I want to restrict the for cycle to only perform task if j is in some range of i (3 units, for example).
I tried the following piece of code:
a <- c(1:100)
b <- c(1:100)
k1 <- length(a)
k2 <- length(b)
for (i in 1:k1){
for (j in 1:k2){
if (j>=i-3 & j<=i+3){
c<-c(a+b)
}
}
}
What I pretended was
if i=1, j={1,2,3}, if i=6, j={1,2,3,4,5,6}
This doesn´t really work since, j and i will end up running from 1 to 100.
If I understand, the problem is that you are looping through 100 combinations of j, when only three to seven are actually useful.
If this is correct, you can loop through seven iterations of j and filter for values that are positive and within bounds:
width <- 3
for (i in seq_along(a)) {
for (j in (i-width):(i+width)) {
if (j > 0 && j <= length(b)) {
# Do something
}
}
}
When you # Do something in your code, I would advise not assigning to a variable named c.
Related
I have the following exercise to be solved in R. Under the exercise, there is a hint towards the solution.
Exercise: If there are no ties in the data set, the function above will produce breakpoints with h observations in the interval between two consecutive breakpoints (except the last two perhaps). If there are ties, the function will by construction return unique breakpoints, but there may be more than h observations in some intervals.
Hint:
my_breaks <-function(x, h = 5) {
x <-sort(x)
breaks <- xb <- x[1]
k <- 1
for(i in seq_along(x)[-1])
{if(k<h)
{k <- k+1}
else{
if(xb<x[i-1]&&x[i-1]<x[i])
{xb <- x[i-1]
breaks <-c(breaks, xb)
k <- 1
}
}
}
However, I am having a hard time understanding the above function particularly the following lines
for(i in seq_along(x)[-1])
{if(k<h)
{k <- k+1}
Question:
How is the for loop supposed to act in k if k is previously defined as 1 and i is different than k? How are the breakpoints chosen according to the h=5 gap if the for loop is not acting on x? Can someone explain to me how this function works?
Thanks in advance!
First, note that your example is incomplete. The return value and the final brace are missing there. Here is the correct version.
my_breaks <-function(x, h = 5) {
x <- sort(x)
breaks <- xb <- x[1]
k <- 1
for(i in seq_along(x)[-1]){
if(k<h) {
k <- k+1
} else {
if(xb<x[i-1]&&x[i-1]<x[i]){
xb <- x[i-1]
breaks <-c(breaks, xb)
k <- 1
}
}
}
breaks
}
Let's check if it works.
my_breaks(c(1,1,1:5,8:10), 2)
#[1] 1 2 4 8
my_breaks(c(1,1,1:5,8:10), 5)
#[1] 1 3
As you can see, everything is fine. And what is seq_along(x)[-1]? We could write this equation as 2:length(x). So the for loop goes through each element of the vector x in sequence, skipping the first element.
What is the k variable for? It counts the distance to take into account the h parameter.
I am trying to write a function to calculate h-point. the function is defined over a rank frequency data frame.
consider the following data.frame :
DATA <-data.frame(frequency=c(49,48,46,38,29,24,23,22,15,12,12,10,10,9,9), rank=c(seq(1, 15)))
and the formula for h-point is :
if {there is an r = f(r), h-point = r }
else { h-point = f(i)j-f(j)i / j-i+f(i)-f(j) }
where f(i) and f(j) are corresponding frequencies for ith and jth ranks and i and j are adjacent ranks that i<f(i) and j>f(j).
NOW, i have tried the following codes :
fr <-function(x){d <-DATA$frequency[x]
return(d)}
for (i in 1:length(DATA$rank)) {
j <- i+1
if (i==fr(i))
return(i)
else(i<fr(i) && j>fr(j)) {
s <-fr(i)*j-fr(j)*i/j-i+fr(i)-fr(j)
return(s)
}}
I also tried:
for (i in 1:length(DATA$rank)) {
j <- i+1
if (i==fr(i))
return(i)
if (i<fr(i) while(j>fr(j))) {
s <-fr(i)*j-fr(j)*i/j-i+fr(i)-fr(j)
return(s)
}}
and neither of them works. for the DATA ,the desired result would be i=11 and j=12, so:
h-point=12×12 - 10×11 / 12 - 11 + 12 - 10
can you please tell me what I`m doing wrong here?
You could do:
h_point <- function(data){
x <- seq(nrow(data))
f_x <- data[["frequency"]][x]
h <- which(x == f_x)
if(length(h)>1) h
else{
i <- which(x<f_x)
j <- which(x>f_x)
s <- which(outer(i,j,"-") == -1, TRUE)
i <- i[s[,1]]
j <- j[s[,2]]
cat("i: ",i, "j: ", j,"\n")
f_x[i]*j - f_x[j]*i / (i-j + f_x[i]-f_x[j])
}
}
h_point(DATA)
i: 11 j: 12
[1] 34
I think I have figured out what you are trying to achieve. My loop will go through DATA and break at any point if rank == frequency for a given row. If might be more prudent to explicitly test this with DATA$rank[i] == fr(i) rather than relying on i, in case tied ranks etc.
The second if statement calculates h-point (s) for rows i and j if row i has rank that is lower than freq and row j has a rank that is higher.
Is this what you wanted?
DATA <-data.frame(frequency=c(49,48,46,38,29,24,23,22,15,12,12,10,10,9,9), rank=c(seq(1, 15)))
fr <-function(x){d <-DATA$frequency[x]
return(d)}
for(i in 1:nrow(DATA)){
j <- i+1
if (i==fr(i)){
s <- list(ij=c(i=i,j=j), h=i)
break
}else if(i <fr(i) && j>fr(j)){
s <-list(ij=c(i=i,j=j),h=fr(i)*j-fr(j)*i/j-i+fr(i)-fr(j))
}}
I am not sure the formula is correct, in your loop you had j-i but in explanation it was i-j. Not sure if the entire i-j+fr(i)-fr(j) is the denominator and similarly for the numerator. Simple fixes.
I have an event A that is triggered when the majority of coin tosses in a series of tosses comes up heads. I have an unfair coin and I'd like to see how the likelihood of A changes as the number of tosses change and the probability in each toss changes.
This is my function assuming 3 tosses
n <- 3
#victory requires majority of tosses heads
#tosses only occur in odd intervals
k <- seq(n/2+.5,n)
victory <- function(n,k,p){
for (i in p) {
x <- 0
for (i in k) {
x <- x + choose(n, k) * p^k * (1-p)^(n-k)
}
z <- x
}
return(z)
}
p <- seq(0,1,.1)
victory(n,k,p)
My hope is the victory() function would:
find the probability of each of the outcomes where the majority of tosses are heads, given a particular value p
sum up those probabilities and add them to a vector z
go back and do the same thing given another probability p
I tested this with n <- 3, k <- c(2,3) and p <- (.5,.75) and the output was 0.75000, 0.84375. I know that the output should've been 0.625, 0.0984375.
I wasn't able to get exactly the result you wanted, but maybe can help you along a bit.
When looping in R the vector you are looping through remains unchanged and value you are using to loop changes. For example see the differences in these loops:
test <- seq(0,1,length.out = 5)
for ( i in test){
print(test)
}
for ( i in test){
print(i)
}
for ( i in 1:length(test)){
print(test[i])
}
when you are iterating you are firstly setting i to the first number in p, then to the first number in k and then using the unchanged vectors.
You are also assigning to z in the first loop of p and then writing over it in the second loop.
Try using the below - I am still not getting the answer you say but it might help you find where the error is (printing out along the way or using debug(victory) might also be helpful
victory <- function(n,k,p){
z <-list()
for (i in 1:length(p)) {
x <- 0
for (j in 1:length(k)) {
x <- x + choose(n, k[j]) * p[i]^k[j] * (1-p[i])^(n-k[j])
}
z[i] <- x
}
return(z)
}
I have a vector of positive integers of unknown length. Let's call it vector a with elements a[1], a[2], ...
I want to perform calculations on vector b where for all i, 0 <= b[i] <= a[i].
The following does not work:
for(b in 0:a)
{
# calculations
}
The best I have come up with is:
probabilities <- function(a,p)
{
k <- a
k[1] <- 1
h <- rep(0,sum(a)+1)
for(i in 2:length(a))
{
k[i] <- k[i-1]*(a[i-1]+1)
}
for(i in 0:prod(a+1))
{
b <- a
for(j in 1:length(a))
{
b[j] <- (floor(i/k[j]) %% (a[j]+1))
}
t <- 1
for(j in 1:length(a))
{
t <- t * choose(a[j],b[j])*(p[j])^(b[j])*(1-p[j])^(a[j]-b[j])
}
h[sum(b)+1] <- h[sum(b)+1] + t
}
return(h)
}
In the middle of my function is where I create b. I start off by setting b equal to a (so that it is the same size). Then, I replace all of the elements of b with different elements that are rather tricky to calculate. This seems like an inefficient solution. It works, but it is fairly slow as the numbers get large. Any ideas for how I can cut down on process time? Essentially, what this does for b is the first time through, b is all zeros. Then, it is 1, 0,0,0,... The first element keeps incrementing until it reaches a[1], then b[2] increments and b[1] is set to 0. Then b[1] starts incrementing again.
I know the math is sound, I just do not trust that it is efficient. I studied combinatorics for a few years, but have never studied computational complexity theory, so coming up with a fast algorithm is a bit beyond my realm of knowledge. Any ideas would be helpful!
I am new in programming, and I just start learning R language. I am trying to do a bubble sort, but it shows the following error message. Can anyone help me solve the problem?
x <-sample(1:100,10)
n <- length(x)
example <- function(x)
{
for (i in 1:n-1)
{
while (x[i] > x[i+1])
{
temp <- x[i+1]
x[i+1] <- x[i]
x[i] <- temp
}
i <- i+1
}
}
example(x)
Error in while (x[i] > x[i + 1]) { : argument is of length zero
x<-sample(1:100,10)
example <- function(x){
n<-length(x)
for(j in 1:(n-1)){
for(i in 1:(n-j)){
if(x[i]>x[i+1]){
temp<-x[i]
x[i]<-x[i+1]
x[i+1]<-temp
}
}
}
return(x)
}
res<-example(x)
#input
x
#output
res
It is working fine with little modification of your code. In 'R' it is better to use sort() function.
x <-sample(1:100,10)
x
res<-sort(x)
res
You have some inaccuracies in your algorithm of sorting. I've made changes to make it work.
set.seed(1)
x <-sample(1:100,10)
x
# [1] 27 37 57 89 20 86 97 62 58 6
example <- function(x)
{
n <- length(x) # better insert this line inside the sorting function
for (k in n:2) # every iteration of the outer loop bubbles the maximum element
# of the array to the end
{
i <- 1
while (i < k) # i is the index for nested loop, no need to do i < n
# because passing j iterations of the for loop already
# places j maximum elements to the last j positions
{
if (x[i] > x[i+1]) # if the element is greater than the next one we change them
{
temp <- x[i+1]
x[i+1] <- x[i]
x[i] <- temp
}
i <- i+1 # moving to the next element
}
}
x # returning sorted x (the last evaluated value inside the body
# of the function is returned), we can also write return(x)
}
example(x)
# [1] 6 20 27 37 57 58 62 86 89 97
BTW, R language has a lot of functions and methods for doing things. This example function can be a learning example, but I advice to use existing function sort for solving real problems.
In R language you should try to avoid loops and make usage of vectorized functions to make the code faster.
It gives you that error message because he cannot compare a value that is out of his bounds which is the case for you at (x[i] > x[i + 1]). Try this if you want to sort your array in a decreasing order:
for (i in 1:n){
j = i
while((j>1)){
if ((X[j]> X[j-1])){
temp = X[j]
X[j] = X[j-1]
X[j-1] = temp
}
j = j-1
}
}
For an increasing order you just have to switch around the > sign in the while loop.