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I am trying to exercise a simulation of Sierpinski triangle in R with affine transformation and Iterated Function System (IFS). And hopefully, I can further exercise how the simulation of Barnsley's fern can also be done. For those who know Chinese, this video is my starting point of this exercise.
Here is a short introduction of the simulation process:
Create an equilateral triangle, name the vertices A, B, C
Create a random initial point lying inside the triangle ABC
Sample A, B, C with equal chances
If the outcome is A, then move the initial point to the midpoint of A and itself
Repeat step 3, and move the last point to the midpoint of the outcome point and itself.
By doing this repeatedly, we should see the path of the points looks like a Sierpinski triangle.
I wonder how the assignment of variable works inside a self-defined function. I would like to create an object (a matrix or a dataframe) to store the path of simulated points and keep updating the object to keep track of how the points move.
the following is my current codes:
# create the triangle
triangle <- matrix(c(A = c(-1,0),
B = c(1, 0),
C = c(0, sqrt(3))),
byrow = TRUE, nrow = 3, ncol = 2)
colnames(triangle) <- c("X", "Y") # axis name
rownames(triangle) <- c("A", "B", "C")
# sample an initial point inside the triangle ABC
sampleInit <- function(){
X <- runif(1, min = -1, max = 1)
Y <- runif(1, min = 0, max = sqrt(3))
if( (Y >= 0) && (Y <= (sqrt(3)*X + sqrt(3))) && (Y <= -sqrt(3)*X+sqrt(3)) ){
return(cbind(X, Y))
} else {
sampleInit()
}
}
### graph: plot the triangle and the initial point together
graphics.off()
plot(triangle, xlim = c(-1, 1), ylim = c(0, sqrt(3)))
par(new = TRUE)
plot(sampleInit(), xlim = c(-1, 1), ylim = c(0, sqrt(3)), col = "red")
### a three-sided dice: determine the direction to move along
diceRoll <- function(){
return(sample(c("A", "B", "C"), size = 1, prob = c(1/3, 1/3, 1/3)))
}
## path
stepTrace <- as.data.frame(sampleInit())
move <- function(diceOutCome, stepTrace){
lastStep <- tail(stepTrace, 1)
if(diceOutCome == "A"){
X <- (-1 + lastStep[,1])/2
Y <- (0 + lastStep[,2])/2
} else if(diceOutCome == "B"){
X <- (1 + lastStep[,1])/2
Y <- (0 + lastStep[,2])/2
} else if(diceOutCome == "C"){
X <- (0 + lastStep[,1])/2
Y <- (sqrt(3) + lastStep[,2])/2
}
lastStep <- cbind(X, Y)
stepTrace <- rbind(stepTrace, lastStep)
}
move(diceRoll(), stepTrace)
View(stepTrace)
Sorry for the long story and not jumping to the key question directly. My question is that stepTrace (the object I would like to store the path) didn't get updated as I execute the last two lines.
What I imagined was the assignment process in move() updates the dataframe stepTrace, however it turns out it doesn't. I check my code in the debugger, and found out that stepTrace did get updated inside the function call, but it didn't pass the new assigned value outside the function call. That's why I would like to ask how does the assignment process works in R. What is the difference between the this kind of process and other general purpose languages such as Java? (What I imagined to do this exercise in Java would not encounter this kind of assignment issue. Correct me if I am wrong since I am still new to Java)
Similar problems bother me when I tried to assign variables inside a loop. I know there is a base function assign that helps to resolve is issue, but I just don't know what is the mechanism behind it.
I tried to google my question, but I am not sure which keyword I should use, and I didn't find direct answers to my question. Any comment, keyword or external resource to the documentation is appreciated!
In short, your move function does what you want, but it is not advisable to write it like that. In its current form, stepTrace is updated in the function's local environment, but not in the global environment, where your stepTrace lives. They are not the same stepTrace. To fix it, you can run stepTrace <- move(diceRoll(), stepTrace), but beware of the second circle. For a cleaner approach, remove the last stepTrace assignment from move.
From ?return: If the end of a function is reached without calling return, the value of the last evaluated expression is returned.
Consider the following examples:
x <- 5
a <- b <- c <- d <- 1
f1 <- function(x) x + 1
f2 <- function(x) return(x + 1)
f3 <- function(x) x <- x + 1
f4 <- function(x) x <<- x + 1
f1(1)
f2(1)
f3(1) # your problem
f4(1) # x gets replaced with x in f4, 2 in global environment.
a <- b <- c <- d <- 1
a <- f1(1)
b <- f2(1)
c <- f3(1)
d <- f4(1)
f3 and f4 are generally considered bad practice because of side effects, i.e. they (can) modify a non-local variable, f2 might trigger a discussion. For f3, see the result of
c(f3(1))
#> [1] 2
Given our experiment of calling f3(1) by itself, we'd expect a vector of length 0 (?). Consider removing any assignment as the last operation within your functions, and avoid naming your function arguments the same as the objects you intend to change.
#DonaldSeinen explained how to fix your code in his answer. I'll try to point you to documentation for more details.
First, you don't need to go to external documentation. An Introduction to R and The R Language Definition manuals are included in R distributions. The Introduction describes what's going on in lots of detail in section 10.7, "Scope". There's a different description in the Language Definition in section 3.5, "Scope of Variables".
Some people find the language in those manuals to be too technical. An easier to read external reference that gets it right is Wickham's Advanced R, readable online at https://adv-r.hadley.nz/. Scoping is discussed in chapters 6 and 7, especially sections 6.4 and 7.2.
I am trying to write my first function in R to calculate emittance using Plank's function for different temperatures. I can do it manually as below for temperatures from 200 to 310 K.
pi <- 3.141593
h <- 6.626068963e-34
c <- 2.99792458e+8
lambda <- 4 * 1e-6
k <- 1.38e-23
t <- c (200:310)
a <- (2*pi*(c^2)*h)/(lambda^5)
b <- exp((h*c)/(lambda*k*t))
B <- a * (1/(b-1))
Where B is the vector of values I want.
Now here is an effort to write a function in R:
P_function <- function(t, pi = 3.141593, h = 6.626068963e-34, c = 2.99792458e+8,
lambda = 4 * 1e-6, k = 1.38e-2) {
((2*pi*(c^2)*h)/(lambda^5)) *((1/(exp((h*c)/(lambda*k*t))-1)))
}
Now for different values of t (200-300K), how do I implement this function?
Couple of problems. First, pi is already a defined constant at better precision than you are using.
> rm(pi) # remove your copy
> pi
[1] 3.141593 # default for console printing is only 8 digits
> print(pi, digits=18)
[1] 3.14159265358979312 # but there is more "depth" to be had
Second, it makes no sense to put scientific constants in the parameter list. Since they're constant they can be defined in the body. Parameter lists are for items that might vary from situation to situation.
newPfun <- function(t) { h <- 6.626068963e-34
c <- 2.99792458e+8
lambda <- 4 * 1e-6
k <- 1.38e-23
a <- (2*pi*(c^2)*h)/(lambda^5) #pi is already defined
b <- exp((h*c)/(lambda*k*t))
B <- a * (1/(b-1))
return(B) }
This is just your original code "packaged" to accept a vector of temperatures. (And I'm pretty sure that's not the right spelling the scientist's name.)
Not sure where your second function is flawed. Perhaps a mismatched parenthesis. After trying to duplicate the results with a single expression and failing multiple times, I'm now wondering if it's really a problem with numerical overflow (or underflow).
#========
#DATABASE
#========
database <- matrix(c(51,43,-22,-92,28,-21,68,22,9,-20,-30,-1,-10,10,-10,-5,10,-2,30,-3,-5),ncol=3,byrow=TRUE)
colnames(database ) <- c("A","B","C")
database <- as.data.frame(database )
x<-1
y<-1
z<-1
database$RES<-c(1,0,0,0,1,0,1)
database$SCORE<- database$A*x+database$B*y+database$C*z
database$PREV<- ifelse(database$SCORE>1,1,0)
#========
#TARGET
#========
t<-table(database$RES, database$PREV)
P<-(t[1]+t[4])/nrow(database)
This is an example of my database (60k rows), I want to find values for x y z (in the code I put "1" just for convenience to run the script but I want to find them!) to have maximum value of P. The target P must be 1 or closed to 1.
I didnt find what I'm looking for in thread with similiar title.
In excel is pretty simple but can't find more than 1 parameter.
Thanx in advance.
I'm not satisfied with this answer, but maybe this is something that can at least get you started.
The optim() function finds the optimum set of answers for the problem you're trying to solve, but it looks to me, at least with the toy data, that it finds itself into a local maxima. You'd have to run it several times to find the best parameters, for me it occurs when P = 0.8571429, and even then the x, y, z values can vary quite significantly, which would indicate that there are several equally optimal solutions for this particular data.
database <- matrix(c(51,43,-22,-92,28,-21,68,22,9,-20,-30,-1,-10,10,-10,-5,10,-2,30,-3,-5),ncol=3,byrow=TRUE)
colnames(database ) <- c("A","B","C")
database <- as.data.frame(database )
database$RES <- c(1,0,0,0,1,0,1)
find_best <- function(data, x) {
SCORE <- data$A*x[1]+data$B*x[2]+data$C*x[3]
PREV <- ifelse(SCORE>1,1,0)
t <- table(data$RES, PREV)
P <- (t[1]+t[4])/nrow(data)
P
}
result <- optim(c(1, 1, 1), find_best, data = database, method = "SANN", control = list(fnscale = -1))
result$value
[1] 0.8571429 # The P value
result$par
[1] 2.396844 -4.460343 -7.137460 # These are your sought after x, y, z parameters.
I have a function called "p". In this function, there is an object called "marginal" which I need to make recognizable and retrievable outside this function when called. Right now, when I call the object "marginal" outside the function, I get an wrror message that:
object 'marginal' not found
Question: How can I make "marginal" recognizable outside the function?
p <- function(t, N1, N2=NULL, delta) {
efN = ifelse(is.null(N2), N1, N1*N2/(N1+N2))
df = ifelse(is.null(N2), N1 - 1, N1 + N2 - 2)
prior <- function(delta) dnorm(delta, 0, 1)
likelihood <- function(delta) dt(t, df, delta*sqrt(efN) )
marginal <- integrate(function(x) prior(x)*likelihood(x), -Inf, Inf)[[1]]
post <- function(x) prior(x)*likelihood(x) / marginal
return(post(delta))
list(marginal) ## What to use instead of list to object "marginal" recognizable
# outside the function?
}
marginal
object 'marginal' not found
The best solution for you would be to refactor your current function, possibly into smaller functions, in such a way that you can easily get the value of marginal separately. Then, use it to complete your current calculation.
That being said, if you want a quick fix you could resort to using the parent scope assignment operator <<-, e.g.
marginal <- NULL
p <- function(t, N1, N2=NULL, delta) {
efN = ifelse(is.null(N2), N1, N1*N2/(N1+N2))
df = ifelse(is.null(N2), N1 - 1, N1 + N2 - 2)
prior <- function(delta) dnorm(delta, 0, 1)
likelihood <- function(delta) dt(t, df, delta*sqrt(efN) )
# note the assignment being done here carefully
marginal <<- integrate(function(x) prior(x)*likelihood(x), -Inf, Inf)[[1]]
post <- function(x) prior(x)*likelihood(x) / marginal
return(post(delta))
}
# now marginal contain the value assigned in the function call above
However, it is usually not recommended to take such an approach. I only offer it as a quick fix, with the strong suggestion that you rethink your code design.
To access multiple values from inside a function, make sure all the values you want to access get returned by putting them in a list. Then return the list, assign the results of the function call, and access the values you want:
multi_return = function() {
x = 3
y = 4
res = list(x = x, y = y)
return(res)
}
results = multi_return()
y = results$y
y
# Output:
## [1] 4
This kind of approach is generally preferred over having functions make global assignments, as then calling a function can have unpredictable side effects.
I am trying to split my data set using two parameters, the fraction of missing values and "maf", and store the sub-data sets in a list. Here is what I have done (it's not working). Any help will be appreciated,
Thanks.
library(BLR)
library(missForest)
data(wheat)
X2<- prodNA(X, 0.4) ### creating missing values
dim(X2)
fd<-t(X2)
MAF<-function(geno){ ## markers are in the rows
geno[(geno!=0) & (geno!=1) & (geno!=-1)] <- NA
geno <- as.matrix(geno)
## calc_Freq for alleles
n0 <- apply(geno==0,1,sum,na.rm=T)
n1 <- apply(geno==1,1,sum,na.rm=T)
n2 <- apply(geno==-1,1,sum,na.rm=T)
n <- n0 + n1 + n2
## calculate allele frequencies
p <- ((2*n0)+n1)/(2*n)
q <- 1 - p
maf <- pmin(p, q)
maf}
frac.missing <- apply(fd,1,function(z){length(which(is.na(z)))/length(z)})
maf<-MAF(fd)
lst<-matrix()
for (i in seq(0.2,0.7,by =0.2)){
for (j in seq(0,0.2,by =0.005)){
lst=fd[(maf>j)|(frac.missing < i),]
}}
It sounds like you want the results that the split function provides.
If you have a vector, "frac.missing" and "maf" is defined on the basis of values in "fd" (and has the same length as the number of rows in fd"), then this would provide the split you are looking for:
spl.fd <- split(fd, list(maf, frac.missing) )
If you want to "group" the fd values basesd on of maf(fd) and frac.missing within the bands specified by your for-loop, then the same split-construct may do what your current code is failing to accomplish:
lst <- split( fd, list(cut(maf(fd), breaks = seq(0,0.2,by =0.005) ,
include.lowest=TRUE),
cut(frac.missing, breaks = seq(0.2,0.7,by =0.2),
right=TRUE,include.lowest=TRUE)
)
)
The right argument accomodates the desire to have the splits based on a "<" operator whereas the default operation of cut presumes a ">" comparison against the 'breaks'. The other function that provides similar facility is by.
the below codes give me exactly what i need:
Y<-t(GBS.binary)
nn<-colnames(Y)
fd<-Y
maf<-as.matrix(MAF(Y))
dff<-cbind(frac.missing,maf,Y)
colnames(dff)<-c("fm","maf",nn)
dff<-as.data.frame(dff)
for (i in seq(0.1,0.6,by=0.1)) {
for (j in seq(0,0.2,by=0.005)){
assign(paste("fm_",i,"maf_",j,sep=""),
(subset(dff, maf>j & fm <i))[,-c(1,2)])
} }