Calculate the 95% CI of the ratio of two proportions, with application to diagnostic likelihood ratios

ratio.of.two.prop <- function(x, n)
{
    ## Purpose: Calculate the 95% CI of the ratio of two proportions, with application to diagnostic likelihood ratios. 
    ## Arguments:
    ##   x: a vector of two integers (numerator of the two proportions). 
    ##   n: a vector of two integers (denominator of the two proportions).
    ## Return: The point estimate and the 95% confidence interval of the ratio of two proportions, where the ratio is the first proportion divided by the second proportion. 
    ## Author: Feiming Chen
    ## ________________________________________________

    if (x[1] == 0 || x[1] == n[1] || x[2] == 0 || x[2] == n[2]) {
        x <- x + 0.5
        n <- n + 1
    }

    p <- x / n

    se <- sqrt((1 - p[1]) / (p[1] * n[1]) + (1 - p[2]) / (p[2] * n[2]))
    
    pp <- p[1] / p[2]                   # point estimate
    LB <- exp(log(pp) - 1.96 * se)
    UB <- exp(log(pp) + 1.96 * se)

    c(pp, LB, UB)
}
if (F) {                                # Unit Test
    ratio.of.two.prop(c(431, 30), c(460, 146))
    ## [1] 4.5599 3.3116 6.2786
    ratio.of.two.prop(c(0, 146), c(460, 146))
    ## [1] 0.00108830 0.00006817 0.01737423   
}