Creating many sets of random data

user_17170943 0
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I wrote a function to test if a mean is within a confidence interval, and wanted to test out this function using many sets of data.

checkInt <- function(x, mean=20){
  lower <- t.test(x)$conf.int[1]
  upper <- t.test(x)$conf.int[2]
  Within_interval <- ifelse(between(mean,lower,upper),1 , 0)
  width <- upper - lower
  return(matrix(Within_interval,width))
}

So I create a set of random exp data and wanted to apply my function on each row of this data, but keeps giving me weird results

many_set <- t((replicate(5, rexp(n=300, rate=1/20))))
t(apply(many_set , 1 , checkInt))
r
2 Answers

First, as far as I know, you can't return a matrix in your apply function checkInt, since your typeof(many_set) == 'double'. So, you need to calculate the width value outside, since it doesn't depend on the mean, and return a scalar variable.

Second, if you want to specify a matrix via apply to retrieve , one of the options would be to return a function within a function.

library(dplyr)

checkInt <- function(lower, upper){
  return(function(mean = 20) {
    Within_interval <- ifelse(between(mean,lower,upper), 1, 0)
    
    return(Within_interval)
  })
}

replicate <- replicate(5, rexp(n=300, rate=1/20))
many_set <- t((replicate(5, rexp(n=300, rate=1/20))))

lower <- t.test(many_set)$conf.int[1]
upper <- t.test(many_set)$conf.int[2]
width <- upper - lower

t(apply(many_set , 1 , checkInt(lower, upper)))
user_37856180

Do what @dcarlson suggested in the comment and use c() rather than matrix(). Moreover you can avoid the ifelse by transforming boolean of between as as.integer().

checkInt <- function(x, meanv=20){
  lower <- t.test(x)$conf.int[1]
  upper <- t.test(x)$conf.int[2]
  Within_interval <- as.integer(data.table::between(meanv, lower, upper))
  width <- upper - lower
  return(c(Within_interval, width))
}

set.seed(42)
many_set <- t(replicate(5, rexp(n=300, rate=1/20)))
t(apply(many_set, 1, checkInt))
#      [,1]     [,2]
# [1,]    0 5.333380
# [2,]    1 4.444028
# [3,]    1 4.995460
# [4,]    1 4.294835
# [5,]    1 4.369884
user_65740380
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