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rdplyrplyrstatistics-bootstrap

plyr + simpleboot: NA in probability vector


I am using simplebootpackage (https://cran.r-project.org/web/packages/simpleboot/index.html) to obtain confidence intervals.

This is my function:

lb_weighted_median_dplyr <- function(x,v) {
  set.seed(1234)
  b <- one.boot(x, weights = v, FUN = function(x,w) matrixStats::weightedMedian(x, w = v, na.rm = TRUE), R = 100, student = FALSE)
  round(perc(b, 0.025), 0)
}

What the function does is to calculate the lower bound of the confidence interval when I run

ddply(wage_by_gender_2015, .(sex,region), summarise, FUN = lb_weighted_median_dplyr(wage, exp_region))

Where wage is a numeric column and exp_region is another numeric column that contains weights.

I don't have data for some regions, therefore the function fails with some regions and returns

Error in eval(substitute(expr), envir, enclos) : NA in probability vector

How can I bypass that error and obtain NA as the lower bound for a region without data?

A dplyr equivalent approach that also returns NA in probability vector is

grouped <- group_by(wage_by_gender_2015, sex, region)
dplyr::summarise(grouped, FUN = lb_weighted_median_dplyr(wage, exp_region))

Relevant sample of the data here: http://users.dcc.uchile.cl/~mvargas/casen/wage_by_gender_2015.RData


Solution

  • wage_by_gender_2015 <- data.frame(sex    = rep(c("male", "female"),100),
                                      region = rep(c("north", "south", "east",
                                                     "west"), 50),
                                      exp_region = abs(rnorm(100)),
                                      wage       = abs(rnorm(100))
    )
    
    wage_by_gender_2015$exp_region[10] <- NA
    ddply(wage_by_gender_2015, .(sex,region), summarise, FUN = lb_weighted_median_dplyr(wage, exp_region))
    
     Error in sample.int(length(x), replace = TRUE, ...) :    NA in probability vector
    
    # impute
    wage_by_gender_2015$exp_region <- RRF::na.roughfix(wage_by_gender_2015$exp_region)
    
    ddply(wage_by_gender_2015, .(sex,region), summarise, FUN = lb_weighted_median_dplyr(wage, exp_region))
    
        sex region FUN
    1 female  south   0
    2 female   west   0
    3   male   east   1
    4   male  north   0
    

    As mentioned in the comment I would've used your sample data but it was missing sex.