Inspired by the leetcode challenge for two sum, I wanted to solve it in R. But while trying to solve it by brute-force I run in to an issue with my for loop.
So the basic idea is that given a vector of integers, which two integers in the vector, sums up to a set target integer.
First I create 10000 integers:
set.seed(1234)
n_numbers <- 10000
nums <- sample(-10^4:10^4, n_numbers, replace = FALSE)
The I do a for loop within a for loop to check every single element against eachother.
# ensure that it is actually solvable
target <- nums[11] + nums[111]
test <- 0
for (i in 1:(length(nums)-1)) {
for (j in 1:(length(nums)-1)) {
j <- j + 1
test <- nums[i] + nums[j]
if (test == target) {
print(i)
print(j)
break
}
}
}
My problem is that it starts wildly printing numbers before ever getting to the right condition of test == target. And I cannot seem to figure out why.
I think there are several issues with your code:
First, you don't have to increase your j manually, you can do this within the for-statement. So if you really want to increase your j by 1 in every step you can just write:
for (j in 2:(length(nums)))
Second, you are breaking only the inner-loop of the for-loop. Look here Breaking out of nested loops in R for further information on that.
Third, there are several entries in nums that gave the "right" result target. Therefore, your if-condition works well and prints all combination of nums[i]+nums[j] that are equal to target.
Related
I'm very new to r and programming so please stay with me :)
I am trying to use iterations to find the value of infinite iterations to the 4th decimal place. I.e. where the 4th decimal does not change. so 1.4223, where 3 does not change anymore so the result to 3 decimal place is 1.422.
The link above shows an example of a similar problem that I am faced with. My question is how do I create a for-loop that goes to infinity and find the value where the 4th decimal point stops changing?
I have tried using while loops but I am not sure how to stop it from just looping forever. I need some if statement like below:
result <- 0
i <- 1
d <- 1e-4
while(TRUE)
{
result <- result + (1/(i^2))
if(abs(result) < d)
{
break
}
i <- i + 1
}
result
Here's an example: to do the infinite loop, use while(TRUE) {}, and as you suggested use an if clause and break to stop when necessary.
## example equation shown
## fun <- function(x,n) {
## (x-1)^(2*n)/(n*(2*n-1))
## }
## do it for f(x)=1/x^2 instead
## doesn't have any x-dependence, but leave it in anyway
fun <- function(x,n) {
1/n^2
}
n <- 1
## x <- 0.6
tol <- 1e-4
ans <- 0
while (TRUE) {
next_term <- fun(x,n)
ans <- ans + next_term
if (abs(next_term)<tol) break
n <- n+1
}
When run this gives ans=1.635082, n=101.
R also has a rarely used repeat { } keyword, but while(TRUE) will probably be clearer to readers
there are more efficient ways to do this (i.e. calculating the numerator by multiplying it by (x-1)^2 each time)
it's generally a good idea to test for a maximum number of iterations as well so that you don't set up a truly infinite loop if your series doesn't converge or if you have a bug in your code
I haven't solved your exact problem (chose a smaller value of tol), but you should be able to adjust this to get an answer
as discussed in the answer to your previous question, this isn't guaranteed, but should generally be OK; you can check (I haven't) to be sure that the particular series you want to evaluate has well-behaved convergence
Code is at end of post.
I've been trying to troubleshoot a program to generate cyclic subgroups of GL2. For some reason it's having issues when the generator has an element that was computed from a fraction whose decimal representation has infinitely many digits (but only if the elements are defined as arguments). Even if the elements of some power of that generator are identical to the identity matrix's elements I still can't get identical() to return TRUE.
I've managed to figure out that there's some issue with how the value of the generator is actually being assigned, but I can't figure out why. To reiterate, I know that the issue is arising out of the bit of my code that actually turns the fraction-string into a number (and I've verified this in the shell environment). I just don't understand why this is happening, though it seems obvious that it has to do with the infinity of digits in the decimal representation.
Calling the program like Rscript code.R 0 1/3 3 0 gives me a case where the order of the generator is infinite (when it should be equal to 2). Rscript code.R 0 1/2 2 0 gives me the correct subgroup with the correct order. I also end up with the correct order when I define the generator with components {0,1/3,3,0} without all of the string manipulation. In either case I still get a matrix whose elements are ij = 1 if i=j and ij=0 if i =/= j for even powers of the generator. So what am I missing, here?
library(expm)
args = commandArgs(trailingOnly=TRUE)
if (length(args) != 4){
stop("Usage: Rscript GL2_cyclic 11 12 21 22", call.=FALSE)
}
for(i in 1:length(args)){
if(length(grep("/",args[i])) != 0){
temp <- as.integer(unlist(strsplit(args[i],"/")))
args[i] <- temp[1]/temp[2]
}
}
args <- as.numeric(args)
gen <- matrix(args,nrow=2,ncol=2)
id <- matrix(c(1,0,0,1),nrow=2,ncol=2)
main <- function(A){
i <- 1
matrixlist <- list()
if(det(A) == 0){
stop("Matrix is not invertible", call.=FALSE)
}
else{
while(!(identical(A%^%i,id))){
matrixlist[[i]] <- A%^%i
i <- i+1
}
if(identical(A%^%i,id)){
matrixlist[[i]] <- A%^%i
}
print(matrixlist)
sprintf("Order of generator = %d",i)
}
}
main(gen)
OK, so my initial thought (numeric precision) was off.
The reason is that args is a character vector. So when you save temp[1]/temp[2] (a number) back into args, it gets coerced to a character.
This means it'll get cut off at whatever your display width is, e.g. for me args[2] will be "0.333333333333333". Of course when this is converted back to numeric that's the number you get, which is not the same as 1/3. (I found this out by adding a print(args) before the as.numeric to see what was happening).
So, the solution is to create a different vector - a numeric one - to hold the numeric values of the arguments (to prevent them from being coerced to string).
e.g.
# convert to numeric, suppressing warnings for fractions which won't convert
n.args <- suppressWarnings(as.numeric(args))
Then in your loop, save the numeric value to n.args not args.
for(i in 1:length(args)){
if(length(grep("/",args[i])) != 0){
temp <- as.integer(unlist(strsplit(args[i],"/")))
n.args[i] <- temp[1]/temp[2] # <--- here
}
}
In the rest of your script use n.args not `args.
Then it should be OK.
(Aside:
The line after the while loop seems to double up a bit with the while loop itself. I think by adjusting the start/end conditions you may be able to avoid this.
you are calculating A %^% i many times for the same i. For 2x2 matrices I guess this is not very expensive, but it might become so with larger ones. Consider storing A %^% i for each loop and then just referring to the stored value, rather than re-calculating. You already store it in matrixlist[[i]], so just use that rather than A %^% i for subsequent calls.
e.g in main.
i <- 1
matrixlist <- list(A)
if (det(A) == 0) {
stop("Matrix is not invertible", call.=FALSE)
} else {
while (!identical(matrixlist[[i]], id)) {
i = i + 1
matrixlist[[i]] <- A %^% i
}
print(matrixlist)
sprintf("Order of generator = %d", i)
}
For an assignment I had to create a random vector theta, a vector p containing for each element of theta the associated probability, and another random vector u. No problems thus far, but I'm stuck with the next instruction which I report below:
Generate a vector r1 that has a 1 in position i if pi ≥ ui and 0 if pi < ui. The
vector r1 is a Rasch item given the latent variable theta.
theta=rnorm(1000,0,1)
p=(exp(theta-1))/(1+exp(theta-1))
u=runif(1000,0,1)
I tried the following code, but it doesn't work.
r1<-for(i in 1:1000){
if(p[i]<u[i]){
return("0")
} else {
return("1")}
}
You can use the ifelse function:
r1 <- ifelse(p >= u, 1, 0)
Or you can simply convert the logical comparison into a numeric vector, which turns TRUE into 1 and FALSE into 0:
r1 <- as.numeric(p >= u)
#DavidRobinson gave a nice working solution, but let's look at why your attempt didn't work:
r1<-for(i in 1:1000){
if(p[i]<u[i]){
return("0")
} else {
return("1")}
}
We've got a few problems, biggest of which is that you're confusing for loops with general functions, both by assigning and using return(). return() is used when you are writing your own function, with function() <- .... Inside a for loop it isn't needed. A for loop just runs the code inside it a certain number of times, it can't return something like a function.
You do need a way to store your results. This is best done by pre-allocating a results vector, and then filling it inside the for loop.
r1 <- rep(NA, length(p)) # create a vector as long as p
for (i in 1:1000) {
if (p[i] < u[i]) { # compare the ith element of p and u
r1[i] <- 0 # put the answer in the ith element of r1
} else {
r1[i] <- 1
}
}
We could simplify this a bit. Rather than bothering with the if and the else, you could start r1 as all 0's, and then only change it to a 1 if p[i] >= u[i]. Just to be safe I think it's better to make the for statement something like for (i in 1:length(p)), or best yet for (i in seq_along(p)), but the beauty of R is how few for loops are necessary, and #DavidRobinson's vectorized suggestions are far cleaner.
I'm very new to R, and much more new to programming in R. I have the following question and its answer (which is not mine). I've trying to understand why some values, from where they are obtained, why they are used, etc.
Question: Make the vector 3 5 7 9 11 13 15 17 with a for loop. Start
with x=numeric() and fill this vector with the for loop
I know I have to create x=numeric() so I can fill it with the result obtained from the loop.
The answer from a classmate was:
> x <- numeric()
> for(i in 1:8){
if(i==1){ ## Why ==1 and not 0, or any other value
x[i] <- 3
}else{
x[i] <- x[i-1]+2 ### And why i-1
}
I'm having similar problems in questions like:
Make a for loop that adds the second element of a vector to the first,
subtracts the third element from the result, adds the fourth again and
so on for the entire length of the vector
So far, I created the vector and the empty vector
> y = c(5, 10, 15, 20, 25, 30)
> answer <- 0
And then, when I try to do the for loop, I get stuck here:
for(i in 1:length(y)){
if(i...){ ### ==1? ==0?
answer = y[i] ###and here I really don't know how to continue.
}else if()
}
Believe me when I tell you I've read several replies to questions here, like in How to make a vector using a for loop, plus pages and pages about for loop, but cannot really figure how to solve these (and other) problems.
I repeat, I'm very new, so I'm struggling trying to understand it. Any help would be much appreciated.
First, I will annotate the loop to answer what the loop is doing.
# Initialize the vector
x <- numeric()
for(i in 1:8){
# Initialize the first element of the vector, x[1]. Remember, R indexes start at 1, not 0.
if(i==1){
x[i] <- 3
} else {
# Define each additional element in terms of the previous one (x[i - 1]
# is the element of x before the current one.
x[i] <- x[i-1]+2 ### And why i-1
}
}
A better solution that uses a loop and grows it (like the instructions state) is something like this:
x <- numeric()
for(i in 1:8){
x[i] <- 2 * i + 1
}
This is still not a good way to do things because growing a vector inside a loop is very slow. To fix this, you can preallocate the vector by telling numeric the length of the vector you want:
x <- numeric(8)
The best way to solve this would be:
2 * 1:8 + 1
using vectorized operations.
To help you solve your other problem, I suggest writing out each step of the loop as a table. For example, for my solution, the table would be
i | x[i]
------------------
1 | 2 * 1 + 1 = 3
2 | 2 * 2 + 1 = 5
and so on. This will give you an idea of what the for loop is doing at each iteration.
This is intentionally not an answer because there are better ways to solve the alternating sign summation problem than a for-loop. I suppose there could be value in getting comfortable with for-loops but the vectorized approaches in R should be learned as well. R has "argument recycling" for many of its operations, including the "*" (multiplication) operation: Look at:
(1:10)*c(1,-1)
Then take an arbitrary vector, say vec and try:
sum( vec*c(1,-1) )
The more correct answer after looking at that result would be:
vvec[1] + sum( vec[-1]*c(1,-1) )
Which has the educational advantage of illustrating R's negative indexing. Look up "argument recycling" in your documentation. The shorter objects are automagically duplicatied/triplicated/however-many-needed-cated to exactly match the length of the longest vector in the mathematical or logical expression.
I have got a column with different numbers (from 1 to tt) and would like to use looping to perform a count on the occurrence of these numbers in R.
count = matrix(ncol=1,nrow=tt) #creating an empty matrix
for (j in 1:tt)
{count[j] = 0} #initiate count at 0
for (j in 1:tt)
{
for (i in 1:N) #for each observation (1 to N)
{
if (column[i] == j)
{count[j] = count[j] + 1 }
}
}
Unfortunately I keep getting this error.
Error in if (column[i] == j) { :
missing value where TRUE/FALSE needed
So I tried:
for (i in 1:N) #from obs 1 to obs N
if (column[i] = 1) print("Test")
I basically got the same error.
Tried to do abit research on this kind of error and alot have to said about "debugging" which I'm not familiar with.
Hopefully someone can tell me what's happening here. Thanks!
As you progress with your learning of R, one feature you should be aware of is vectorisation. Many operations that (in C say) would have to be done in a loop, can be don all at once in R. This is particularly true when you have a vector/matrix/array and a scalar, and want to perform an operation between them.
Say you want to add 2 to the vector myvector. The C/C++ way to do it in R would be to use a loop:
for ( i in 1:length(myvector) )
myvector[i] = myvector[i] + 2
Since R has vectorisation, you can do the addition without a loop at all, that is, add a scalar to a vector:
myvector = myvector + 2
Vectorisation means the loop is done internally. This is much more efficient than writing the loop within R itself! (If you've ever done any Matlab or python/numpy it's much the same in this sense).
I know you're new to R so this is a bit confusing but just keep in mind that often loops can be eliminated in R.
With that in mind, let's look at your code:
The initialisation of count to 0 can be done at creation, so the first loop is unnecessary.
count = matrix(0,ncol=1,nrow=tt)
Secondly, because of vectorisation, you can compare a vector to a scalar.
So for your inner loop in i, instead of looping through column and doing if column[i]==j, you can do idx = (column==j). This returns a vector that is TRUE where column[i]==j and FALSE otherwise.
To find how many elements of column are equal to j, we just count how many TRUEs there are in idx. That is, we do sum(idx).
So your double-loop can be rewritten like so:
for ( j in 1:tt ) {
idx = (column == j)
count[j] = sum(idx) # no need to add
}
Now it's even possible to remove the outer loop in j by using the function sapply:
sapply( 1:tt, function(j) sum(column==j) )
The above line of code means: "for each j in 1:tt, return function(j)", an returns a vector where the j'th element is the result of the function.
So in summary, you can reduce your entire code to:
count = sapply( 1:tt, function(j) sum(column==j) )
(Although this doesn't explain your error, which I suspect is to do with the construction or class of your column).
I suggest to not use for loops, but use the count function from the plyr package. This function does exactly what you want in one line of code.