I need to record the maximum value of 5 observations, simulated 10,000 times. This is in order to find the mean of these maximum values.
My current code looks like this:
mc.sim1 = function(){
mu = 20; sigma = 4
(rnorm(5, mu, sigma))
}
m = replicate(10000, mc.sim1())
I'm at a loss to explain how to record the maximum of each iteration, rather than the maximum of the whole simulated set using the 'max' function.
If to follow your code, use dim to understand what m is.
it is a matrix, with a row for each observation of the simulation, and a column for each replication.
a really simple and intuitive loop can look like
recorded_max=numeric(dim(m)[2])
for (i in 1:dim(m)[2])
{
recorded_max[i]=max(m[,i])
}
Just a simple sapply would work
mc.sim1 = function(...){
mu = 20; sigma = 4
max(rnorm(5, mu, sigma))
}
maxdraw <- sapply(seq.int(10000), mc.sim1)
At the end maxdraw will be a vector of length 10,000 with the maximum of the five random normal variables from each iteration.
I would recommend using apply to calculate the maximum value per column (since each column represents an iteration in m):
colMax <- apply(m, 2, max)
Also note #shadow's comment below about the colMaxs function from the matrixStats package.
Related
#Start: Initialize values
#For each block lengths (BlockLengths) I will run 10 estimates (ThetaL). For each estimate, I simulate 50000 observarions (Obs). Each estimate is calculated on the basis of the blocklength.
Index=0 #Initializing Index.
ThetaL=10 #Number of estimations of Theta.
Obs=50000 #Sample size.
Grp=vector(length=7) #Initializing a vector of number of blocks. It is dependent on block lengths (see L:15)
Theta=matrix(data=0,nrow=ThetaL,ncol=7) #Initializing a matrix of the estimates of Thetas. There are 10 for each block length.
BlockLengths<-c(10,25,50,100,125,200,250) #Setting the block lengths
for (r in BlockLengths){
Index=Index+1
Grp[Index]=Obs/r
for (k in 1:ThetaL){
#Start: Constructing the sample
Y1<-matrix(data=0,nrow=Obs,ncol=2)
Y1[1,]<-runif(2,0,1)
Y1[1,1]<--log(-(Y1[1,1])^2 +1)
Y1[1,2]<--log(-(Y1[1,2])^2 +1)
for (i in 2:Obs)
{
Y1[i,1]<-Y1[i-1,2]
Y1[i,2]<-runif(1,0,1)
Y1[i,2]<--log(-(Y1[i,2])^2 +1)
}
X1 <- vector(length=Obs)
for (i in 1:Obs){
X1[i]<-max(Y1[i,])
}
#End: Constructing the sample
K=0 #K will counts number of blocks with at least one exceedance
for (t in 1:Grp[Index]){ #For loop from 1 to number of groups
a=0
for (j in (1+r*(t-1)):(t*r)){ #Loop for the sample within each group
if (X1[j]>quantile(X1,0.99)){ #If a value exceeds high threshold, we add 1 to some variable a
a=a+1
}
}
if(a>=1){ #For the group, if a is larger than 1, we have had a exceedance.
K=K+1 #Counts number of blocks with at least one exceedance.
}
}
N<-sum(X1>=quantile(X1,0.99)) #Summing number of exceedances
Theta[k,Index]<- (1/r) * ((log(1-K/Grp[Index])) / (log(1-N/Obs))) #Estimate
#Theta[k,Index]<-K/N
}
}
I have been running the above code without errors and it took me about 20 minutes, but I want to run the code for larger sample and more repetitions, which makes the run time absurdly large. I tried to only have the necessary part inside the loops to optimize it a little. Is it possible to optimize it even further or should I use another programming language as I've read R is bad for "for loop". Will vectorization help? In case, how can I vectorize the code?
First, you can define BlockLengths before Grp and Theta as both of them depend on it's length:
Index = 0
ThetaL = 2
Obs = 10000
BlockLengths = c(10,25)
Grp = vector(length = length(BlockLengths))
Theta = matrix(data = 0, nrow = ThetaL, ncol = length(BlockLengths))
Obs: I decreased the size of the operation so that I could run it faster. With this specification, your original loop took 24.5 seconds.
Now, for the operation, there where three points where I could improve:
Creation of Y1: the second column can be generated at once, just by creating Obs random numbers with runif(). Then, the first column can be created as a lag of the second column. With only this alteration, the loop ran in 21.5 seconds (12% improvement).
Creation of X1: you can vectorise the max function with apply. This alteration saved further 1.5 seconds (6% improvement).
Calculation of K: you can, for each t, get all the values of X1[(1+r*(t-1)):(t*r)], and run the condition on all of them at once (instead of using the second loop). The any(...) does the same as your a>=1. Furthermore, you can remove the first loop using lapply vectorization function, then sum this boolean vector, yielding the same result as your combination of if(a>=1) and K=K+1. The usage of pipes (|>) is just for better visualization of the order of operations. This by far is the more important alteration, saving more 18.4 seconds (75% improvement).
for (r in BlockLengths){
Index = Index + 1
Grp[Index] = Obs/r
for (k in 1:ThetaL){
Y1 <- matrix(data = 0, nrow = Obs, ncol = 2)
Y1[,2] <- -log(-(runif(Obs))^2 + 1)
Y1[,1] <- c(-log(-(runif(1))^2 + 1), Y1[-Obs,2])
X1 <- apply(Y1, 1, max)
K <- lapply(1:Grp[Index], function(t){any(X1[(1+r*(t-1)):(t*r)] > quantile(X1,0.99))}) |> unlist() |> sum()
N <- sum(X1 >= quantile(X1, 0.99))
Theta[k,Index] <- (1/r) * ((log(1-K/Grp[Index])) / (log(1-N/Obs)))
}
}
Using set.seed() I got the same results as your original loop.
A possible way to improve more is substituting the r and k loops with purrr::map function.
data = read.csv(file= "~/Downloads/data.csv")
temp=(data$temp)
n=75
N=length(temp)
s=sample(1:N, n)
ybar=mean(temp[s])
I want to run the sample 100 times where n is 75. Then calculate average of each sample, and subtract each average from a set number (50).
Maybe a short loop is the way to go. Notice the bracket on the left-side of the equals sign in the last line of code - it's the key to using a loop for your calculation!
# set a seed - always a good idea when using randomness like 'sample()'
set.seed(123)
# pre-allocate an "empty" vector to fill in with results
ybar_vec = vector(length=n)
# do your calculation "n" times
for(i in 1:n) {
s = sample(N)
ybar_vec[i] = 50 - mean(temp[s]) # store i^th calc as i^th element of ybar_vec
}
I need to generate random numbers with rbinom but I need to exclude 0 within the range.
How can I do it?
I would like something similar to:
k <- seq(1, 6, by = 1)
binom_pdf = dbinom(k, 322, 0.1, log = FALSE)
but I need to get all the relative dataset, because if I do the following:
binom_ran = rbinom(100, 322, 0.1)
I get values from 0 to 100.
Is there any way I can get around this?
Thanks
Let`s suppose that we have the fixed parameters:
n: number of generated values
s: the size of the experiment
p: the probability of a success
# Generate initial values
U<-rbinom(n,s,p)
# Number and ubication of zero values
k<-sum(U==0)
which.k<-which(U==0)
# While there is still a zero, . . . generate new numbers
while(k!=0){
U[which.k]<-rbinom(k,s,p)
k<-sum(U==0)
which.k<-which(U==0)
# Print how many zeroes are still there
print(k)
}
# Print U (without zeroes)
U
In addition to the hit and miss approach, if you want to sample from the conditional distribution of a binomial given that the number of successes is at least one, you can compute the conditional distribution then directly sample from it.
It is easy to work out that if X is binomial with parameters p and n, then
P(X = x | X > 0) = P(X = x)/(1-p)
Hence the following function will work:
rcond.binom <- function(k,n,p){
probs <- dbinom(1:n,n,p)/(1-p)
sample(1:n,k,replace = TRUE,prob = probs)
}
If you are going to call the above function numerous times with the same n and p then you can just precompute the vector probs and simply use the last line of the function whenever you need it.
I haven't benchmarked it, but I suspect that the hit-and-miss approach is preferable when k is small, p not too close to 0, and n large, but for larger k larger, p closer to 0, and n smaller then the above might be preferable.
I have two input matrices, dt(10,3) & wt(3,3), that i need to use to find the optimal decision matrix (same dimension), Par(10,3) so as to maximize an objective function. Below R code would give some direction into the problem (used Sample inputs here) -
#Input Matrices
dt <- matrix(runif(300),100,3)
wt <- matrix(c(1,0,0,0,2,0,0,0,1),3,3) #weights
#objective function
Obj <- function(Par) {
P = matrix(Par, nrow = 10, byrow=F) # Reshape
X = t((dt%*%wt)[,1])%*%P[,1]
Y = t((dt%*%wt)[,2])%*%P[,2]
Z = t((dt%*%wt)[,3])%*%P[,3]
as.numeric(X+Y+Z) #maximize
}
Now I am struggling to apply the following constraints to the problem :
1) Matrix, Par can only have binary values (0 or 1)
2) rowSums(Par) = 1 (Basically a row can only have 1 in one of the three columns)
3) colSums(Par[,1]) <= 5, colSums(Par[,2]) <= 6, & colSums(Par[,3]) <= 4
4) X/(X+Y+Z) < 0.35, & Y/(X+Y+Z) < 0.4 (X,Y,Z are defined in the objective function)
I tried coding the constraints in constrOptim, but not sure how to input binary & integer constraints. I am reading up on lpSolve, but not able to figure out. Any help much appreciated. Thanks!
I believe this is indeed a MIP so no issues with convexity. If I am correct the model can look like:
This model can be easily transcribed into R. Note that LP/MIP solvers do not use functions for the objective and constraints (opposed to NLP solvers). In R typically one builds up matrices with the LP coefficients.
Note: I had to make the limits on the column sums much larger (I used 50,60,40).
Based on Erwin's response, I am able to formulate the model using lpSolve in R. However still struggling to add the final constraint to the model (4th constraint in my question above). Here's what I am able to code so far :
#input dimension
r <- 10
c <- 3
#input matrices
dt <- matrix(runif(r*c),r,c)
wt <- matrix(c(1,0,0,0,2,0,0,0,1),3,3) #weights
#column controller
c.limit <- c(60,50,70)
#create structure for lpSolve
ncol <- r*c
lp.create <- make.lp(ncol=ncol)
set.type(lp.create, columns=1:ncol, type = c("binary"))
#create objective values
obj.vals <- as.vector(t(dt%*%wt))
set.objfn(lp.create, obj.vals)
lp.control(lp.create,sense='max')
#Add constraints to ensure sum of parameters for every row (rowSum) <= 1
for (i in 1:r){
add.constraint(lp.create, xt=c(1,1,1),
indices=c(3*i-2,3*i-1,3*i), rhs=1, type="<=")
}
#Add constraints to ensure sum of parameters for every column (colSum) <= column limit (defined above)
for (i in 1:c){
add.constraint(lp.create, xt=rep(1,r),
indices=seq(i,ncol,by=c), rhs=c.limit[i], type="<=")
}
#Add constraints to ensure sum of column objective (t((dt%*%wt)[,i])%*%P[,i) <= limits defined in the problem)
#NOT SURE HOW TO APPLY A CONSTRAINT THAT IS DEPENDENT ON THE OBJECTIVE FUNCTION
solve(lp.create)
get.objective(lp.create) #20
final.par <- matrix(get.variables(lp.create), ncol = c, byrow=T) # Reshape
Any help that can get me to the finish line is much appreciated :)
Thanks
Forgive me if this has been asked before (I feel it must have, but could not find precisely what I am looking for).
Have can I draw one element of a vector of whole numbers (from 1 through, say, 10) using a probability function that specifies different chances of the elements. If I want equal propabilities I use runif() to get a number between 1 and 10:
ceiling(runif(1,1,10))
How do I similarly sample from e.g. the exponential distribution to get a number between 1 and 10 (such that 1 is much more likely than 10), or a logistic probability function (if I want a sigmoid increasing probability from 1 through 10).
The only "solution" I can come up with is first to draw e6 numbers from the say sigmoid distribution and then scale min and max to 1 and 10 - but this looks clumpsy.
UPDATE:
This awkward solution (and I dont feel it very "correct") would go like this
#Draw enough from a distribution, here exponential
x <- rexp(1e3)
#Scale probs to e.g. 1-10
scaler <- function(vector, min, max){
(((vector - min(vector)) * (max - min))/(max(vector) - min(vector))) + min
}
x_scale <- scaler(x,1,10)
#And sample once (and round it)
round(sample(x_scale,1))
Are there not better solutions around ?
I believe sample() is what you are looking for, as #HubertL mentioned in the comments. You can specify an increasing function (e.g. logit()) and pass the vector you want to sample from v as an input. You can then use the output of that function as a vector of probabilities p. See the code below.
logit <- function(x) {
return(exp(x)/(exp(x)+1))
}
v <- c(seq(1,10,1))
p <- logit(seq(1,10,1))
sample(v, 1, prob = p, replace = TRUE)