I have a weird question..
Essentially, I have a function which takes a data frame of dimension Nx(2k) and transforms it into an array of dimension Nx2xk. I then further use that array in various locations in the function.
My issue is this, when k == 2, I'm left with a matrix of degree Nx2, and even worse, if N = 1, I'm stuck with a matrix of degree 1x2.
I would like to write myArray[thisRow,,] to select that slice of the array, but this falls short for the N = 1, k = 2 case. I tried myArray[thisRow,,,drop = FALSE] but that gives an 'incorrect number of dimensions' error. This same issue arrises for the Nx2 case.
Is there a work around for this issue, or do I need to break my code into cases?
Sample Code Shown Below:
thisFunction <- function(myDF)
{
nGroups = NCOL(myDF)/2
afMyArray = myDF
if(nGroups > 1)
{
afMyArray = abind(lapply(1:nGroups, function(g){myDF[,2*(g-1) + 1:2]}),
along = 3)
}
sapply(1:NROW(myDF),
function(r)
{
thisSlice = afMyArray[r,,]
*some operation on thisSlice*
})
}
Thanks,
James
Related
I am trying to define a function with a for loop and inside a conditional in R studio. Yesterday I was able with the help of another thread to devise this piece of code. The problem is that I want to sum the vector elements ma for any possible x, so that is inside the function l. This is a simpler case which I am trying to solve to adapt the original model. However, I do not know how to proceed.
ma<-rep(0,20)
l <- function(x, ma) {
for(i in seq_along(ma)) {
if(i %% 2 == 1) {
ma[i] <- i + x
} else {
ma[i] <- 0
}
}
return(ma)
}
My problem is that I would like to have the sum of i+x+0+i+x... for any possible x. I mean a function of the kind for any possible x.
Question:
Can someone explain to me how to implement such a function in R?
Thanks in advance!
I am going to update the original function:
Theta_alpha_s<-function(s,alpha,t,Basis){
for (i in seq_along(Basis)){
if(i%% 2==1) {Basis[i]=s*i^{-alpha-0.5}*sqrt(2)*cos(2*pi*i*t)}
else{Basis[i]=s*i^{-alpha-0.5}*sqrt(2)*sin(2*pi*i*t)}
}
return(Basis)
}
If you don't want to change the values in Basis, you can create a new vector in the function (here result) that you will return:
l = function(s,alpha,t,Basis){
is.odd = which(Basis %% 2 == 1)
not.odd = which(Basis %% 2 == 0)
result = rep(NA, length(Basis))
result[is.odd] = s*is.odd^{-alpha-0.5}*sqrt(2)*cos(2*pi*is.odd*t)
result[not.odd] = s*not.odd^{-alpha-0.5}*sqrt(2)*sin(2*pi*not.odd*t)
#return(result)
return(c(sum(result[is.odd]), sum(result[not.odd])))
}
I am pretty new to R, so there is definitely some improvement to my code needed. What I want is to do quicksort on an array of n elements, count the number of comparisons made and output the sorted array after k comparisons.
So far, I have reused the code for a quicksort algorithm found here:
quickSort <- function(arr) {
# Pick a number at random.
mid <- sample(arr, 1)
print(arr)
print(mid)
# Place-holders for left and right values.
left <- c()
right <- c()
# Move all the smaller values to the left, bigger values to the right.
lapply(arr[arr != mid], function(d) {
count <<- count + 1
stopifnot(count <= k)
if (d < mid) {
left <<- c(left, d)
}
else {
right <<- c(right, d)
}
})
if (length(left) > 1) {
left <- quickSort(left)
}
if (length(right) > 1) {
right <- quickSort(right)
}
# Finally, return the sorted values.
c(left, mid, right)
}
I am currently struggling with several things:
How can I get not only the partial vector that is currently being sorted but also the full vector?
Did I put the right stopping condition in the right place?
An example of what I want:
given an array (2,4,1,3,5) and the first pivot element 3, after four comparisons I would want the output to be (2,1,3,4,5).
Any help would be greatly appreciated!
I would like to find the value of "p" below (which is between 0 and 1), knowing the following equations:
RI_26 = min(IR,na.rm=FALSE)
RI_min = 100-(sse*SUM/((1+p)*Dotation2017*100))^(1/p)
where RI_26 is the minimum of resources index of my 26 area. It is a constant in my case. In RI_min, sse and Dotations2017 are 2 constants and p is a unknown. I know that RI_26 should be equal to RI_min.
It would be easy to solve it, but SUM (which is present in RI_min) is as well unknown as it is a function of p as following:
`sum.function = function(p){
SUM <- c(0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0)
for(i in 1:length(Canton))
if(IR[i] < 100) {
SUM[i] <- (100-IR[i])^(1+p)*Pop[i]
SUM[27] <- SUM[27]+SUM[i]
}
SUM <- round(SUM,0)
return(SUM[27])
}
SUM = sum.function(p)
SUM returns a number (or vector 1X1). To deal with it, I would like to find the value of p that satisfied:
RI_26/RI_min = 1
To do so, I would like to do a loop, beginning with p = 0 and then increasing the value of p by 0.01 until it reaches 1. The loop should return the value of p_star when the constraint is True (RI_26/RI_min = 1.00).
I don't have any idea how to do this but it could look like the following code:
p.function = function(){
for(...)
if(RI_26/RI_min = 1.000000) {
p_star <- p
}
return(p_star)
}
So the function will return the value of p_star when RI_26/RI_min = 1.000000. What am I suppose to write in my function: p.function to increment "p" and have the result that I want? Any idea?
for (i in seq(0, 1, by = 0.1)) {
"Your code here"
}
I want to use the data of [x] to fill in [test] based on certain sequence:
x = matrix(rnorm(330),165,2)
origins = 130:157
horizon = 8
col = 1:2
test = array(0, c(length(origins)*length(col), horizon))
for( origin in origins){
for (c in col){
test[which(origin==origins), ] = x[(origin+1):(origin+8), c]
}
}
However, this code only helps extract the second column of [x] to fill in the first 28 rows of [test]. The following picture is only a part of a complete [test] table, showing the ineffective filling from row 29 to row 56.
enter image description here
Anyone who can help me fill in them completely? Thank you very much.
Here is a possible solution, but it is still not clear what you want the result to be. better to make much smaller data and show desired result.
The left hand side of the assignment, in the original code, does not vary with c, so each time through the loop for c the same rows of test will be overwrittem,
x = matrix(rnorm(330),165,2)
origins = 130:157
horizon = 8
col = 1:2
test = array(0, c(length(origins)*length(col), horizon))
for( origin in origins){
for (c in col){
# the left hand side must vary somehow with c as well.
test[(which(origin==origins)-1) + (c - 1) * length(origins) + 1, ] = x[(origin+1):(origin+8), c]
}
}
I am working on the NSGA2 package on R (library mco).
My NSGA2 code takes forever to run, so I am wondering:
1) Is there a way to limit the precision of the solution values (say, maybe up to 3 decimal places) instead of infinite?
2) How do I set an equality constraint (the ones online all seemed to be about >= or <= than =)? Not sure if I'm doing it right.
My entire relevant code for reference, for easy tracing: https://docs.google.com/document/d/1xj7OPng11EzLTTtWLdRWMm8zJ9f7q1wsx2nIHdh3RM4/edit?usp=sharing
Relevant sample part of code reproduced here:
VTR = get.hist.quote(instrument = 'VTR',
start="2010-01-01", end = "2015-12-31",
quote = c("AdjClose"),provider = "yahoo",
compress = "d")
ObjFun1 <- function (xh){
f1 <- sum(HSVaR_P(merge(VTR, CMI, SPLS, KSS, DVN, MAT, LOE, KEL, COH, AXP), xh, 0.05, 2))
tempt = merge(VTR, CMI, SPLS, KSS, DVN, MAT, LOE, KEL, COH, AXP)
tempt2 = tempt[(nrow(tempt)-(2*N)):nrow(tempt),]
for (i in 1:nrow(tempt2))
{
for (j in 1:ncol(tempt2))
{
if (is.na(tempt2[i,j]))
{
tempt2[i,j] = 0
}
}
}
f2 <- ((-1)*abs(sum((xh*t(tempt2)))))
c(f1=f1,f2=f2)
}
Constr <- function(xh){
totwt <- (1-sum(-xh))
totwt2 <- (sum(xh)-1)
c(totwt,totwt2)
}
Solution1 <- nsga2(ObjFun1, n.projects, 2,
lower.bounds=rep(0,n.projects), upper.bounds=rep(1,n.projects),
popsize=n.solutions, constraints = Constr, cdim=1,
generations=generations)
The function HSVaR_P returns matrix(x,2*500,1).
Even when I set generations = 1, the code does not seem to run. Clearly there should be some error in the code, somewhere, but I am not entirely sure about the mechanics of the NSGA2 algorithm.
Thanks.