[R] Box plot with 5th and 95th percentiles instead of 1.5 * IQR: problems implementing an existing solution...

[Olaf]

Hello, 
you may just first change the boxplot.stats directly:

boxplot.stats <-  function (x, coef = NULL, do.conf = TRUE, do.out =
                                 TRUE)
{
  nna <- !is.na(x)
  n <- sum(nna)
  stats <- quantile(x, c(.05,.25,.5,.75,.95), na.rm = TRUE)
  iqr <- diff(stats[c(2, 4)])
  out <- x < stats[1] | x > stats[5]
  conf <- if (do.conf)
    stats[3] + c(-1.58, 1.58) * diff(stats[c(2, 4)])/sqrt(n)
  list(stats = stats, n = n, conf = conf, out = x[out & nna])
} 


then use it plotting your data. 
And after that change them back :

boxplot.stats <- function (x, coef = 1.5, do.conf = TRUE, do.out = TRUE) 
{
  if (coef < 0) 
    stop("'coef' must not be negative")
  nna <- !is.na(x)
  n <- sum(nna)
  stats <- stats::fivenum(x, na.rm = TRUE)
  iqr <- diff(stats[c(2, 4)])
  if (coef == 0) 
    do.out <- FALSE
  else {
    out <- if (!is.na(iqr)) {
      x < (stats[2L] - coef * iqr) | x > (stats[4L] + coef * 
                                            iqr)
    }
    else !is.finite(x)
    if (any(out[nna], na.rm = TRUE)) 
      stats[c(1, 5)] <- range(x[!out], na.rm = TRUE)
  }
  conf <- if (do.conf) 
    stats[3L] + c(-1.58, 1.58) * iqr/sqrt(n)
  list(stats = stats, n = n, conf = conf, out = if (do.out) x[out & 
                                                                nna] else
numeric())
}


it works fine for me.





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