Using modelrs bootstrap in R for medians

Question

I have found that the following works

iris %>% 
    select(Sepal.Length) %>% 
    modelr::bootstrap(100) %>% 
    mutate(mean = map(strap, mean))

but the below does not

iris %>% 
    select(Sepal.Length) %>% 
    modelr::bootstrap(100) %>% 
    mutate(median = map(strap, median)) 

The only difference is that the second line of code uses the median.

The error I get is

Error in mutate_impl(.data, dots) : Evaluation error: unimplemented type 'list' in 'greater' .

Answer 1

The code looks like it's working, but if you unnest it, you're actually just getting a lot of NAs because you're trying to take the mean of a resample object, which is a classed list with a reference to the data resampled and the indices for the particular resample. Taking the mean of such a list is not useful, so returning NA with a warning is helpful behavior. To get the code to work, coerce the resample to a data frame, which you can operate upon as usual within map's anonymous function.

For a direct route, extract the data and take the mean, simplifying the list to a numeric vector with map_dbl:

library(tidyverse)
set.seed(47)

iris %>% 
    select(Sepal.Length) %>% 
    modelr::bootstrap(100) %>% 
    mutate(sepal_mean = map_dbl(strap, ~mean(as_data_frame(.x)$Sepal.Length))) 
#> # A tibble: 100 x 3
#>             strap   .id sepal_mean
#>            <list> <chr>      <dbl>
#>  1 <S3: resample>   001   5.844000
#>  2 <S3: resample>   002   6.016000
#>  3 <S3: resample>   003   5.851333
#>  4 <S3: resample>   004   5.869333
#>  5 <S3: resample>   005   5.840667
#>  6 <S3: resample>   006   5.825333
#>  7 <S3: resample>   007   5.824000
#>  8 <S3: resample>   008   5.790000
#>  9 <S3: resample>   009   5.858000
#> 10 <S3: resample>   010   5.810000
#> # ... with 90 more rows

Translating this approach to median works fine:

iris %>% 
    select(Sepal.Length) %>% 
    modelr::bootstrap(100) %>% 
    mutate(sepal_median = map_dbl(strap, ~median(as_data_frame(.x)$Sepal.Length)))
#> # A tibble: 100 x 3
#>             strap   .id sepal_median
#>            <list> <chr>        <dbl>
#>  1 <S3: resample>   001          5.9
#>  2 <S3: resample>   002          5.8
#>  3 <S3: resample>   003          5.8
#>  4 <S3: resample>   004          5.7
#>  5 <S3: resample>   005          5.7
#>  6 <S3: resample>   006          5.8
#>  7 <S3: resample>   007          5.8
#>  8 <S3: resample>   008          5.7
#>  9 <S3: resample>   009          5.8
#> 10 <S3: resample>   010          5.7
#> # ... with 90 more rows

If you'd like both median and mean, you could repeatedly coerce the resample to a data frame, or store it in another column, but neither approach is very efficient. It's better to return a list of data frames with map that can be unnested:

iris %>% 
    select(Sepal.Length) %>% 
    modelr::bootstrap(100) %>% 
    mutate(stats = map(strap, ~summarise_all(as_data_frame(.x), funs(mean, median)))) %>% 
    unnest(stats)
#> # A tibble: 100 x 4
#>             strap   .id     mean median
#>            <list> <chr>    <dbl>  <dbl>
#>  1 <S3: resample>   001 5.744667   5.60
#>  2 <S3: resample>   002 5.725333   5.70
#>  3 <S3: resample>   003 5.808667   5.70
#>  4 <S3: resample>   004 5.809333   5.70
#>  5 <S3: resample>   005 5.964000   5.85
#>  6 <S3: resample>   006 5.931333   5.95
#>  7 <S3: resample>   007 5.838667   5.80
#>  8 <S3: resample>   008 5.926000   5.95
#>  9 <S3: resample>   009 5.855333   5.75
#> 10 <S3: resample>   010 5.888667   5.70
#> # ... with 90 more rows

Answer 2

Updated syntax is:

iris %>% 
    select(Sepal.Length) %>% 
    modelr::bootstrap(100) %>% 
    mutate(stats = map(strap, ~summarise_all(as_tibble(.x), list(mean = mean, median = median)))) %>% 
    unnest(stats)

because as_data_frame and funs are deprecated