[R] Issues when setting k to 1L

[Rui Barradas]

Hello,

Inliner.

Às 02:45 de 25/04/2022, Avi Gross via R-help escreveu:
> Try: average_prob_frame_6 <- rowMeans(prob_frame_6, na.rm=TRUE)
> Any NA propagates into calculations to make it NA.
> Then again, should there be any NA?

Yes, na.rm = TRUE is a good idea.
And no, there should be no NA's in prob_frame_6.


Rui Barradas

> 
> 
> 
> -----Original Message-----
> From: Paul Bernal <paulbernal07 using gmail.com>
> To: Rui Barradas <ruipbarradas using sapo.pt>; R <r-help using r-project.org>
> Sent: Sun, Apr 24, 2022 8:56 pm
> Subject: [R] Issues when setting k to 1L
> 
> Dear Rui,
> 
> I made the modification to k <- 1L (see the code below), but I get the
> following odd result (maybe I am forgetting to do something):
> print(final_frame_6)
>          p1 NA NA NA NA NA NA NA NA NA average_prob_frame_6
> 1 0.437738 NA NA NA NA NA NA NA NA NA                  NA
>> print(paste("The average probability of success when doing 1,000,000
> single trials is:", average_prob_frame_6))
> [1] "The average probability of success when doing 1,000,000 single trials
> is: NA"
>>
> 
> #single 1,000,000 trials
> 
> # these two are equal
> cnames0 <-
> c("Jan","Feb","Mar","Apr","May","Jun","Jul","Aug","Sep","Oct","Nov","Dec")
> cnames <- month.abb
> identical(cnames0, cnames)
> # [1] TRUE
> 
> # performing 1,000,000 simulations 10 times
> num_trials_6 <- 1e6
> dice_rolls_6 <- num_trials_6*12
> num_dice_6 <- 1
> dice_sides_6 <- 6
> 
> set.seed(2022)
> 
> prob_frame_6 <- as.data.frame(matrix(ncol = 10L, nrow = 1L))
> K <- 1L
> for(k in seq_len(K)){
>    #
>    dice_simul_6 <- sample(dice_sides_6, dice_rolls_6, replace = TRUE)
>    # constructing matrix containing results of all dice rolls by month
>    prob_matrix_6 <- matrix(dice_simul_6, ncol = 12, byrow = TRUE)
> 
>    # naming each column by it's corresponding month name
>    colnames(prob_matrix_6) <- month.abb
> 
>    # calculating column  which will have a 1
>    # if trial was successful and a 0 otherwise
>    success <- integer(num_trials_6)
>    for(i in seq_len(num_trials_6)){
>      success[i] <- as.integer(all(1:6 %in% prob_matrix_6[i, ]))
>    }
> 
>    #calculating probability of success
> 
>    p6 <- mean(success)
>    prob_frame_6[1, k] <- p6
> }
> 
> colnames(prob_frame_6) <- sprintf("p%d", seq_len(K))
> average_prob_frame_6 <- rowMeans(prob_frame_6)
> final_frame_6 <- cbind(prob_frame_6, average_prob_frame_6)
> 
> write.csv(final_frame_6, "OneMillion_Trials_Ten_Times_Results.csv")
> 
> print(final_frame_6)
> print(paste("The average probability of success when doing 1,000,000 single
> trials is:", average_prob_frame_6))
> 
>      [[alternative HTML version deleted]]
> 
> ______________________________________________
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> 
> 	[[alternative HTML version deleted]]
> 
> ______________________________________________
> R-help using r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.