I was wondering how I can modify the code below to repeat the function up to 7 times. I can not use replicate(7, func(f)) since I need a function which gives me some option to choose the number of repetitions. I mean, a function which asks me whether I want to continue or not.
suppose
speed<-cars$speed
dist<-cars$dist
level<-c(1:50)
f<-data.frame(speed,dist,level)
plot(speed, dist, main="Milage vs. Car Weight",
xlab="Weight", ylab="Mileage", pch=18, col="blue")
text(speed, dist, row.names(f), cex=0.6, pos=4, col="red")
This is my function
func = function(A){
id.co1 <- identify(f$speed, f$dist,labels=row.names(f), n = 2, pos = TRUE,plot = TRUE)
xy <- f[c(id.co1$ind[1],id.co1$ind[2]),]
lines(xy, col="red", lwd=5)
lm(dist~speed, xy)
abline(coef(lm(dist~speed, xy)),col="blue")
x.co1 <- f$speed[id.co1$ind[1]:id.co1$ind[2]]
y.co1 <- f$dist[id.co1$ind[1]:id.co1$ind[2]]
m.co1 <- lm(y.co1 ~ x.co1)
}
if i understand correctly, you want to specify how often to repeat execution of a function interactively, not programmatically. You can do this with readline:
I need a function which gives me some option to choose the number of repetitions
# some function
funcA <- function(){
cat("i am funcA\n")
}
# some function that interactively repeats funcA a specified amount of times
doNTimesA <- function() {
Ntimes <- readline("number of repeats: ")
for (i in 1:Ntimes) funcA()
}
doNTimesA()
I mean, a function which asks me whether I want to continue or not
funcB <- function(){
while (TRUE) {
cat("i am funcB\n")
continue <- readline("again? y/n: ")
if (tolower(continue)=="n") break
}
cat("funcB is done")
}
funcB()
edit: for your specific case, you can wrap your function declaration in a while loop that asks you whether you want to continue, as in my example funcB above. updated where it also stores its output:
func <- function(){
#initiate an iteration counter and an outputlist
counter <- 1
output <- list()
#start an infinite loop
while (TRUE) {
#do your thing
id.co1 <- identify(f$speed, f$dist,labels=row.names(f), n = 2, pos = TRUE,plot = TRUE)
xy <- f[c(id.co1$ind[1],id.co1$ind[2]),]
lines(xy, col="red", lwd=5)
lm(dist~speed, xy)
abline(coef(lm(dist~speed, xy)),col="blue")
x.co1 <- f$speed[id.co1$ind[1]:id.co1$ind[2]]
y.co1 <- f$dist[id.co1$ind[1]:id.co1$ind[2]]
m.co1 <- lm(y.co1 ~ x.co1)
#store the result at counter index in the outputlist
output[[counter]] <- list(xco =x.co1, yco=y.co1, lm =m.co1)
counter <- counter+1
#prompt for next iteration
continue <- readline("again? y/n: ")
#break the loop if the answer is 'n' or 'N'
if (tolower(continue)=="n") break
}
#return your output
output
}
What happens now is that after every iteration, the function asks if you want to rerun the function: continue <- readline("again? y/n: ") and checks whether you have answered N or n. You can add more checks on the answer if you like; if you answer anything but N or n now, the loop will run again.
If you run all <- func(), after you're done you can access each iterations' result using all[[1]], all[[2]], etc.
Please note that it's generally frowned upon to manipulate objects outside your function environment, so it would be cleaner to pull the initial plot generation into your function.
Related
I am trying to learn how to implement control structures such as FOR and while loops.
I created a function the simulates betting in a famous Brazilian lotto.
In the lotto, a player bets on 6 unique integers from a 1:60 vector (called your_bet).
The function samples 6 values from the 1 to 60 universe ("result") and tests how many values in result match your_bet, printing out:
your_bet
result
total points (out of 6 possible)
one of three possible commentaries on the result of the bet.
Code as follows:
```
LOTTO<-function(your_bet=sample(1:60, size=6, replace=FALSE)){
result<-sample(1:60, size=6, replace=FALSE)
logical_vector<-(your_bet %in% result)
total_points<-sum(as.integer(logical_vector))
print(paste(c("Your bet:", as.character(your_bet))), collapse="")
print(paste(c("Result", as.character(result))), collapse="")
print(paste(c("Total points", as.character(total_points))), collapse="")
if (total_points==6)
print("You are a millonaire")
else if (total_points==5)
print("5 points, you are rich!")
else print("good luck next time")
}
```
I then tried to implement a loop that would make the function go over and over again in a loop until total points>= a given target (here as target_points), modifying the function as bellow.
```
LOTTO<-function(your_bet=sample(1:60, size=6, replace=FALSE), stubborn_until_x_points=FALSE,
target_points)#inserted stubborn_until_x_points and target_points arguments{
result<-sample(1:60, size=6, replace=FALSE)
logical_vector<-(your_bet %in% result)
total_points<-sum(as.integer(logical_vector))
print(paste(c("Your bet:", as.character(your_bet))), collapse="")
print(paste(c("Result", as.character(result))), collapse="")
print(paste(c("Total points", as.character(total_points))), collapse="")
if (total_points==6)
print("You are a millonaire")
else if (total_points==5)
print("5 points, you are rich!")
else print("good luck next time")
if (stubborn_until_x_points==TRUE)#Inserted WHILE loop here{
while(total_points < target_points){
LOTTO(your_bet, stubborn_until_x_points=TRUE, target_points)}
}
}
```
This did make the function repeat in a loop, but for some reason, it keeps looping even when the condition is met.
Observation - I realized that when the condition is met on the first run, it actually stops, but after entering a loop, it goes on forever.
I canĀ“t find what is wrong. Any ideas?
Thanks in advance.
I haven't dug quite deep into the function, but the only reason that this should not break at any point is when target_points > n_samples (6 here).
The problem in this case is quite obvious and simple to fix. Reduce target_points to be less than 6 or add n_samples (6 atm.) and make it greater than target_points. Rather than this I suspect the main problem lies within the recursive function though. R is rather restrictive when it comes to recursion, if one tries to do a simple recursion for example
i <- 0
f <- function(){
i <<- i + 1
f()
}
f()
i
# 896 on my pc
one can see that we cannot use recursion for very deep recursive functions (in R). This throws the very unhelpful error
Error: C stack usage 7974196 is too close to the limit
To alleviate this, one simply has to remove the recursion (simply in italian because sometimes it is not simple). In this case we just move the while loop to contain the main body of the function, and use an if statement to break early if necessary.
Below is a slightly modified version of the function (note that sample_n and number_range has been added as arguments).
In this function the while loop has been moved to contain the main body, and the result is instead printed at the end (using the variable res to figure out the result). At the end of the loop I use a if(isFALSE(stubborn_until_x_points))break statement to exit early if necessary.
LOTTO <- function(your_bet,
sample_n = 6,
number_range = 1:60,
stubborn_until_x_points = FALSE,
target_points){
if(missing(target_points) || target_points > sample_n)
stop('missing target_points or target_points too large')
total_points <- -Inf # Always smaller than target_points
i <- 0
res <- 0
# If you want a new bet every iteration.
# Add this at the end of the loop, but remove the 'if'
if(missing(your_bet))
your_bet <- sample(number_range, size=sample_n, replace=FALSE)
while(total_points < target_points){
result <- sample(number_range, size=sample_n, replace=FALSE)
logical_vector <- your_bet %in% result
total_points <- sum(logical_vector)
if (total_points==6){
res <- 1
}else if (total_points==5){
res <- 2
}
i <- i + 1
if(isFALSE(stubborn_until_x_points))
break
}
if(res == 1)
cat(res <- 'You\'re a millionaire!\n', sep = '\n')
else if(res == 2)
cat(res <- '5 points, you are rich!\n', sep = '\n')
else
cat(res <- 'Better luck next time.', sep = '\n')
c(result = res, number_of_tries = i)
}
The function is called as before, but now also returns the number of attempts and the result obtained from trials as shown below.
LOTTO(target_points = 6, stubborn_until_x_points = TRUE)
You're a millionaire!
#Output:
result number_of_tries
"You're a millionaire!\n" "8297820"
Removing recursion, including the body of the function inside the while loop, assigning -Inf to the initial total_points and adding the break statement were much usefull.
Addapting the answer by #Oliver yielded exactly what I was looking for:
LOTTO<-function(your_bet=sample(1:60, size=6, replace=FALSE), stubborn_until_x_points=FALSE,
target_points=0){
total_points<--Inf
while(total_points < target_points){
result<-sample(1:60, size=6, replace=FALSE)
logical_vector<-(your_bet %in% result)
total_points<-sum(as.integer(logical_vector))
print(paste(c("Your bet:", as.character(your_bet))), collapse="")
print(paste(c("Result", as.character(result))), collapse="")
print(paste(c("Total points", as.character(total_points))), collapse="")
if (total_points==6)
print("You are a millonaire")
else if (total_points==5)
print("5 points, you are rich!")
else print("good luck next time")
if (isFALSE(stubborn_until_x_points==TRUE))
break
}
His answer, however, yields interesting additional results, and better control over inadequate argument inputs
Consider the function below:
f = function(i) i^2
Now we find the output of f for an input vector of length 1000 (or equivalently to run f for 1000 replications) by:
output = c()
for (i in 1:1000) output[i] = f(i)
In the case of running time-consuming functions, we might like to know which replication we are. So we can use:
output = c()
for (i in 1:1000) {output[i] = f(i); cat("Replicate=", i, "\n")}
This gives the replicate number at the end of each replication. Now what if we use sapply instead of for:
output = sapply(1:1000, function(i) f(i))
How can we see which replicate we are while using sapply? Note that I tried adding cat("Replicate=", i, "\n") in the definition of f. This shows the replicate number, but only at the end of the entire run and not at the end of each replicate.
you say you have tried it, but this code works just great for me:
result <- sapply(1:1000, function(x) {
print(x) # cat works too
return(x^2)
})
You may have forgotten the curly brackets! :-)
I'm trying to improve the speed of my code, which is trying to optimise a value using 3 variables which have large ranges. The most likely output uses values in the middle of the ranges, so it is wasting time starting from the lowest possible value of each variable. I want to start from the middle value and iterate out! The actual problem has thousands of lines with numbers from 150-650. C,H and O limits will be defined somewhat based on the starting number, but will always be more likely at a central value in the defined range. Is there a way to define the for loop to work outwards like I want? The only, quite shabby, way I can think of is to simply redefine the value within the loop from a vector (e.g. 1=20, 2=21, 3=19, etc). See current code below:
set_error<-2.5
ct<-c(325.00214,325.00952,325.02004,325.02762,325.03535,325.03831,325.04588, 325.05641,325.06402,325.06766,325.07167,325.07454,325.10396)
FormFun<-function(x){
for(C in 1:40){
for(H in 1:80){
for(O in 1:40){
test_mass=C*12+H*1.007825+O*15.9949146-1.0072765
error<-1000000*abs(test_mass-x)/x
if(error<set_error){
result<-paste("C",C,"H",H,"O",O,sep ="")
return(result)
break;break;break;break
}
}
}
}
}
old_t <- Sys.time()
ct2<-lapply(ct,FormFun)
new_t <- Sys.time() - old_t # calculate difference
print(new_t)
Use vectorization and create a closure:
FormFun1_fac <- function(gr) {
gr <<- gr
function(x, set_error){
test_mass <- with(gr, C*12+H*1.007825+O*15.9949146-1.0072765)
error <- 1000000 * abs(test_mass - x) / x
ind <- which(error < set_error)[1]
if (is.na(ind)) return(NULL)
paste0("C", gr[ind, "C"],"H", gr[ind, "H"],"O", gr[ind, "O"])
}
}
FormFun1 <- FormFun1_fac(expand.grid(C = 1:40, H = 1:80, O = 1:40))
ct21 <- lapply(ct, FormFun1, set_error = set_error)
all.equal(ct2, ct21)
#[1] TRUE
This saves a grid of all combinations of C, H, O in the function environment and calculates the error for all combinations (which is fast in vectorized code). The first combination that passes the test is returned.
I've written a function that is a simulation, that outputs a vector of 100 elements, and I want to use the *apply functions to run the function many times and store the repeated output in a new vector for each time the simulation is run.
The function looks like:
J <- c(1:100)
species_richness <- function(J){
a <- table(J)
return(NROW(a))
}
simulation <- function(J,gens,ploton=FALSE,v=0.1){
species_richness_output <- rep(NA,gens)
for(rep in 1:gens){
index1 <- sample(1:length(J),1)
if(runif(1,0,1) < v){
J[index1] <- (rep+100)
}
else{
index2 <- sample(1:length(J),1)
while(index1==index2) {
index2 <- sample(1:length(J),1)
}
J[index1] <- J[index2]
}
species_richness_output[rep] <- species_richness(J)
}
species_abundance <- function(J){
a <- table(J)
return(a)
}
abuntable <- species_abundance(J)
print(abuntable)
octaves <- function(abuntable){
oct <- (rep(0,log2(sum(abuntable))))
for(i in 1:length(abuntable)){
oct2 <- floor(log2(abuntable[i])+1)
oct[oct2] <- oct[oct2]+1
}
print(oct)
}
# octaves(c(100,64,63,5,4,3,2,2,1,1,1,1))
if(ploton==TRUE){
hist(octaves(abuntable))
}
print(species_richness(J))
return(J)
}
simulation(J, 10000,TRUE,v=0.1)
So that's my function, it takes J a vector I defined earlier, manipulates it, then returns:
the newly simulated vector J of 100 elements
a function called octave that categorises the new vector
a histogram corresponding to the above "octave"
I have tried a number of variations: using lapply, mapply
putting args=args_from_original_simulation
simulation_repeated <- c(mapply(list, FUN=simulation(args),times=10000))
but I keep getting an error with the match.fun part of the mapply function
Error in match.fun(FUN) :
'simulation(J, 10000, FALSE, 0.1)' is not a function, character or symbol
This is despite the simulation I have written showing as being saved as a function in the workspace.
Does anyone know what this error is pointing to?
In this line:
simulation_repeated <- c(mapply(list, FUN=simulation(args),times=10000))
You are not giving a function to mapply. You are (essentially) passing the result of calling simulation(args) and simulation does not return a function.
I've been trying to write a program in R that implements Newton's method. I've been mostly successful, but there are two little snags that have been bothering me. Here's my code:
Newton<-function(f,f.,guess){
#f <- readline(prompt="Function? ")
#f. <- readline(prompt="Derivative? ")
#guess <- as.numeric(readline(prompt="Guess? "))
a <- rep(NA, length=1000)
a[1] <- guess
a[2] <- a[1] - f(a[1]) / f.(a[1])
for(i in 2:length(a)){
if(a[i] == a[i-1]){
break
}
else{
a[i+1] <- a[i] - f(a[i]) / f.(a[i])
}
}
a <- a[complete.cases(a)]
return(a)
}
I can't get R to recognize the functions f and f. if I try using readline() to prompt for user input. I get the error "Error in Newton() : could not find function "f."" However, if I comment out the readlines (as above), define f and f. beforehand, then everything works fine.
I've been trying to make R calculate the derivative of a function. The problem is that the class object with which R can take symbolic derivatives is expression(), but I want to take the derivative of a function() and have it give me a function(). In short, I'm having trouble with type conversion between expression() and function().
I have an ugly but effective solution for going from function() to expression(). Given a function f, D(body(f)[[2]],"x") will give the derivative of f. However, this output is an expression(), and I haven't been able to turn it back into a function(). Do I need to use eval() or something? I've tried subsetting, but to no avail. For instance:
g <- expression(sin(x))
g[[1]]
sin(x)
f <- function(x){g[[1]]}
f(0)
sin(x)
when what I want is f(0) = 0 since sin(0) = 0.
EDIT: Thanks all! Here's my new code:
Newton<-function(f,f.,guess){
g<-readline(prompt="Function? ")
g<-parse(text=g)
g.<-D(g,"x")
f<-function(x){eval(g[[1]])}
f.<-function(x){eval(g.)}
guess<-as.numeric(readline(prompt="Guess? "))
a<-rep(NA, length=1000)
a[1]<-guess
a[2]<-a[1]-f(a[1])/f.(a[1])
for(i in 2:length(a)){
if(a[i]==a[i-1]){break
}else{
a[i+1]<-a[i]-f(a[i])/f.(a[i])
}
}
a<-a[complete.cases(a)]
#a<-a[1:(length(a)-1)]
return(a)
}
This first problem arises because readline reads in a text string, whereas what you need is an expression. You can use parse() to convert the text string to an expression:
f <-readline(prompt="Function? ")
sin(x)
f
# [1] "sin(x)"
f <- parse(text = f)
f
# expression(sin(x))
g <- D(f, "x")
g
# cos(x)
To pass in values for the arguments in the function call in the expression (whew!), you can eval() it in an environment containing the supplied values. Nicely, R will allow you to supply those values in a list supplied to the envir= argument of eval():
> eval(f, envir=list(x=0))
# [1] 0
Josh has answered your question
For part 2 you could have used
g <- expression( sin(x) )
g[[1]]
# sin(x)
f <- function(x){ eval( g[[1]] ) }
f(0)
# [1] 0
f(pi/6)
# [1] 0.5
BTW, having recently written a toy which calculates fractal patterns based on root convergence of Newton's method in the complex plane, I can recommend you toss in some code like the following (where the main function's argument list includes "func" and "varname" ).
func<- gsub(varname, 'zvar', func)
funcderiv<- try( D(parse(text=func), 'zvar') )
if(class(funcderiv) == 'try-error') stop("Can't calculate derivative")
If you're more cautious, you could include a an argument "funcderiv" , and wrap my code in
if(missing(funcderiv)){blah blah}
Ahh, why not: here's my complete function for all to use and enjoy:-)
# build Newton-Raphson fractal
#define: f(z) the convergence per Newton's method is
# zn+1 = zn - f(zn)/f'(zn)
#record which root each starting z0 converges to,
# and to get even nicer coloring, record the number of iterations to get there.
# Inputs:
# func: character string, including the variable. E.g., 'x+ 2*x^2' or 'sin(x)'
# varname: character string indicating the variable name
# zreal: vector(preferably) of Re(z)
# zim: vector of Im(z)
# rootprec: convergence precision for the NewtonRaphson algorithm
# maxiter: safety switch, maximum iterations, after which throw an error
#
nrfrac<-function(func='z^5 - 1 ', varname = 'z', zreal= seq(-5,5,by=.1), zim, rootprec=1.0e-5, maxiter=1e4, drawplot=T, drawiterplot=F, ...) {
zreal<-as.vector(zreal)
if(missing(zim)) zim <- as.vector(zreal)
# precalculate F/F'
# check for differentiability (in R's capability)
# and make sure to get the correct variable name into the function
func<- gsub(varname, 'zvar', func)
funcderiv<- try( D(parse(text=func), 'zvar') )
if(class(funcderiv) == 'try-error') stop("Can't calculate derivative")
# Interesting "feature" of deparse : default is to limit each string to 60 or64
# chars. Need to avoid that here. Doubt I'd ever see a derivative w/ more
# than 500 chars, the max allowed by deparse. To do it right,
# need sum(nchar(funcderiv)) as width, and even then need to do some sort of
# paste(deparse(...),collapse='') to get a single string
nrfunc <- paste(text='(',func,')/(',deparse(funcderiv, width=500),')', collapse='')
# first arg to outer() will give rows
# Stupid Bug: I need to REVERSE zim to get proper axis orientation
zstart<- outer(rev(zim*1i), zreal, "+")
zindex <- 1:(length(zreal)*length(zim))
zvec <- data.frame(zdata=as.vector(zstart), zindex=zindex, itermap=rep(0,length(zindex)), badroot=rep(0,length(zindex)), rooterr=rep(0,length(zindex)) )
#initialize data.frame for zout.
zout=data.frame(zdata=rep(NA,length(zstart)), zindex=rep(NA,length(zindex)), itermap=rep(0,length(zindex)), badroot=rep(0,length(zindex)), rooterr=rep(0,length(zindex)))
# a value for rounding purposes later on; yes it works for rootprec >1
logprec <- -floor(log10(rootprec))
newtparam <- function(zvar) {}
body(newtparam)[2] <- parse(text=paste('newz<-', nrfunc, collapse=''))
body(newtparam)[3] <- parse(text=paste('return(invisible(newz))'))
iter <- 1
zold <- zvec # save zvec so I can return original values
zoutind <- 1 #initialize location to write solved values
while (iter <= maxiter & length(zold$zdata)>0 ) {
zold$rooterr <- newtparam(zold$zdata)
zold$zdata <- zold$zdata - zold$rooterr
rooterr <- abs(zold$rooterr)
zold$badroot[!is.finite(rooterr)] <- 1
zold$zdata[!is.finite(rooterr)] <- NA
# what if solvind = FFFFFFF? -- can't write 'nothing' to zout
solvind <- (zold$badroot >0 | rooterr<rootprec)
if( sum(solvind)>0 ) zout[zoutind:(zoutind-1+sum(solvind)),] <- zold[solvind,]
#update zout index to next 'empty' row
zoutind<-zoutind + sum(solvind)
# update the iter count for remaining elements:
zold$itermap <- iter
# and reduce the size of the matrix being fed back to loop
zold<-zold[!solvind,]
iter <- iter +1
# just wonder if a gc() call here would make any difference
# wow -- it sure does
gc()
} # end of while
# Now, there may be some nonconverged values, so:
# badroot[] is set to 2 to distinguish from Inf/NaN locations
if( zoutind < length(zindex) ) { # there are nonconverged values
# fill the remaining rows, i.e. zout.index:length(zindex)
zout[(zoutind:length(zindex)),] <- zold # all of it
zold$badroot[] <- 2 # yes this is safe for length(badroot)==0
zold$zdata[]<-NA #keeps nonconverged values from messing up results
}
# be sure to properly re-order everything...
zout<-zout[order(zout$zindex),]
zout$zdata <- complex(re=round(Re(zout$zdata),logprec), im=round(Im(zout$zdata),logprec) )
rootvec <- factor(as.vector(zout$zdata), labels=c(1:length(unique(na.omit(as.vector(zout$zdata))))))
#convert from character, too!
rootIDmap<-matrix(as.numeric(rootvec), nr=length(zim))
# to colorize very simply:
if(drawplot) {
colorvec<-rainbow(length(unique(as.vector(rootIDmap))))
imagemat<-rootIDmap
imagemat[,]<-colorvec[imagemat] #now has color strings
dev.new()
# all '...' arguments used to set up plot
plot(range((zreal)),range((zim)), t='n',xlab='real',ylab='imaginary',... )
rasterImage(imagemat, range(zreal)[1], range(zim)[1], range(zreal)[2], range(zim)[2], interp=F)
}
outs <- list(rootIDmap=rootIDmap, zvec=zvec, zout=zout, nrfunc=nrfunc)
return(invisible(outs))
}