time-to-botec/nim/samples.nim

85 lines
2.0 KiB
Nim

import std/math
import std/sugar
import std/random
import std/sequtils
randomize()
## Distribution functions
## Normal
## <https://en.wikipedia.org/wiki/Box%E2%80%93Muller_transform#Basic_form>
proc ur_normal(): float =
let u1 = rand(1.0)
let u2 = rand(1.0)
let z = sqrt(-2.0 * ln(u1)) * sin(2 * PI * u2)
return z
proc normal(mean: float, sigma: float): float =
return (mean + sigma * ur_normal())
proc lognormal(logmean: float, logsigma: float): float =
let answer = pow(E, normal(logmean, logsigma))
return answer
proc to(low: float, high: float): float =
let normal95confidencePoint = 1.6448536269514722
let loglow = ln(low)
let loghigh = ln(high)
let logmean = (loglow + loghigh)/2
let logsigma = (loghigh - loglow) / (2.0 * normal95confidencePoint)
return lognormal(logmean, logsigma)
## echo ur_normal()
## echo normal(10, 20)
## echo lognormal(2, 4)
## echo to(10, 90)
## Manipulate samples
proc mixture(fs: seq[proc (): float{.nimcall.}], ps: seq[float], n: int): seq[float] =
assert fs.len == ps.len
var ws: seq[float]
var sum = 0.0
for i, p in ps:
sum = sum + p
ws.add(sum)
ws = ws.map(w => w/sum)
var samples: seq[float]
let rs = toSeq(1..n).map(_=>rand(1.0))
for i in 0..(n-1):
let r = rs[i]
var j = ws.len - 1
for k, w in ws:
if r < w:
j = k
break
## only occasion when ^ doesn't assign j
## is when r is exactly 1
## which would correspond to choosing the last item in ws
## which is why j is initialized to ws.len - 1
let f = fs[j]
samples.add(f())
return samples
## Actual model
let n = 1000000
let p_a = 0.8
let p_b = 0.5
let p_c = p_a * p_b
let weights = @[ 1.0 - p_c, p_c/2.0, p_c/4.0, p_c/4.0 ]
let fs = @[ proc (): float = 0.0, proc (): float = 1.0, proc (): float = to(1.0, 3.0), proc (): float = to(2.0, 10.0)]
let result = mixture(fs, weights, n)
let mean_result = foldl(result, a + b, 0.0) / float(result.len)
# echo result
echo mean_result