What i am trying to accomplish in the forloop and am still having difficulty with:
prb[i]=(prb[i-1]*ER)+b[i]
prb[1]=(prb[0]*ER)+b[1]
prb[2]=(prb[1]*ER)+b[2]
and then output prb[1,2,3....] from the left hand side of the equation.
Also, defining SS to reflect prb at the previous time step (i.e. (prb-1))
I have attempted to save the results from my forloop in the empty vectors. However, the values outputted into the vectors are the same values(i.e. from the first iteration) and it doesn't appear to be having the additive affect I am attempting. Somethings appears to be wrong with the logic of my code. I would like for the values from b[i] to be used in b[i+1] for the next run of the forloop. Does anyone have any ideas or solutions to problem? Best!
Matthew
#Parameters
c=0.2
A=5
d=8
d0=5
s=0.5
e=0.1
p=0.6
ER=e/A
#Colonization Equation Probabilities
C2 = c*A*exp(-d*s/d0) #ML to SS
#Empty Vectors
l=vector(mode="numeric", length=100) #open vectors to store the different probability values from the forloop
b=vector(mode="numeric", length=100)
prb=(l*ER) + b #total probability of SS being colonized
#Island States
ML=1
SS=prb
n.I=c(ML, SS)
#Forloop and Conditional Statements
for(i in 2:101) {
(SS[i]=prb[i-1])
(prb[i]=(l[i]*ER)+b[i])
if (SS < 1 ) {
(l[i]=prb[i-1])
} else if(SS < 1){
b[i]=C2
}
}
I copied your code and ran it. My observations are:
First of all, in R, vector indices start from 1 not 0.
So SS=prb[0:1] yields prb[0] which is just 0.
Since to execute the line for populating prb is conditional on SS being >= 1, you can never run this line with this condition. When prb[1] is not 1 or larger, you cannot alter it so there is a vicious circle.
Even if you can run it by satisfying the condition (a different one of course), since for the first iteration prb[0] is non existent so prb[0]*ER just yields numeric(0), which is an empty vector item!
So first you should modify your vector indices, then you should modify your condition so that it allows for a jump start.
Related
Inspired by the leetcode challenge for two sum, I wanted to solve it in R. But while trying to solve it by brute-force I run in to an issue with my for loop.
So the basic idea is that given a vector of integers, which two integers in the vector, sums up to a set target integer.
First I create 10000 integers:
set.seed(1234)
n_numbers <- 10000
nums <- sample(-10^4:10^4, n_numbers, replace = FALSE)
The I do a for loop within a for loop to check every single element against eachother.
# ensure that it is actually solvable
target <- nums[11] + nums[111]
test <- 0
for (i in 1:(length(nums)-1)) {
for (j in 1:(length(nums)-1)) {
j <- j + 1
test <- nums[i] + nums[j]
if (test == target) {
print(i)
print(j)
break
}
}
}
My problem is that it starts wildly printing numbers before ever getting to the right condition of test == target. And I cannot seem to figure out why.
I think there are several issues with your code:
First, you don't have to increase your j manually, you can do this within the for-statement. So if you really want to increase your j by 1 in every step you can just write:
for (j in 2:(length(nums)))
Second, you are breaking only the inner-loop of the for-loop. Look here Breaking out of nested loops in R for further information on that.
Third, there are several entries in nums that gave the "right" result target. Therefore, your if-condition works well and prints all combination of nums[i]+nums[j] that are equal to target.
So far I've tried the following code but it didn't work in R-studio; it just hangs there.
Am I doing something wrong? This is my first real R code project so I'd love suggestions!
new.rref <- function(M,fractions=FALSE)
{
#M is a matrix.
#Require numeric matricies.
if ((!is.matrix(M)) || (!is.numeric(M)))
stop("Sorry pal! Data not a numeric matrix.")
#Specify and differentiate between rows and columns.
r=nrow(M)
c=ncol(M)
#Now establish a continuous loop (*needed help on this one)
#According to the help documents I've read, this has to do with a
#computerized version of the Gaussian Reducing Algorithm
#While 1<r and 1<c, must set first column entries in which
#1:r < 1 equal to zero. This while loop is used to loop the
#algorithm until a specific condition is met -- in this case,
#until elements in the first column to which 1:r < 1
#are set to zero.
while((1<=r) & (1<=c))
new <- M[,1]
new[1:r < y.position] <- 0
# Now here's the fun part :)
#We need to find the maximum leading coefficient that lies
#at or below the current row.
new1 <- which.max(abs(new))
#We will assign these values to the vector "LC"
LC <- col[which]
#Now we need to allow for row exchange!
#Basically tells R that M[c(A,B),] = M[c(B,A),].
if (which > 1) { M[c(1,which),]<-A[c(which,1),] }
#Now we have to allow for the pivot, "sweep", and restoration
#of current row. I totally didn't know how to do this so I
#used and changed some code from different documentations.
#PIVOT (friends reference)
M[1,]<-M[1,]/LC
new2 <-M[1,]
#CLEAN
M <- M - outer(M[,x.position],new2)
#RESTORE
A[1,]<-new2
#Last, but certantly not least, we're going to round the matrix
#off to a certain value. I might have did this wrong.
round(M)
return(M)
print(M)
}
Edit: I added the first line, for some reason it got deleted.
Edit 2: Say you have a matrix M=matrix(c(2,3,4,7), nrow=2, ncol=2, byrow=TRUE); new.rref(M) needs to produce the reduced row echelon form of matrix M. I already did the math; new.rref(M) should be equal to matrix(c(1,0,0,1), nrow=2, ncol=2, byrow=T
I am normally a maple user currently working with R, and I have a problem with correctly indexing variables.
Say I want to define 2 vectors, v1 and v2, and I want to call the nth element in v1. In maple this is easily done:
v[1]:=some vector,
and the nth element is then called by the command
v[1][n].
How can this be done in R? The actual problem is as follows:
I have a sequence M (say of length 10, indexed by k) of simulated negbin variables. For each of these simulated variables I want to construct a vector X of length M[k] with entries given by some formula. So I should end up with 10 different vectors, each of different length. My incorrect code looks like this
sims<-10
M<-rnegbin(sims, eks_2016_kasko*exp(-2.17173), 840.1746)
for(k in 1:sims){
x[k]<-rep(NA,M[k])
X[k]<-rep(NA,M[k])
for(i in 1:M[k]){x[k][i]<-runif(1,min=0,max=1)
if(x[k][i]>=0 & x[i]<=0.1056379){
X[k][i]<-rlnorm(1, 6.228244, 0.3565041)}
else{
X[k][i]<-rlnorm(1, 8.910837, 1.1890874)
}
}
}
The error appears to be that x[k] is not a valid name for a variable. Any way to make this work?
Thanks a lot :)
I've edited your R script slightly to get it working and make it reproducible. To do this I had to assume that eks_2016_kasko was an integer value of 10.
require(MASS)
sims<-10
# Because you R is not zero indexed add one
M<-rnegbin(sims, 10*exp(-2.17173), 840.1746) + 1
# Create a list
x <- list()
X <- list()
for(k in 1:sims){
x[[k]]<-rep(NA,M[k])
X[[k]]<-rep(NA,M[k])
for(i in 1:M[k]){
x[[k]][i]<-runif(1,min=0,max=1)
if(x[[k]][i]>=0 & x[[k]][i]<=0.1056379){
X[[k]][i]<-rlnorm(1, 6.228244, 0.3565041)}
else{
X[[k]][i]<-rlnorm(1, 8.910837, 1.1890874)
}
}
This will work and I think is what you were trying to do, BUT is not great R code. I strongly recommend using the lapply family instead of for loops, learning to use data.table and parallelisation if you need to get things to scale. Additionally if you want to read more about indexing in R and subsetting Hadley Wickham has a comprehensive break down here.
Hope this helps!
Let me start with a few remarks and then show you, how your problem can be solved using R.
In R, there is most of the time no need to use a for loop in order to assign several values to a vector. So, for example, to fill a vector of length 100 with uniformly distributed random variables, you do something like:
set.seed(1234)
x1 <- rep(NA, 100)
for (i in 1:100) {
x1[i] <- runif(1, 0, 1)
}
(set.seed() is used to set the random seed, such that you get the same result each time.) It is much simpler (and also much faster) to do this instead:
x2 <- runif(100, 0, 1)
identical(x1, x2)
## [1] TRUE
As you see, results are identical.
The reason that x[k]<-rep(NA,M[k]) does not work is that indeed x[k] is not a valid variable name in R. [ is used for indexing, so x[k] extracts the element k from a vector x. Since you try to assign a vector of length larger than 1 to a single element, you get an error. What you probably want to use is a list, as you will see in the example below.
So here comes the code that I would use instead of what you proposed in your post. Note that I am not sure that I correctly understood what you intend to do, so I will also describe below what the code does. Let me know if this fits your intentions.
# define M
library(MASS)
eks_2016_kasko <- 486689.1
sims<-10
M<-rnegbin(sims, eks_2016_kasko*exp(-2.17173), 840.1746)
# define the function that calculates X for a single value from M
calculate_X <- function(m) {
x <- runif(m, min=0,max=1)
X <- ifelse(x > 0.1056379, rlnorm(m, 6.228244, 0.3565041),
rlnorm(m, 8.910837, 1.1890874))
}
# apply that function to each element of M
X <- lapply(M, calculate_X)
As you can see, there are no loops in that solution. I'll start to explain at the end:
lapply is used to apply a function (calculate_X) to each element of a list or vector (here it is the vector M). It returns a list. So, you can get, e.g. the third of the vectors with X[[3]] (note that [[ is used to extract elements from a list). And the contents of X[[3]] will be the result of calculate_X(M[3]).
The function calculate_X() does the following: It creates a vector of m uniformly distributed random values (remember that m runs over the elements of M) and stores that in x. Then it creates a vector X that contains log normally distributed random variables. The parameters of the distribution depend on the value x.
I have written the following loop in R. I want the loop to keep running until there is a case where Y is greater than or equal to 3. I then want to display T which is the number of experiments run until this outcome occurs for the first time.
T=0
while(Y<3)
{
X=rbinom(1,5,0.5)
Y=rbinom(1,X,0.5)
Outcome=Y
T=T+1
}
T
I am very new to R and I'm unsure how to change what i've done to achieve what I need.
You can use a do until construction:
while(TRUE){
# Do things
if(Y >= 3) break()
}
You don't need a loop for this. The following uses R's vectorization and is much more efficient.
set.seed(42) #for reproducibility
n <- 1e4 #increase n and rerun if condition is never satisfied
X=rbinom(n,5,0.5)
Y=rbinom(n,X,0.5)
#has condition been satisfied?
any(Y>3)
#TRUE
#first value that satisfies the condition
which.max(Y>3)
#[1] 141
If you don't know how many times you are to perform the loop, while loop is used in that case. It performs the looping until your desired condition is satisfied. You can try the following codes.
T=0 #Index variable
Y=2 #Initial value that starts the while looping
At first while loop inspect this initial Y=2, if it satisfies the condition then the lopping starts until the condition gets dissatisfied.
while(Y<3) #Initialization of your looping
{
X=rbinom(1,5,0.5)
Y=rbinom(1,X,0.5)
T=T+1
}
T
The first time the while checks its condition it does not find Y. Try to initialize Y to something less than 3.
Y <- 0
I am trying to set up a Gibbs sampler in R where I update my value at each step.
I have a function in R that I want to maximise for 2 values; my previous value and a new one.
So I know the maximum outcome from the function applied to both values. But then how do I select the best input without doing it manually? (I need to do a lot of iterations). Here is an idea of the code and the variables:
g0<-function(k){sample(0:1,k,replace=T)}
this is a k dimensional vector with entries 1 or 0 uniformly. Initial starting point for my chain. If i=1 then include the i'th variable in the design matrix.
X1 design matrix
Xg<-function(g){
Xg<-cbind(X1[,1]*g[1],X1[,2]*g[2],X1[,3]*g[3],X1[,4]*g[4],X1[,5]*g[5],X1[,6]*g[6],X1[,7]*g[7])
return(Xg[,which(!apply(Xg,2,FUN = function(x){all(x == 0)}))])
}
Xg0<-Xg(g0)
reduced design matrix for g0
c<-1:100000
mp<-function(g){
mp<-sum((1/(c*(c+1)^-((q+1)/2)))*
(t(Y)%*%Y-(c/(c+1))*t(Y)%*%Xg(g)%*%solve(t(Xg(g))%*%Xg(g))%*%t(Xg(g))%*%Y)^(-27/2))
return(mp)
}
this is my function.
Therefore if I have mp(g) and mp(g*), for 2 inputs g and g*, such that the max is mp(g*) how can I return g*?
Thanks for any help and if you have any queries just ask. sorry about the messy code as well; I have not used this site before.
Like this:
inputs <- list(g, g2)
outputs <- sapply(inputs, mp)
best.input <- inputs[which.max(outputs)]