I looked up on the web, but couldn’t find anything helpful.

I’m basically looking for a way to measure how ‘evenly’ a value is distributed. As in, an

‘evenly’distributed distribution likeX:

and an

‘unevenly’distributed distributionYof roughly the same mean and standard deviation:

But is there any evenness measure m, such that m(X) > m(Y)? If there isn’t, what would be the best way to create a measure like this?

(Images screenshot from Khan Academy)

**Answer**

A standard, powerful, well-understood, theoretically well-established, and frequently implemented measure of “evenness” is the **Ripley K function and its close relative, the L function.** Although these are typically used to evaluate two-dimensional spatial point configurations, the analysis needed to adapt them to one dimension (which usually is not given in references) is simple.

### Theory

The K function estimates the mean proportion of points within a distance $d$ of a typical point. For a uniform distribution on the interval $[0,1]$, the true proportion can be computed and (asymptotically in the sample size) equals $1 – (1-d)^2$. The appropriate one-dimensional version of the L function subtracts this value from K to show *deviations from uniformity.* We might therefore consider normalizing any batch of data to have a unit range and examining its L function for deviations around zero.

### Worked Examples

**To illustrate**, I have simulated $999$ independent samples of size $64$ from a uniform distribution and plotted their (normalized) L functions for shorter distances (from $0$ to $1/3$), thereby creating an envelope to estimate the sampling distribution of the L function. (Plotted points well within this envelope cannot be significantly distinguished from uniformity.) Over this I have plotted the L functions for samples of the same size from a U-shaped distribution, a mixture distribution with four obvious components, and a standard Normal distribution. The histograms of these samples (and of their parent distributions) are shown for reference, using line symbols to match those of the L functions.

The sharp separated spikes of the **U-shaped distribution** (dashed red line, leftmost histogram) create clusters of closely spaced values. This is reflected by a very large slope in the L function at $0$. The L function then decreases, eventually becoming negative to reflect the gaps at intermediate distances.

The sample from the **normal distribution** (solid blue line, rightmost histogram) is fairly close to uniformly distributed. Accordingly, its L function does not depart from $0$ quickly. However, by distances of $0.10$ or so, it has risen sufficiently above the envelope to signal a slight tendency to cluster. The continued rise across intermediate distances indicates the clustering is diffuse and widespread (not confined to some isolated peaks).

The initial large slope for the sample from the **mixture distribution** (middle histogram) reveals clustering at small distances (less than $0.15$). By dropping to negative levels, it signals separation at intermediate distances. Comparing this to the U-shaped distribution’s L function is revealing: the slopes at $0$, the amounts by which these curves rise above $0$, and the rates at which they eventually descend back to $0$ all provide information about the nature of the clustering present in the data. Any of these characteristics could be chosen as a single measure of “evenness” to suit a particular application.

These examples show how an L-function can be examined to evaluate departures of the data from uniformity (“evenness”) and how *quantitative* information about the scale and nature of the departures can be extracted from it.

(One can indeed plot the entire L function, extending to the full normalized distance of $1$, to assess large-scale departures from uniformity. Ordinarily, though, assessing the behavior of the data at smaller distances is of greater importance.)

### Software

`R`

code to generate this figure follows. It starts by defining functions to compute K and L. It creates a capability to simulate from a mixture distribution. Then it generates the simulated data and makes the plots.

```
Ripley.K <- function(x, scale) {
# Arguments:
# x is an array of data.
# scale (not actually used) is an option to rescale the data.
#
# Return value:
# A function that calculates Ripley's K for any value between 0 and 1 (or `scale`).
#
x.pairs <- outer(x, x, function(a,b) abs(a-b)) # All pairwise distances
x.pairs <- x.pairs[lower.tri(x.pairs)] # Distances between distinct pairs
if(missing(scale)) scale <- diff(range(x.pairs))# Rescale distances to [0,1]
x.pairs <- x.pairs / scale
#
# The built-in `ecdf` function returns the proportion of values in `x.pairs` that
# are less than or equal to its argument.
#
return (ecdf(x.pairs))
}
#
# The one-dimensional L function.
# It merely subtracts 1 - (1-y)^2 from `Ripley.K(x)(y)`.
# Its argument `x` is an array of data values.
#
Ripley.L <- function(x) {function(y) Ripley.K(x)(y) - 1 + (1-y)^2}
#-------------------------------------------------------------------------------#
set.seed(17)
#
# Create mixtures of random variables.
#
rmixture <- function(n, p=1, f=list(runif), factor=10) {
q <- ceiling(factor * abs(p) * n / sum(abs(p)))
x <- as.vector(unlist(mapply(function(y,f) f(y), q, f)))
sample(x, n)
}
dmixture <- function(x, p=1, f=list(dunif)) {
z <- matrix(unlist(sapply(f, function(g) g(x))), ncol=length(f))
z %*% (abs(p) / sum(abs(p)))
}
p <- rep(1, 4)
fg <- lapply(p, function(q) {
v <- runif(1,0,30)
list(function(n) rnorm(n,v), function(x) dnorm(x,v), v)
})
f <- lapply(fg, function(u) u[[1]]) # For random sampling
g <- lapply(fg, function(u) u[[2]]) # The distribution functions
v <- sapply(fg, function(u) u[[3]]) # The parameters (for reference)
#-------------------------------------------------------------------------------#
#
# Study the L function.
#
n <- 64 # Sample size
alpha <- beta <- 0.2 # Beta distribution parameters
layout(matrix(c(rep(1,3), 3, 4, 2), 2, 3, byrow=TRUE), heights=c(0.6, 0.4))
#
# Display the L functions over an envelope for the uniform distribution.
#
plot(c(0,1/3), c(-1/8,1/6), type="n",
xlab="Normalized Distance", ylab="Total Proportion",
main="Ripley L Functions")
invisible(replicate(999, {
plot(Ripley.L(x.unif <- runif(n)), col="#00000010", add=TRUE)
}))
abline(h=0, lwd=2, col="White")
#
# Each of these lines generates a random set of `n` data according to a specified
# distribution, calls `Ripley.L`, and plots its values.
#
plot(Ripley.L(x.norm <- rnorm(n)), col="Blue", lwd=2, add=TRUE)
plot(Ripley.L(x.beta <- rbeta(n, alpha, beta)), col="Red", lwd=2, lty=2, add=TRUE)
plot(Ripley.L(x.mixture <- rmixture(n, p, f)), col="Green", lwd=2, lty=3, add=TRUE)
#
# Display the histograms.
#
n.breaks <- 24
h <- hist(x.norm, main="Normal Sample", breaks=n.breaks, xlab="Value")
curve(dnorm(x)*n*mean(diff(h$breaks)), add=TRUE, lwd=2, col="Blue")
h <- hist(x.beta, main=paste0("Beta(", alpha, ",", beta, ") Sample"),
breaks=n.breaks, xlab="Value")
curve(dbeta(x, alpha, beta)*n*mean(diff(h$breaks)), add=TRUE, lwd=2, lty=2, col="Red")
h <- hist(x.mixture, main="Mixture Sample", breaks=n.breaks, xlab="Value")
curve(dmixture(x, p, g)*n*mean(diff(h$breaks)), add=TRUE, lwd=2, lty=3, col="Green")
```

**Attribution***Source : Link , Question Author : Ketan , Answer Author : whuber*