r - for loop returns all zeroes when segmenting a beta distribution












0















I am trying to segment a beta distribution, but the for loop returns a vector of zeroes:



my_vector <- vector("numeric", 11L)
prob <- seq(from = 0.0, to = 1, by = 0.1)
for (i in prob ) {
my_vector[i] <- dbeta(i, shape1=7, shape2=2)
}
my_vector
[1] 0 0 0 0 0 0 0 0 0 0 0









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  • 1





    my_vector <- dbeta(prob, shape1=7, shape2=2) You can not indexing with doubles (my_vector[i])! e.g. my_vector[0.1]

    – jogo
    Nov 25 '18 at 17:33













  • @jogo thank you!

    – Nottolina
    Nov 25 '18 at 17:52
















0















I am trying to segment a beta distribution, but the for loop returns a vector of zeroes:



my_vector <- vector("numeric", 11L)
prob <- seq(from = 0.0, to = 1, by = 0.1)
for (i in prob ) {
my_vector[i] <- dbeta(i, shape1=7, shape2=2)
}
my_vector
[1] 0 0 0 0 0 0 0 0 0 0 0









share|improve this question


















  • 1





    my_vector <- dbeta(prob, shape1=7, shape2=2) You can not indexing with doubles (my_vector[i])! e.g. my_vector[0.1]

    – jogo
    Nov 25 '18 at 17:33













  • @jogo thank you!

    – Nottolina
    Nov 25 '18 at 17:52














0












0








0








I am trying to segment a beta distribution, but the for loop returns a vector of zeroes:



my_vector <- vector("numeric", 11L)
prob <- seq(from = 0.0, to = 1, by = 0.1)
for (i in prob ) {
my_vector[i] <- dbeta(i, shape1=7, shape2=2)
}
my_vector
[1] 0 0 0 0 0 0 0 0 0 0 0









share|improve this question














I am trying to segment a beta distribution, but the for loop returns a vector of zeroes:



my_vector <- vector("numeric", 11L)
prob <- seq(from = 0.0, to = 1, by = 0.1)
for (i in prob ) {
my_vector[i] <- dbeta(i, shape1=7, shape2=2)
}
my_vector
[1] 0 0 0 0 0 0 0 0 0 0 0






r for-loop






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share|improve this question











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asked Nov 25 '18 at 17:22









NottolinaNottolina

7511




7511








  • 1





    my_vector <- dbeta(prob, shape1=7, shape2=2) You can not indexing with doubles (my_vector[i])! e.g. my_vector[0.1]

    – jogo
    Nov 25 '18 at 17:33













  • @jogo thank you!

    – Nottolina
    Nov 25 '18 at 17:52














  • 1





    my_vector <- dbeta(prob, shape1=7, shape2=2) You can not indexing with doubles (my_vector[i])! e.g. my_vector[0.1]

    – jogo
    Nov 25 '18 at 17:33













  • @jogo thank you!

    – Nottolina
    Nov 25 '18 at 17:52








1




1





my_vector <- dbeta(prob, shape1=7, shape2=2) You can not indexing with doubles (my_vector[i])! e.g. my_vector[0.1]

– jogo
Nov 25 '18 at 17:33







my_vector <- dbeta(prob, shape1=7, shape2=2) You can not indexing with doubles (my_vector[i])! e.g. my_vector[0.1]

– jogo
Nov 25 '18 at 17:33















@jogo thank you!

– Nottolina
Nov 25 '18 at 17:52





@jogo thank you!

– Nottolina
Nov 25 '18 at 17:52












1 Answer
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If you want to do it using the for loop, you may do it like this:



my_vector <- vector("numeric", 0)
prob <- seq(from = 0.0, to = 1, by = 0.1)
for (i in prob) {
my_vector <- c(my_vector, dbeta(i, shape1=7, shape2=2))
}

my_vector
[1] 0.0000000 0.0000504 0.0028672 0.0285768 0.1376256 0.4375000 1.0450944 1.9765032 2.9360128 2.9760696 0.0000000


Or you better follow the suggestion of @ jogo by doing it like this:



prob <- seq(from = 0.0, to = 1, by = 0.1)
my_vector <- dbeta(prob, shape1=7, shape2=2)





share|improve this answer

























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    1 Answer
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    1














    If you want to do it using the for loop, you may do it like this:



    my_vector <- vector("numeric", 0)
    prob <- seq(from = 0.0, to = 1, by = 0.1)
    for (i in prob) {
    my_vector <- c(my_vector, dbeta(i, shape1=7, shape2=2))
    }

    my_vector
    [1] 0.0000000 0.0000504 0.0028672 0.0285768 0.1376256 0.4375000 1.0450944 1.9765032 2.9360128 2.9760696 0.0000000


    Or you better follow the suggestion of @ jogo by doing it like this:



    prob <- seq(from = 0.0, to = 1, by = 0.1)
    my_vector <- dbeta(prob, shape1=7, shape2=2)





    share|improve this answer






























      1














      If you want to do it using the for loop, you may do it like this:



      my_vector <- vector("numeric", 0)
      prob <- seq(from = 0.0, to = 1, by = 0.1)
      for (i in prob) {
      my_vector <- c(my_vector, dbeta(i, shape1=7, shape2=2))
      }

      my_vector
      [1] 0.0000000 0.0000504 0.0028672 0.0285768 0.1376256 0.4375000 1.0450944 1.9765032 2.9360128 2.9760696 0.0000000


      Or you better follow the suggestion of @ jogo by doing it like this:



      prob <- seq(from = 0.0, to = 1, by = 0.1)
      my_vector <- dbeta(prob, shape1=7, shape2=2)





      share|improve this answer




























        1












        1








        1







        If you want to do it using the for loop, you may do it like this:



        my_vector <- vector("numeric", 0)
        prob <- seq(from = 0.0, to = 1, by = 0.1)
        for (i in prob) {
        my_vector <- c(my_vector, dbeta(i, shape1=7, shape2=2))
        }

        my_vector
        [1] 0.0000000 0.0000504 0.0028672 0.0285768 0.1376256 0.4375000 1.0450944 1.9765032 2.9360128 2.9760696 0.0000000


        Or you better follow the suggestion of @ jogo by doing it like this:



        prob <- seq(from = 0.0, to = 1, by = 0.1)
        my_vector <- dbeta(prob, shape1=7, shape2=2)





        share|improve this answer















        If you want to do it using the for loop, you may do it like this:



        my_vector <- vector("numeric", 0)
        prob <- seq(from = 0.0, to = 1, by = 0.1)
        for (i in prob) {
        my_vector <- c(my_vector, dbeta(i, shape1=7, shape2=2))
        }

        my_vector
        [1] 0.0000000 0.0000504 0.0028672 0.0285768 0.1376256 0.4375000 1.0450944 1.9765032 2.9360128 2.9760696 0.0000000


        Or you better follow the suggestion of @ jogo by doing it like this:



        prob <- seq(from = 0.0, to = 1, by = 0.1)
        my_vector <- dbeta(prob, shape1=7, shape2=2)






        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited Nov 26 '18 at 15:11

























        answered Nov 26 '18 at 1:15









        TeeKeaTeeKea

        3,22851932




        3,22851932
































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