PTXQC (version 1.0.4)

qualUniform: Compute deviation from uniform distribution

Description

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').

Value

Value between [0, 1]

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

Examples

# NOT RUN {
 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)
 
# }