qualUniform {PTXQC}

Compute deviation from uniform distribution

Description

The score ranges between 0 (worst score) and 1 (best score). Input 'x' is a vector of counts (or probabilities) for equally spaced bins in a histogram. A uniform distribution (e.g. c(3,3,3) will get a score of 1. The worst possible case (e.g. c(4,0,0)), will get a score of 0, and a linear increasing function (e.g. c(1,2,3)) will get something in between (0.585 here)

Usage

qualUniform(x, weight = vector())

Arguments

x

Vector of numeric intensity/count values (e.g. ID's per RT bin); bins are assumed to have equal widths

weight

Vector of weights for values in 'x' (same length as 'x').

Details

In addition, bin values can be weighted (e.g. by their confidence). The total sum of weights is normalized to 1 internally.

The distance function used is the square root of the absolute difference between a uniform distribution and the input 'x' (summed for each element of 'x'). This distance is normalized to the worst possible input (e.g. one bin with 100

Value

Value between [0, 1]

Examples

 stopifnot(qualUniform(c(3,3,3))==1)
 stopifnot(qualUniform(c(4,0,0))==0)         

 ## how 'uniform' is a vector where only a single index has weight?-- answer: very
 stopifnot(qualUniform(c(4,0,0), c(1,0,0))==1)   
 stopifnot(qualUniform(c(4,0,0), c(0,1,0))==1)     
 stopifnot(qualUniform(c(0,4,0))==0)              
 stopifnot(abs(qualUniform(c(3,2,1))-0.58578) < 0.0001)
 stopifnot(abs(qualUniform(c(1,2,3))-0.58578) < 0.0001)
 stopifnot(qualUniform(c(1,2,3), c(0,1,0))==1)   
 stopifnot(abs(qualUniform(c(1,2,3))-0.58578) < 0.0001)
 stopifnot(abs(qualUniform(c(1,2,3), c(0,1,1))- 0.590316) < 0.0001)
 stopifnot(abs(qualUniform(c(2,3), c(1,1))-0.552786) < 0.0001)
 stopifnot(abs(qualUniform(1:120)-0.38661) < 0.0001)