y <- as.integer(readline(prompt ="Enter a number: "))
factorial = 1
if (y< 0){
print("Error")
} else if (y== 0)
{
print("1")
} else
{
for(i in 1:y) {
factorial = factorial * i
}
return(factorial)
}
wondering why this is giving:
Error in if (y< 0) { : missing value where TRUE/FALSE needed
is it cause the first line has data type NA_integer?
There are three possible ways to pass values to the if statement.
y <- 1
if (y > 0) print("more")
This one works as expected.
y <- 1:3
if (y > 0) print("ignores all but 1st element")
As the warning message will tell you, only the first element was used to evaluate it. You could use any or all to make this right.
y <- NA
if (y > 0) print("your error")
This case actually gives you your error. I would wager a bet that y is somehow NA. You will probably need to provide a reproducible example (with data and the whole shebang) if you'll want more assistance. Note also that it helps visually structure your code to improve readability.
Related
First post ever, excuse my ignorance as I'm relatively new to programming. I'm trying to create a function that samples a number given by x and returns "number too small" below 5 and "number too big" if its above 10. I already figured this part out, however my issue is that I want to add an optional len argument that will display the length of x if used and if not will still function as intended. I'm sure its possible if I just do a bunch of if else statements with every possible scenario, but is there a more organized way to achieve my goal. Was thinking that I can make another if/else that returns null if they opt out of the second len argument but unsure on how i would code it as I already have two if or else statements.
vector.maker.num2 <-function(x){
generator <-sample(1:20, x)
if (x < 5) {
print(c(generator, "warning, number too low"))
}
if (x > 10) {
print(c(generator, "warning, number too high"))
}
}
vector.maker.num2(4)
It sounds like what you want is the missing function.
vector.maker.num2 <-function(x, len){
generator <-sample(1:20, x)
if (!missing(len)) {
print(length(x))#Will usually print 1, because x is a vector of length 1.
}
if (x < 5) {
print(c(generator, "warning, number too low"))
}
if (x > 10) {
print(c(generator, "warning, number too high"))
}
}
vector.maker.num2(4)
I am trying reproduce the Table 1 results from the page 12 using the equation given the page 13.To access the journal article please click https://arxiv.org/pdf/math/0605322.pdf. The corresponding equation is given below.
My r code is give below. Am I programmed correctly?
mytest=function(n,s,c1){
t = sum(s)
k=which.max(s[19:n]>=c1)
if(k==1 && s[19]<c1)
return(c(n,0))
else
return(c(k,1))
}
for (n in c(100,200,400)){
for (i in c(-0.5, -1.0)){
a1=0
c1 = 20
asn1=0
for (m in 1:1000){
g=c(dnorm(n,0,1))
f=c(dnorm(n,i,1))
s = log(g/f)
test=mytest(n,s,c1)
a1=a1+test[2]
asn1=asn1+test[1]
}
}
out <- list(power= a1/m, asn=asn1/m)
return(out)
}
But I am getting the following errors.
Error in if (k == 1 && s[19] < c1) return(c(n, 0)) else return(c(k, 1)) :
missing value where TRUE/FALSE needed
The first time you call mytest, you have n=100, i=-0.5 which yields s=NaN. Therefore, you get an error on line if(k==1 && s[19]<c1) given that s[19]=NaN.
Here's a workaround, but you need to make sure it does what you expect/wish :
mytest=function(n,s,c1){
if(is.na(s)) return(c(c1,1)) # skips if NaN
t = sum(s)
k=which.max(s[19:n]>=c1)
if(k==1 && s[19]<c1)
return(c(n,0))
else
return(c(k,1))
}
I made a function to to compute the sum of I(Xi
my.ecdf<- function(x,y) {
if(!is.null(dim(y)))
stop("y has more than one dimension")
n<-length(x)
i<-1:n
p<-if(x[i]<y) 1 else {
0
}
(sum(p))/n
}
But when I run it with input (rnorm(11),6), I get this error:
Warning message:
In if (x[i] < y) 1 else { :
the condition has length > 1 and only the first element will be used
Any ideas? I'm new to r so sorry if it's something obvious. (Also I don't want to use the for loop)
There are a number of issues in your code:
1) Whats the point of x[1:length(x)] in the if statement? Right now these are meaningless and can be dropped:
n<-length(x)
i<-1:n
x[i]
2) If statement accepts a logical argument not a vector of logical, you can consider adding all() any() etc like
if(all(x < y)) 1 else {0}
or use ifelse() statement for the assignment
3) Finally from what I can understand you overcomplicate things and the whole thing can be written as one-liner:
sum(x < y)/length(x)
This is a logical vector of the same length as y
is.null(dim(y))
You're using it as a logical test. An object with a length greater than 1 can't be unambiguously interpreted by the if statement. Consider if (TRUE FALSE FALSE TRUE) <do something>. When should you do that thing?
If you want to make sure y doesn't have more than one dimension, do
if(length(dim(y)) > 1){
stop("message")
}
I did some programming work on R language to do the bubble sort. Sometimes it works perfectly without any error message, but sometimes, it shows "Error in if (x[i] > x[i + 1]) { : argument is of length zero". Can any one help me check whats wrong with it? I have attached my code below
example <- function(x) {
n <- length(x)
repeat {
hasChanged <- FALSE
n <- n - 1
for(i in 1:n) {
if ( x[i] > x[i+1] ) {
temp <- x[i]
x[i] <- x[i+1]
x[i+1] <- temp
hasChanged <- TRUE
cat("The current Vector is", x ,"\n")
}
}
if ( !hasChanged ) break;
}
}
x <-sample(1:10,5)
cat("The original Vector is", x ,"\n")
example(x)
The error occurs because you are iteratively decreasing n. Depending on the original vector's order (or lack thereof), n can reach the value of 1 after the last change. In that case, a further reduction of n in the next iteration step addresses the value x[0], which is undefined.
With a minimal correction your code will work properly, without giving error messages. Try to replace the line
if ( !hasChanged ) break;
with
if ( !hasChanged | n==1 ) break
Basically you have two termination criteria: Either nothing has been changed in the previous iteration or n is equal to one. In both cases, a further iteration won't change the vector since it is already ordered.
By the way, in R programming you don't need a semicolon at the end of a command. It is tolerated/ignored by the interpreter, but it clutters the code and is not considered good programming style.
Hope this helps.
Please can anyone advise how I can turn the following statement into one that will do the same thing but NOT using ifelse please?
<-ifelse(y>=50, 0.2*x+0.8*y, ifelse(y<50 & x>70, y+10, ifelse(y<50 & x<70, y)))
x=80
y=60
So I the final code should give an answer of 64 - selecting the first condition. I will then test it to ensure the other 3 conditions give the correct result for varying values of x and y
Thanks a lot.
This should work:
finalmark <- (x * 0.2 + y * 0.8) * (y >= 50) + (y + 10 * (x > 70)) * (y < 50)
Something like this?
if(y>=50){
0.2*x+0.8*y
}else{
if(y<50 & x>70){
y+10
}else{
if(y<50 & x<70){
y
}else{
"OMG I did not expect this scenario"
}
}
}
try: y=45; x=70 to see why I have the last condition.
If y is a number then, once you've tested for y > = 50 then y must be less than 50 so don't keep testing for that. Similarly, once you've found x > 70 then you don't need the last ifelse. You don't have a return for x = 70. My guess is that you want to test for a <= or >= situation there.
ifelse(y>=50, 0.2*x+0.8*y, ifelse(x>70, y+10, y))
in scalar that's
if(y >= 50){
0.2*x+0.8*y
}else if(x > 70){
y+10
}else y
Given you seem to be having a hard time in general writing the logic I suggest you post a more complete question. It's possible (probable) that you're doing something here that you really don't want to do.
There are several approaches you can take. Below are a few examples of building a function 'f', so that 'f(x,y)' meets your criteria listed in the question using logic other than 'ifelse' statements.
Note: I'm also adding in one amendment to the original post, since 'x=70' would break the logic. I'm adding 'x>=70' to the second criterion.
Option 1: Use a standard 'if / else if / else' logic block. Personally, I like this option, because it's easily readable.
f <- function(x, y){
if (y>= 50){
return(0.2*x+0.8*y)
} else if (y < 50 & x >= 70){
return(y+10)
} else {
return(y)
}
}
Option 2: Combine your two logical tests (there are really only two) into a string, and use a switch. Note that the final and unnamed option is treated as an 'else'.
f <- function(x, y){
return(
switch(paste(x >= 70, y >= 50, sep=""),
TRUEFALSE = y + 10,
FALSEFALSE = y,
0.2*x+0.8*y
)
)
}
Option 3: Order your 'if' statements to reduce logical comparisons. This is the sort of thing to do if you have a large data set or very limited memory. This is slightly harder to troubleshoot, since you have to read the whole block to fully understand it. Option 1 is better if you don't have memory or cycle limitations.
f <- function(x, y){
if (y >= 50){
return(0.2*x+0.8*y)
} else {
if (x >=70){
return(y+10)
} else {
return(y)
}
}
}
There are other options, but these are the simplest that come readily to mind.