I am totally new to R. Hopefully you can help. I am trying to simulate from a Hawkes process using R. The main idea is that-first of all I simulated some events from a homogeneous Poisson process. Then each of these events will create their own children using a non homogeneous Poisson process. The code is like as below:
SimulateHawkesprocess<-function(n,tmax,lambda,lambda2){
times<-Simulatehomogeneousprocess(n,lambda)
count<-1
while(count<n){
newevent<-times[count] + Simulateinhomogeneousprocess(lambda2,tmax,lambdamax=NA)
times<-c(times,newevent)
count<-count+1
n<-length(times)
}
return(times)
}
But the r code is producing this infinite loop(probably because of the last line: (n<-length(times))). How can I overcome this problem? How can I put a stopping condition?
This is not a R specific problem. You need to get your algorithm working correctly first. Compare the code you have written against what you want to do. If you need help with the algorithm then tag the question as such. Moreover the function call to Simulateinhomogeneousprocess is very inconsistent. Some insight into that function would help. What is that function returning, a number or a vector?
Within the loop you are increasing the value of n by at least 1 each time so you never reach the end.
newevent<-times[count] + Simulateinhomogeneousprocess(lambda2,tmax,lambdamax=NA)
This creates a non empty variable
times<-c(times,newevent)
Increases the "times" vector by at least 1 (since newevent is non-empty)
count<-count+1
n<-length(times)
You increase the count by 1 but also increase the n value by atleast 1 thus creating a never ending loop. One of these things has to change for the loop to stop.
Related
I'm trying to figure out how to get a for loop setup in R when I want it to run two or more parameters at once. Below I have posted a sample code where I am able to get the code to run and fill a matrix table with two values. In the 2nd line of the for loop I have
R<-ARMA.var(length(x_global_sample),ar=c(tt[i],-.7))
And what I would like to do is replace the -.7 with another tt[i], example below, so that my for loop would run through the values starting at (-1,-1), then it would be as follows (-1,-.99),
(-1,-.98),...,(1,.98),(1,.99),(1,1) where the result matrix would then be populated by the output of Q and sigma.
R<-ARMA.var(length(x_global_sample),ar=c(tt[i],tt[i]))
or something similar to
R<-ARMA.var(length(x_global_sample),ar=c(tt[i],ss[i]))
It may be very possible that this would be better handled by two for loops however I'm not 100% sure on how I would set that up so the first parameter would be fixed and the code would run through the sequence of the second parameter, once that would get finished the first parameter would now increase by one and fix itself at that increase until the second parameter does another run through.
I've posted some sample code down below where the ARMA.var function just comes from the ts.extend package. However, any insight into this would be great.
Thank you
tt<-seq(-1,1,0.01)
Result<-matrix(NA, nrow=length(tt)*length(tt), ncol=2)
for (i in seq_along(tt)){
R<-ARMA.var(length(x_global_sample),ar=c(tt[i],-.7))
Q<-t((y-X%*%beta_est_d))%*%solve(R)%*%(y-X%*%beta_est_d)+
lam*t(beta_est_d)%*%D%*%beta_est_d
RSS<-sum((y-X%*%solve(t(X)%*%solve(R)%*%X+lam*D)%*%t(X)%*%solve(R)%*%y)^2)
Denom<-n-sum(diag(X%*%solve(t(X)%*%solve(R)%*%X+lam*D)%*%t(X)%*%solve(R)))
sigma<-RSS/Denom
Result[i,1]<-Q
Result[i,2]<-sigma
rm(Q)
rm(R)
rm(sigma)
}
Edit: I realize that what I have posted above is quite unclear so to simplify things consider the following code,
x<-seq(1,20,1)
y<-seq(1,20,2)
Result<-matrix(NA, nrow=length(x)*length(y), ncol=2)
for(i in seq_along(x)){
z1<-x[i]+y[i]
z2<-z1+y[i]
Result[i,1]<-z1
Result[i,2]<-z2
}
So the results table would appear as follow as the following rows,
Row1: 1+1=2, 2+1=3
Row2: 1+3=4, 4+3=7
Row3: 1+5=6, 6+5=11
Row4: 1+7=8, 8+7=15
And this pattern would continue with x staying fixed until the last value of y is reached, then x would start at 2 and cycle through the calculations of y to the point where my last row is as,
RowN: 20+19=39, 39+19=58.
So I just want to know if is there a way to do it in one loop or if is it easier to run it as 2 loops.
I hope this is clearer as to what my question was asking, and I realize this is not the optimal way to do this, however for now it is just for testing purposes to see how long my initial process takes so that it can be streamlined down the road.
Thank you
I have a model code written in Matlab, and I am trying to translate the code to R. I am almost done, however, I did not manage to convert some simple codes.
These are below:
Assume that I have a row of cells (lets say 50), and the first 10 cells are saturated to water. The rest are under saturated. The below code finds the last saturated cell in the row.
idx_sat_last = find(Exc(t,:)>0, 1, 'last' );
If a cell is saturated, it creates an excess water, so Excess(t,:) > 0 statement is understandable. However, I do not understand rest of the code.
The 2nd code is below. The story of the code is:
If the cell is saturated it creates an excess, else it is a deficit. I do not understand the "includenan" statement.
InSurf(t+1,j)=min(Excess,Deficit(j),'includenan');
Is there anyone who knows how to translate these codes to R?
Thanks in advance..
Maybe you can try this (given a list Exc)
idx_sat_last <- tail(which(Exc[[t]]>0),1)
I scripted a simple for-loop to iterate over each row of a data set to calculate the distance between two coordinates. The code uses the 'geosphere' package and the 'distm' function which takes two sets of coordinates and returns the distance in meters (which I convert to miles by multiplying by 0.00062137).
Here is my loop:
##For loop to find distance in miles for each coordinate pair
miles <- 0
for (i in i:3303) {
miles[i] <- distm(x = c(clean.zips[i,4], clean.zips[i,3]), y = c(clean.zips[i,7], clean.zips[i,6]))[,1] * 0.00062137
}
However, when I run it I receive an error:
Error: object 'i' not found
The thing is, I've run this code before and it worked. Other times, I get this error. I'm not changing any code, it just seems to randomly work only some of the times. I feel the loop must be constructed correctly if it does what I want on occasion, but why would it only work sometimes?
OK, I'm not certain what justifies the down votes on this, but guess I apologize to whomever thought that necessary.
The issue seems to have just been starting the indexing with an actual numeric value like Zheyuan suggested (i.e. using '1:3303' rather than 'i:3303'). I feel like I've created loops before using 'i in i:xxx' without first defining 'i' but maybe not. Anyway, it's solved and thank you!
So I am taking a course that requires learning R and I am struggling with one of the questions:
In this question, you will practice calling one function from within another function. We will estimate the probability of rolling two sixes by simulating dice throws. (The correct probability to four decimal places is 0.0278, or 1 in 36).
(1) Create a function roll.dice() that takes a number ndice and returns the result of rolling ndice number of dice. These are six-sided dice that can return numbers between 1 and 6. For example roll.dice(ndice=2) might return 4 6. Use the sample() function, paying attention to the replace option.
(2) Now create a function prob.sixes() with parameter nsamples, that first sets j equal to 0, and then calls roll.dice() multiple times (nsample number of times). Every time that roll.dice() returns two sixes, add one to j. Then return the probability of throwing two sixes, which is j divided by nsamples.
I am fine with part one, or at least I think so, so this is what I have
roll.dice<-function(ndice)
{
roll<-sample(1:6,ndice,TRUE)
return(roll)
}
roll.dice(ndice=2)
but I am struggling with part two. This is what I have so far:
prob.sixes<-function(nsamples) {
j<-vector
j<-0
roll.dice(nsamples)
if (roll.dice==6) {
j<-j+1
return(j)
}
}
prob.sixes(nsamples=3)
Sorry for all the text, but can anybody help me?
Your code has a couple of problems that I can see. The first one is the interpretation of the question. The question says:
Now create a function prob.sixes() with parameter nsamples, that first sets j equal to 0, and then calls roll.dice() multiple times (nsample number of times).
Check on your code, are you doing this? Or are you calling roll.dice() a single time? Look for ways to do the same thing (in your case, roll.dice) several times; you may consider the function for. Also, here, you need to store the result of this function on a variable, something like
rolled = roll.dice(2)
Second problem:
Every time that roll.dice() returns two sixes, add one to j.
You are checking if roll.dice==6. But this has two problems. First, roll.dice is a function, not a variable. So it will never be equal to 6. Also, you don't want to check if this variable is equal to six. You should ask whether this variable is equal to a pair of sixes. How can you write "a pair of sixes"?
I wrote a function in R - called "filtre": it takes a dataframe, and for each line it says whether it should go in say bin 1 or 2. At the end, we have two data frames that sum up to the original input, and corresponding respectively to all lines thrown in either bin 1 or 2. These two sets of bin 1 and 2 are referred to as filtre1 and filtre2. For convenience the values of filtre1 and filtre2 are calculated but not returned, because it is an intermediary thing in a bigger process (plus they are quite big data frame). I have the following issue:
(i) When I later on want to use filtre1 (or filtre2), they simply don't show up... like if their value was stuck within the function, and would not be recognised elsewhere - which would oblige me to copy the whole function every time I feel like using it - quite painful and heavy.
I suspect this is a rather simple thing, but I did search on the web and did not find the answer really (I was not sure of best key words). Sorry for any inconvenience.
Thxs / g.
It's pretty hard to know the optimum way of achieve what you want as you do not provide proper example, but I'll give it a try. If your variables filtre1 and filtre2 are defined inside of your function and you do not return them, of course they do not show up on your environment. But you could just return the classification and make filtre1 and filtre2 afterwards:
#example data
df<-data.frame(id=1:20,x=sample(1:20,20,replace=TRUE))
filtre<-function(df){
#example function, this could of course be done by bins<-df$x<10
bins<-numeric(nrow(df))
for(i in 1:nrow(df))
if(df$x<10)
bins[i]<-1
return(bins)
}
bins<-filtre(df)
filtre1<-df[bins==1,]
filtre2<-df[bins==0,]