rep_sample_n() error when prob vector specified

[rudeboybert]

An issue was brought to my attention by @unoe w.r.t. to specifying non-uniform sampling probabilities. Here is a reprex:

library(tidyverse)
library(infer)

tbl <- tibble(out = c('S', 'F'))
size <- 12
replace <- TRUE
prob <- c(1/5, 4/5)

# No problem:
rep_sample_n(tbl = tbl, size = size, replace = replace)
#> # A tibble: 12 x 2
#> # Groups:   replicate [1]
#>    replicate out  
#>        <int> <chr>
#>  1         1 F    
#>  2         1 S    
#>  3         1 S    
#>  4         1 S    
#>  5         1 F    
#>  6         1 S    
#>  7         1 S    
#>  8         1 S    
#>  9         1 F    
#> 10         1 S    
#> 11         1 S    
#> 12         1 F

# Problem when including prob argument:
rep_sample_n(tbl = tbl, size = size, replace = replace, prob = prob)
#> Error: All columns in a tibble must be 1d or 2d objects:
#> * Column `vals` is NULL

I've traced the issue to https://github.com/tidymodels/infer/blob/master/R/rep_sample_n.R#L59. In fact, there is a residual there should be a better way!! comment! In short, rep_sample_n() currently assumes that the variable of interest to sample is

• in the first column of the input data frame
• more importantly, is of type factor

I'm happy to file a PR to fix this, however I'm not quite sure I understand why the following code on lines 69-73 are necessary, instead of just using the prob argument as is

prob <- tibble::tibble(vals = levels(dplyr::pull(tbl, 1))) %>%
  dplyr::mutate(probs = prob) %>%
  dplyr::inner_join(tbl) %>%
  dplyr::select(probs) %>%
  dplyr::pull()

Judging by #82, it seems @mine-cetinkaya-rundel wrote this function back in the day for oilabs. Mine, could you shed some light as to necessity of the above wrangling?

[rudeboybert]

Also, I'm wondering if per tidy principles, we should force/require that the sampling probabilities be specified as a variable prob of the input data frame. This does break with how various base::sample() functions work (like the base::sample.int() engine for rep_sample_n()), but I think this is ultimately safer.

[mine-cetinkaya-rundel]

Indeed, this was based on oilabs::rep_sample_n() (which, by the way, is destined to live in the openintro package going forward) but I didn't implement the probabilities so I'm not the best person to answer:

• Here is the oilabs version: https://github.com/OpenIntroStat/oilabs/blob/master/R/rep_sample_n.R
• I was able to track down where prob was first introduced, and it seems to be here by @andrewpbray: 6bca134#diff-37da32d3b793ee432e9cec69ee5721bb. So maybe he'll be a better person to answer that.

Also, since we're talking about rep_sample_n(), we should introduce a new function called rep_slice_sample() or something like that as sample_n() will be superseeded by slice_sample() in the next release of dplyr. See https://github.com/tidyverse/dplyr/blob/master/R/slice.R and https://github.com/tidyverse/dplyr/blob/master/R/sample.R.

[rudeboybert]

Sounds good. @andrewpbray, could you please let me know your reasoning? I'll then PR rep_sample_n() in infer, which then later on can be ported to oilabs along with the new slice_sample() functions.

[unoe]

Hi all, thanks for your help in understanding how to do this.
I basically wanted to show a randomization distribution for the null that the probability of success (coded as S) is p = 1/5, using rep_sample_n and then summarising with prop = mean(output == 'S').

Currently the code:

• uses the first column of the tibble: dplyr::pull(tbl, 1)

• assumes that it is a factor: levels(dplyr::pull(tbl, 1))

• assumes that the column with the outcomes has name vals.

A temporary workaround, which returns a warning, is to make sure exactly of those points: that the tibble contains a column called vals which is a factor.

tbl <- tibble(vals = factor(c("S","F")))
samples <- rep_sample_n(tbl, size = 12, replace = TRUE, prob = c(1/5, 4/5), reps = 1)
# Warning message:
# Column `vals` joining character vector and factor, coercing into character vector 

[ismayc]

Removing @andrewpbray as the assignee here. @echasnovski or @simonpcouch, can you look into this whenever you get a chance?

[unoe]

Hi everyone,

I hope you are well. I just thought this might help in getting started.
I made this little draft - it might not be the most efficient implementation or perhaps you might not want to load the libraries in there, but it does the job:

rep_sample_n <- function(tbl, size, replace = FALSE, reps = 1, prob = NULL) {
    
    require(magrittr)
    require(dplyr)
    require(purrr)
    
    samples <- 1:reps %>%
        map_dfr(~ tbl %>%
                    slice_sample(n = size, weight_by = prob, replace = replace)) %>%
        mutate(replicate = rep(1:reps, each = size)) %>%
        group_by(replicate)
    
    return(samples)
}

Hope this helps,
Umberto

[simonpcouch]

Starting to give a go at this on the branch linked above—will drop a link to here in the PR once it's ready!

@unoe, your draft was really helpful. :-)

[simonpcouch]

🦋

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