Is there a way to find the median using beta distribution parameters in R?

Question

I am working with CSV dataset called productQuality where every row represents a weld type and the beta distribution parameters (α and β) for that specific weld. I was wondering if there's a way to calculate and list the medians for each weld type? Here's a dput of my dataset:

structure(list(weld.type.ID = 1:33, weld.type = structure(c(29L, 
11L, 16L, 4L, 28L, 17L, 19L, 5L, 24L, 27L, 21L, 32L, 12L, 20L, 
26L, 25L, 3L, 7L, 13L, 22L, 33L, 1L, 9L, 10L, 18L, 15L, 31L, 
8L, 23L, 2L, 14L, 6L, 30L), .Label = c("1,40,Material A", "1,40S,Material C", 
"1,80,Material A", "1,STD,Material A", "1,XS,Material A", "10,10S,Material C", 
"10,160,Material A", "10,40,Material A", "10,40S,Material C", 
"10,80,Material A", "10,STD,Material A", "10,XS,Material A", 
"13,40,Material A", "13,40S,Material C", "13,80,Material A", 
"13,STD,Material A", "13,XS,Material A", "14,40,Material A", 
"14,STD,Material A", "14,XS,Material A", "15,STD,Material A", 
"15,XS,Material A", "2,10S,Material C", "2,160,Material A", "2,40,Material A", 
"2,40S,Material C", "2,80,Material A", "2,STD,Material A", "2,XS,Material A", 
"4,80,Material A", "4,STD,Material A", "6,STD,Material A", "6,XS,Material A"
), class = "factor"), alpha = c(281L, 196L, 59L, 96L, 442L, 98L, 
66L, 30L, 68L, 43L, 35L, 44L, 23L, 14L, 24L, 38L, 8L, 8L, 5L, 
19L, 37L, 38L, 6L, 11L, 29L, 6L, 16L, 6L, 16L, 3L, 4L, 9L, 12L
), beta = c(7194L, 4298L, 3457L, 2982L, 4280L, 3605L, 2229L, 
1744L, 2234L, 1012L, 1096L, 1023L, 1461L, 1303L, 531L, 233L, 
630L, 502L, 328L, 509L, 629L, 554L, 358L, 501L, 422L, 566L, 403L, 
211L, 159L, 268L, 167L, 140L, 621L)), class = "data.frame", row.names = c(NA, 
-33L))

Answer 1

According to Wikipedia there is an approximate solution for the median for alpha, beta >1, but no general closed-form solution. Below I implement the brute-force exact solution and the approximate solution:

## I_{1/2}^{-1}(alpha,beta)
med_exact0 <- function(alpha,beta,eps=1e-12) {
    uniroot(function(x) pbeta(x,alpha,beta)-1/2,
            interval=c(eps,1-eps))$root
}
med_exact <- Vectorize(med_exact0, vectorize.args=c("alpha","beta"))
med_approx <- function(alpha,beta) (alpha-1/3)/(alpha+beta-2/3)

edit comments point out that the inverse ('brute force') solution is already implemented in base R as qbeta(p=0.5,...) ! Almost certainly more robust and computationally efficient than my solution ...

I called your data dd :

evals <- with(dd,med_exact(alpha,beta))
avals <- with(dd,med_approx(alpha,beta))
evals2 <- with(dd,qbeta(0.5,alpha,beta))
max(abs((evals-avals)/evals))  ## 0.0057

In the worst case in your data the exact and approximate solutions are off by about 0.6% ...