I need to create a vector with multiple inputs (integers) from user.
The intent is to create a list and verify if it has a mode and where is its median.
I am using this code:
ReadVector <- function()
{
x <- 0
while(x<16) {
n <- readline(prompt="Input one integer: ")
return(as.integer(n))
VectorUser <- c(n)
x <- x+1
}
print(VectorUser)
}
ReadVector()
And I can only get one integer, I dont know if my mistake is in the while loop or(and) in the concatenate command after it. Can you help me?
Does this work for you?
ReadVector <- function()
{
x <- 0
myvector = vector()
while(x<16) {
n <- readline(prompt="Input one integer: ")
myvector = c(myvector,n)
x <- x+1
}
return (as.integer(myvector))
}
You need yo save your values in a vector, and keep it (without returning inside the loop), until you completed it.
Hope it helps
ff=function(){
d=c()
while (TRUE){
int = readline('ENTER to quit > ')
if(nchar(int)==0) {
if(length(d)>0)cat("The numbers you entered are:",d)
else(cat("You did not enter any number!!"));break}
else{
value=suppressWarnings(as.integer(int))
if(!is.na(value)){cat(value);d=c(d,value)} else cat(ran[sample(6,1)])
}}
ff()
Related
I have tried the following but the output brings an argument stating,
Error in append("0") : argument "values" is miss
for (rowz in final_data$Ingridients) {
Cobalt_row<-lst()
if (sum(str_detect(rowz, 'Cobalt'))>0) {
Cobalt_row.append(1)
} else {
Cobalt_row<-append(0)
}
print(Cobalt_row)
}
I intended to loop through the list and generate a boolean of ones and twos depending on
whether or not I had the value.
Please help
Without the data, I can't test it, but this should work:
Cobalt_row<-lst()
k <- 1
for (rowz in final_data$Ingridients) {
Cobalt_row[[k]] <- ifelse(str_detect(rowz, 'Cobalt'), 1, 0)
k <- k+1
}
or even simpler if you need a list:
Cobalt_row <- as.list(as.numeric(str_detect(final_data$Ingredients, "Cobalt")))
I'm working with panel data in R and am endeavoring to build a function that returns every user ID where PCA==1. I've largely gotten this to work, with one small problem: it only returns the values when I end the function with print() but does not do so when I end the function with return(). As I want the ids in a vector so I can later subset the data to only include those IDs, that's a problem. Code reflected below - can anyone advise on what I'm doing wrong?
The version that works (but doesn't do what I want):
retrievePCA<-function(data) {
for (i in 1:dim(data)[1]) {
if (data$PCA[i] == 1) {
id<-data$CPSIDP[i]
print(id)
}
}
}
retrievePCA(data)
The version that doesn't:
retrievePCA<-function(data) {
for (i in 1:dim(data)[1]) {
if (data$PCA[i] == 1) {
id<-data$CPSIDP[i]
return(id)
}
}
}
vector<-retrievePCA(data)
vector
Your problem is a simple misunderstanding of what a function and returning from a function does.
Take the small example below
f <- function(x){
x <- x * x
return x
x <- x * x
return x
}
f(2)
[1] 4
4 is returned, 8 is not. That is because return exits the function returning the specific value. So in your function the function hits the first instance where PCA[i] == 1 and then exits the function. Instead you should create a vector, list or another alternative and return this instead.
retrievePCA<-function(data) {
ids <- vector('list', nrow(data))
for (i in 1:nrow(data)) {
if (data$PCA[i] == 1) {
ids[[i]] <-data$CPSIDP[i]
}
}
return unlist(ids)
}
However you could just do this in one line
data$CPSIDP[data$PCA == 1]
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])))
}
enter image description here
I know there exists function 'unique' which works similar to what I want to make, but I want to make this function.
I want this function finally returns 'result' which contains unique elements of input vector.
But I don't know why this function's result is totally different from my expect.
Why c which is to combine before result and new unique element is not working.
Please tell me how to fix my code.
Thank you.
I think what you expect might be something like below, where result should be an argument of m_uni:
m_uni <- function(x,result = c()) {
if (class(x)=='numeric'| class(x)=='character') {
if (length(x) <= 1){
return(result)
} else {
if (x[length(x)] %in% result) {
x <- x[-length(x)]
m_uni(x,result)
} else {
result <- c(result,x[length(x)])
x <- x[-length(x)]
m_uni(x,result)
}
}
} else {
return('This function only gets numeric or character vector')
}
}
such that
> m_uni(x)
[1] 0 4 5 -2
I want to make a program in which user will be asked to enter the location and based on that location value it should run a particular set of codes. It should wait until user enter the value of location.
readinteger <- function()
{
n <- readline(prompt="Enter your location: ")
n <- as.integer(n)
if (is.na(n))
return(as.integer(n))
}
LC <- readinteger()
if ( LC== 1)
{
print(x)
}
else if ( LC == 2)
{
print(y)
}
else
print(z)
But here it proceeds to if loop directly and then ask to enter the location
This will work fine if you include the if statements to control printing in your readinteger function. That way, readline will behave as expected (wait for input from the user) and then automatically move on to the printing commands.
readinteger <- function()
{
n <- readline(prompt="Enter your location: ")
n <- as.integer(n)
x <- "You are at location one."
y <- "You are at location two."
z <- "You are lost."
if ( n == 1)
{
print(x)
}
else if ( n == 2)
{
print(y)
}
else
print(z)
}
readinteger()
You had a if (is.na(n)) in your code that wasn't doing anything, but my guess is you want to include a check to make sure the user has provided valid input. If so, you may find a while loop helpful, so the user can correct their input if there's an error. For example:
readinteger <- function()
{
n <- NULL
while( !is.integer(n) ){
n <- readline(prompt="Enter your location: ")
n <- try(suppressWarnings(as.integer(n)))
if( is.na(n) ) {
n <- NULL
message("\nYou must enter an integer. Please try again.")
}
}
x <- "You are at location one."
y <- "You are at location two."
z <- "You are lost."
if ( n == 1)
{
print(x)
}
else if ( n == 2)
{
print(y)
}
else
print(z)
}
readinteger()
Ok. I think it's got something to do with running the whole file at one go.
It will work if you run it in Rsudio and run one step at a time, i.e., interactively.
According to the documentation of ?readline : `In non-interactive use the result is as if the response was RETURN and the value is "". In which case, one solution is to call the function again, as shown here : http://www.rexamples.com/4/Reading%20user%20input
I am not 100% sure it works without hitch.
For a more complex solution to your problem , you can see this question: Make readline wait for input in R
(Note: I'm adding this as answer simply for he sake of formatting. It's not really an answer )