I am comparing a 20-day moving average against a 50-day, 100-day 200-day and 333-day. The condition is essentially just
if(20MA > 50MA > 100MA > 200MA > 333MA) {
return TRUE
} else{
FALSE
}
Is there a way in R to handle this without multiple nested if statements?
if(tail(MA_20,n=1) > tail(MA_50,n=1) > tail(MA_100,n=1) > tail(MA_200,n=1) > tail(MA_333,n=1)) {
score[1] <- 1
} else{
score[1] <- -1
}
Use && (see help("&") for other logical operators)
if (tail(MA_20,n=1) > tail(MA_50,n=1) &&
tail(MA_50,n=1) > tail(MA_100,n=1) &&
tail(MA_100,n=1) > tail(MA_200,n=1) &&
tail(MA_200,n=1) > tail(MA_333,n=1)) {
score[1] <- 1
} else {
score[1] <- -1
}
The if statement will return true if all four of those comparisons are true.
The ifelse function might be helpful for you in speeding up your code as it helps vectorize comparisons.
The following way seems more complicated but is more flexible. It first gets all variables named after the pattern "MA_" and sapply the tail to extract the last element. Then uses diff to see if they are in decreasing order.
First make up some data.
library(zoo)
set.seed(1234) # Reproducible results
n <- 1e3
x <- rnorm(n)
MA_20 <- rollmean(x, k = 20)
MA_50 <- rollmean(x, k = 50)
MA_100 <- rollmean(x, k = 100)
MA_333 <- rollmean(x, k = 333)
Now the problem.
score <- NULL
ma <- stringr::str_sort(ls(pattern = "^MA_"), numeric = TRUE)
MA_last <- sapply(ma, function(m) tail(get(m), n = 1))
score[1] <- if(all(diff(MA_last) > 0)) 1 else -1
Related
Consider the below MWE that splits a distance matrix and attempts to compare partitions:
set.seed(1234) # set random seed for reproducibility
# generate random normal variates
x <- rnorm(5)
y <- rnorm(5)
df <- data.frame(x, y) # merge vectors into dataframe
d <- dist(x) # generate distance matrix
splt <- split(d, 1:5) # split data with 5 values in each partition
# compare partitions
for (i in 1:length(splt)) {
for (j in 1:length(splt)) {
if(splt[[i]] != splt[[j]]) {
a <- length(which(splt[[i]] >= min(splt[[j]]))) / length(splt[[i]])
b <- length(which(splt[[j]] <= max(splt[[i]]))) / length(splt[[j]])
}
}
}
# Error in if (splt[[i]] != splt[[j]]) { : the condition has length > 1
The above for loop should compare all unique partitions (i.e., (1, 2), (1, 3), ... ,(4, 5)). However, the condition is greater than 1.
The result for comparing partition 1 (split[[1]]) and partition 2 (split[[2]]) for instance should be a = b = 1.
a <- length(which(splt[[1]] >= min(splt[[2]]))) / length(splt[[1]])
b <- length(which(splt[[2]] <= max(splt[[1]]))) / length(splt[[2]])
I know the solution is to instead use ifelse() but there is no else within the nested loop.
Any ideas on how to proceed?
Is your problem the error message? That is, why R does not like your comparison splt[[i]] == splt[[j]]? The reason is that we get a vector of comparisons:
> splt[[1]] != splt[[2]]
[1] TRUE TRUE
If I understand you correctly, splt[[i]] is equal to splt[[j]] if all entries are equal and different otherwise. If so, change the comparison to be !(all(splt[[i]] == splt[[j]])).
The total loop looks like this:
for (i in 1:length(splt)) {
for (j in 1:length(splt)) {
if (!(all(splt[[i]] == splt[[j]]))) {
a <- length(which(splt[[i]] >= min(splt[[j]]))) / length(splt[[i]])
b <- length(which(splt[[j]] <= max(splt[[i]]))) / length(splt[[j]])
}
}
}
I am learning about loops and I have this code to check if a number is prime or not, but doesn't work. Where is the bug?
x <- 7
y <- seq(1,sqrt(x),by=1)
for(i in 1: sqrt(x)){
if(y[x%%y == 0]) {
print("FALSE")
}else{
print("TRUE")
}
}
This gives me the right solution, but it repeats the answer as many times as number of elements in i. Also I would like to ask how to use function inside a for with if:
i <- c(1: sqrt(x))
y3 <- x%%i == 0
y4 <- y3[-1]
for(value in i){
if(y4 == FALSE) {
print("TRUE")
}else{
print("FALSE")
}
}
version 3, gives me the solution but for evey element in i:
x <- 107
i <- c(1: sqrt(x))
y3 <- c(x%%i == 0)
y4 <- y3[-1]
for(value in i){
if(all(y4==F)) {
print("TRUE")
}else{
print("FALSE")
}
}
Since you mentioned that you must use a loop, the following code will work:
x <- 7
y <- seq(1, ceiling(sqrt(x)), by=1)
# is.factor is a vector which checks whether each element in y is a factor or not
# initialize to F
is.factor = F
# Start at y = 2 because 1 will be a factor
for(i in 2:length(y) ){
# Check whether current value in vector is a factor of x or not
# if not a factor, set value in index to F, otherwise set to T
ifelse( x%%y[i] != 0, is.factor[i] <- F, is.factor[i] <- T)
# If we are at the last element in y, print a result
if(i == length(y)){
# check if we have any factors.
# if we have any factors (i.e. any index in is.factor vector is T), then number is not prime
ifelse( any(is.factor), print("FALSE"), print("TRUE") )
}
}
You can do this-
check_prime <- function(num) {
if (num == 2) {
TRUE
} else if (any(num %% 2:(num-1) == 0)) {
FALSE
} else {
TRUE
}
}
> check_prime(7)
[1] TRUE
I wrote the following code trying to find all the prime numbers from a random generated data set. sadly it seems something went wrong, could anybody help me.
set.seed(20171106)
n <- 10000
num <- sample(1:100000,n,replace=TRUE)
findPrime <- function(x){
apple<-c()
n<-length(x)
for(i in n){
if(any(x[i]%%(1:(x[i]-1))!=0)) apple <-c(apple,x[i])
}
return(apple)
}
To get results:
type:findPrime(num)
This is the warning message:
Warning message:
In if (x[i]%%(1:(x[i] - 1)) == 0) apple <- c(apple, x[i]) :
the condition has length > 1 and only the first element will be used
so how can I fix the problem?
if statements only accept single elements and in your declaration seems to get the whole vector. I have rewritten your function using a ifelse expression wrapped inside a sapply loop.
I hope this works for you.
findPrime <- function(x = 0){
primes <- c()
# Prime finder
primes <- sapply(X = x,FUN = function(x) {
ifelse(any(x %% (1:(x - 1)) != 0), T, F)}
)
# Select primes
primes <- num[primes]
return(primes)
}
findPrime(num)
I have checked another silly mistake... Inside the function change num for x in the select primes step and invert the F, T outcomes. It should look like this:
findPrime <- function(x = 0){
primes <- c()
# Prime finder
primes <- sapply(X = x,FUN = function(x) {
ifelse(any(x %% (2:(x - 1)) == 0), F, T)}
)
# Select primes
primes <- x[primes]
return(primes)
}
I have just tried it and it works fine.
use package "gmp" which has a function "isprime" which returns 0 for non prime numbers and 2 for prime numbers and then subset the data based on the same
say you have a vector a = c(1:10)
a = c(1:10)
b = gmp::isprime(a)
c = cbind(a,b)
c = as.data.frame(c)
c = c[c$b==2,]
a1 = c$a
a1
In your code: for(i in 1:n), there is the error
I am trying to generate a random sample that excludes certain "bad data." I do not know whether the data is "bad" until after I sample it. Thus, I need to make a random draw from the population and then test it. If the data is "good" then keep it. If the data is "bad" then randomly draw another and test it. I would like to do this until my sample size reaches 25. Below is a simplified example of my attempt to write a function that does this. Can anyone please tell me what I am missing?
df <- data.frame(NAME=c(rep('Frank',10),rep('Mary',10)), SCORE=rnorm(20))
df
random.sample <- function(x) {
x <- df[sample(nrow(df), 1), ]
if (x$SCORE > 0) return(x)
#if (x$SCORE <= 0) run the function again
}
random.sample(df)
Here is a general use of a while loop:
random.sample <- function(x) {
success <- FALSE
while (!success) {
# do something
i <- sample(nrow(df), 1)
x <- df[sample(nrow(df), 1), ]
# check for success
success <- x$SCORE > 0
}
return(x)
}
An alternative is to use repeat (syntactic sugar for while(TRUE)) and break:
random.sample <- function(x) {
repeat {
# do something
i <- sample(nrow(df), 1)
x <- df[sample(nrow(df), 1), ]
# exit if the condition is met
if (x$SCORE > 0) break
}
return(x)
}
where break makes you exit the repeat block. Alternatively, you could have if (x$SCORE > 0) return(x) to exit the function directly.
use this after your first sample
while (any(bad <- (x$SCORE <= 0)))
x[bad, ] <- df[sample(nrow(df), sum(bad)), ]
You can just select the rows to sample directly like so (just 5):
> df <- data.frame(NAME=c(rep('Frank',10),rep('Mary',10)), SCORE=rnorm(20))
> df[sample(which(df$SCORE>0), 5),]
NAME SCORE
14 Mary 1.0858854
10 Frank 0.7037989
16 Mary 0.7688913
5 Frank 0.2067499
17 Mary 0.4391216
this is without replacement, for bootstrap put in replace=T.
random.sample <- function(x) {
x <- df[sample(nrow(df), 1), ]
if (x$SCORE > 0) return(x)
Recall(x)# run the function again
}
random.sample(df)
# NAME SCORE
#14 Mary 1.252566
It seems to me that this should work as well:
df$SCORE[ df$SCORE > 0 ][ sample(1:sum(df$SCORE > 0), 1) ]
#[1] 0.6579631
I want to recursively count the log cylces in my function
logCounter <- function(number) {
k <- 0
if(k>=0){
k = k+1
}
result <- log(number)
if (result > 1) {
logCounter(result)
} else {
return(k)
}
}
logCounter(123)#returns 3 because log(log(log(123))) < 1
However, my counter k does not work as I would have inspected. Therefore I really would appreciate your answer!!!
You don't need to use Recall. Try this:
logCounter <- function(number) {
if (number <1) return(0) # A minor edit.
result <- log(number)
if (result > 1) return(logCounter(result)+1)
return(1)
}
The key is to try to compose your function in a way that doesn't require storing intermediate results.
You could do this much more easily without calling the function recursively with a while loop:
logCounter <- function(number) {
k <- 0
result <- number
while(result>1){
k <- k + 1
result <- log(result)
}
return(k)
}
> logCounter(123)
[1] 3
EDIT: If you need to use recursion, consider the Recall function:
logCounter <- function(number, iter=1) {
if(log(number)>1)
out <- Recall(log(number), iter+1)
else
out <- list(log(number),iter)
return(out)
}
> logCounter(123)
[[1]]
[1] 0.4518085
[[2]]
[1] 3