. If extract the variables y1, y2 and y3 then the rates I get are only for the integer maturities i.e. 1,2,3,etc. on the curve. Where do I need to change the code that I get the rates at 0.25, 0.5, 0.75 etc.

#INPUTS
#ve c t o r with the swap r a t e s (
swaps <- c(0.002, 0.00225, 0.003, 0.004)

#ve c t o r o f cash f low
vector.cf <- c(0.002, 0.00225, 0.003, 0.004)

#ve c t o r o f ma t u r i t i e s f o r observed swaps
maturities <- c(1,2,3,4)

#ve c t o r o f p r i c e s
prices <- c(rep(1, length(swaps)))

#time v e c t o r s needed to c r e a t e the cash f low mat rix
tj <- seq(1:length(swaps))
tj
#ve c t o r ma t u r i t i e s to LLP
t <- seq(1:length(vector.cf))
t

#func t i on that c r e a t e s the cash f l ows o f swaps
fCashFlowMatrix <- function(s=c(), tt, ttj, n) {
  M <- matrix(, length(ttj), length(tt))
  for(i in 1:length(ttj)) {
    M[i,] <- c(rep(s[i], n[i]-1), 1+s[1],
               rep(0, length(M[i,])-n[i]))
  }
  return(M)
}

#cash f low mat r ix
cf <- fCashFlowMatrix(swaps, t, tj, maturities)
tcf <- t(cf)
cf
#l a s t l i q u i d point
LLP <- 20

#func t i on f o r c a l c u l a t i n g the Heart o f Wilson in one point
fHeart.v.u <- function(v, u, a){
  heart=a*min(v,u)-exp(-a*max(v,u))*0.5*
    (exp(a*min(v,u))-exp(-a*min(v,u)))
  return(heart)
}


#Heart o f Wilson f o r u as ve c t o r
fHeart.v <- function(v, u=c(), a){
  heart.v <- NULL
  for(i in 1:length(u)){
    heart.v[i]=fHeart.v.u(v, u[i], a)
    
  }
  return(heart.v)
}

#func t i on f o r the d e r i v a t i v e o f the Heart o f Wilson
fGuv <- function(v, u,a){
  guve <- NULL
  if(v<=u){
    guv=a-a*(exp(-a*u))*(cosh(a*v))
  }
  else guv=a*exp(-a*v)*(sinh(a*u))
  guv
}

#func t i on f o r the d e r i v a t i v e o f the Heart o f Wilson
#with u as ve c t o r
fGuv.v <- function(v, u=c(),a){
  guv.v<- NULL
  for(i in 1:length(u)){
    guv.v[i]=fGuv(v,u[i], a)
    
  }
  return(guv.v)
}

#f u n c ti o n f o r the Wilson f u n ti o n i n one poi n t
fWilson <- function(v, u, ufr, alpha){
  wilson=exp(-ufr*(v+u))*fHeart.v.u(v,u,alpha)
  return(wilson)
}
#f u n c ti o n f o r the Wilson ma t rix
fWilsonMatrix <- function(tt=c(), ufr, alpha){
  W=matrix(length(tt), length(tt), length(tt))
  for(k in 1:length(tt)){
    Wv=c()
    for(i in 1:length(tt)){
      Wv[i] <- fWilson(tt[k], tt[i], ufr, alpha)
    }
    W[k,] <- Wv
  }
  return(W)
}


#calculus

UFR = log(1+0.042)
alpha= 0.123760
wilson <- fWilsonMatrix(t, UFR, alpha)
wilson

#func t i on to c a l c u l a t e the parameter z e ta
zeta.function <- function(X=matrix(), W=matrix(), pr, ufr, u){
  mu <- exp(-ufr*u)
  zeta <- (solve(X%*%W%*%t(X)))%*%(pr-X%*%mu)
  return(zeta)
}

#parameters
zeta <- zeta.function(cf, wilson, prices, UFR, t)


#funt i on to compute the quas i􀀀cons tant k :
kfunction <- function(a, u=c(), ufr, X=matrix(), z){
  mu <- exp(-ufr*u)
  dmu <- diag(mu)
  Q <- dmu%*%t(X)
  k <- (1+a*u%*%Q%*%z)/((sinh(a*u))%*%Q%*%z)
  return(k)
}

#check k
k <- kfunction(alpha, t, UFR, cf, zeta)


#check Qzeta
mu <- exp(-UFR*t)
dmu <- diag(mu)
Q <- dmu%*%t(cf)
Q%*%zeta

#func t i on f o r the conve rgenc e point
cpt.function <- function(llp){
  T <- max(llp+40, 60)
  return(T)
}
T <- cpt.function(LLP)

#func t i on f o r the conve rgenc e pe r i od
cpd.function <- function(llp){
  S <- max(40,60-llp)
  return(S)
}
S <- cpd.function(LLP)
  

#func t i on f o r the upper forward i n t e n s i t y
f.function <- function(w, a, K, v){
  f=(w+(a/1-K*exp(a*v)))
  return(f)
}
#check
g.alpha.1 <- abs(f.function(UFR, alpha, k, T)-UFR)

#func t i on g . alpha
g.alpha.function <- function(a){
  wilson <- fWilsonMatrix(t, UFR, a)
  zeta <- zeta.function(cf, wilson, prices, UFR, t)
  k <- kfunction(a, t, UFR, cf, zeta)
  sg <- a/(abs(1-k*exp(a*T)))
  sg
}
#check
g.alpha.2 <- g.alpha.function(alpha)

#e u r i s t i c s o l u t i o n
g.alpha.function(0.05)
a.lower = 0.05
while (g.alpha.function(a.lower)>=0.0001){
  a.lower=a.lower+0.000001
}

#opt imi z ed alpha
alpha.opt <- a.lower
alpha.opt

#p r i c e func t i on t given time v
fPresentValue <- function(ufr, v,u,q,z,alp) {
  pv <- exp(-ufr*v)*(1+fHeart.v(v,u,alp)%*%q%*%z)
  return(pv)
}

#spot i n t e n s i t y func t i on
fYieldIntensity <- function(ufr, v,u,q,z,alp){
  yi = (-log(fPresentValue(ufr,v,u,q,z,alp)))/v
  return(yi)
}
#y i e l d i n t e n s i t y at time 0
one.vec <- rep(1, length(zeta))
y.zero <- UFR-alpha*one.vec%*%Q%*%zeta+alpha*(exp(-alpha*t)%*%Q%*%zeta)


#y i e l d i n t e n s i t y
x1 <- c(0, seq(from=1, to=30, by=1))
x1
y1 <- c()
y1[1] <- y.zero*100
for(i in seq(from=1, to=30, by=1)) {
  y1[i+1] <- c(fYieldIntensity(UFR, x1[i+1], t,Q, zeta, alpha)*100)
}
y1

#annual i z ed y i e l d r a t e func t i on
fAnnualRate <- function(ufr,v,u,q,z,alp){
  ar = (fPresentValue(ufr, v,u,q,z,alp))^(-1/v)-1
  return(ar)
}

#annual r a t e at time 0
rate.zero <- exp(y.zero)-1

#annual r a t e s
y2 <- c()
y2[1] <- rate.zero*100
for(i in seq(from=1, to=30, by=1)){
  y2[i+1] <- c(fAnnualRate(UFR, x1[i+1],t,Q,zeta, alpha)*100)
}

#forward i n t e n s i t y func t i on
fForwardIntensity <- function(ufr,v,u,q,z,alp){
  fif=ufr-((fGuv.v(v,u,alp)%*%q%*%z)/(1+fHeart.v(v,u,alp)%*%q%*%z))
  return(fif)
}


#forward i n t e n s i t y
y3 <- c()
y3[1]<- y.zero*100
for(i in seq(from=1, to=30, by=0.1)){
  y3[i+1] <- c(fForwardIntensity(UFR, x1[i+1],t,Q,zeta,alpha)*100)
}
#pl o t
plot(x1, y3, type="l", pch=19,
     xlim=c(0,30), ylim=c(-0.22,5), xlab="Maturity",
     ylab="Yields(%)", col="red")
lines(x1, y2, col="green")
lines(x1, y1, col="blue")
legend("bottomright", c("forward intensity", "annual rate",
                        "yield to maturity"), bty = "n", lty =c(1,1,1),
       col=c("red", "green", "blue"))
grid()```

Answer 1

Not being able to follow the logic of you code up above, but noticing you have the x and y values for the yield curves, a easy solution is to interpolate the desired values by either linear interpolation or a cubic spline approach.
For example the annual rate (green curve)

Using linear interpolation at Maturity=10.5:

approx(x1, y2, xout=10.1)
$x
[1] 10.5

$y
[1] 1.375156

By using a cubic spline:

spline(x1, y2, xout=10.5)
$x
[1] 10.5

$y
[1] 1.37635

Both approaches provide similar values, from the plots, the splines might be slightly more accurate towards the center of the range.