Logistic curve fitting in epidemiology
The least-squares method (see references below) in statistics is used to fit data to the best of a family of similar curves by finding the parameters for a curve which minimizes the total of the distances from each data point to the curve.
Often, the curve used is a straight line, in which case the method is also called linear regression. If a curve which uses logarithmic growth is fit, the method can be called logistic regression.
A commonly used family of functions used in statistical studies of populations, including the growth of epidemics, are curves akin to the logistic curve:
f(x) = L / (1 + e-k(x-x0))
Though predictions based on fitting to such curves may err, especially if used to extrapolate from incomplete data, curves similar to the logistic curve have had good fits in population studies, including modeling the growth of past epidemics.
The task:
- Task
- Given the following daily world totals since December 31, 2019 for persons
who have become infected with the novel coronavirus Covid-19:
Daily totals: 27, 27, 27, 44, 44, 59, 59, 59, 59, 59, 59, 59, 59, 60, 60, 61, 61, 66, 83, 219, 239, 392, 534, 631, 897, 1350, 2023, 2820, 4587, 6067, 7823, 9826, 11946, 14554, 17372, 20615, 24522, 28273, 31491, 34933, 37552, 40540, 43105, 45177, 60328, 64543, 67103, 69265, 71332, 73327, 75191, 75723, 76719, 77804, 78812, 79339, 80132, 80995, 82101, 83365, 85203, 87024, 89068, 90664, 93077, 95316, 98172, 102133, 105824, 109695, 114232, 118610, 125497, 133852, 143227, 151367, 167418, 180096, 194836, 213150, 242364, 271106, 305117, 338133, 377918, 416845, 468049, 527767, 591704, 656866, 715353, 777796, 851308, 928436, 1000249, 1082054, 1174652
- Use the following variant of the logistic curve as a formula:
f(t) = n0 e(r t) / ((1 + n0 (e(r t) - 1) / K)
Where r is the rate of growth of the infection in the population.
The R0 of an infection (different from r above) is a measure of how many new individuals will become infected for every individual currently infected. It is an important measure of how quickly an infectious disease may spread.
R0 is related to the logistic curve's r parameter by the formula
r ≈ ln(R0) / G
where G, the generation time, is roughly the sum of the incubation time, perhaps 5 days, and the mean contagion period, perhaps 7 days, so, for covid-19, roughly we have
R0 ≈ e12r
Assume the following constants hold in the formula above:
- K is the world population, about 7.8 billion
- n0 is 27, the number of cases found in China at the start of the pandemic.
- Task
- Demonstrate code that finds a least-squares fits of the curve to the data.
- Show the calculated r for the logistic curve.
- Show the final R0 parameter you calculate from the logistic curve r value parameter.
- See also
Go
This uses the Levenberg-Marquardt method. <lang go>package main
import (
"fmt" "github.com/maorshutman/lm" "log" "math"
)
const (
K = 7_800_000_000 // approx world population n0 = 27 // number of cases at day 0
)
var y = []float64{
27, 27, 27, 44, 44, 59, 59, 59, 59, 59, 59, 59, 59, 60, 60, 61, 61, 66, 83, 219, 239, 392, 534, 631, 897, 1350, 2023, 2820, 4587, 6067, 7823, 9826, 11946, 14554, 17372, 20615, 24522, 28273, 31491, 34933, 37552, 40540, 43105, 45177, 60328, 64543, 67103, 69265, 71332, 73327, 75191, 75723, 76719, 77804, 78812, 79339, 80132, 80995, 82101, 83365, 85203, 87024, 89068, 90664, 93077, 95316, 98172, 102133, 105824, 109695, 114232, 118610, 125497, 133852, 143227, 151367, 167418, 180096, 194836, 213150, 242364, 271106, 305117, 338133, 377918, 416845, 468049, 527767, 591704, 656866, 715353, 777796, 851308, 928436, 1000249, 1082054, 1174652,
}
func f(dst, p []float64) {
for i := 0; i < len(y); i++ { t := float64(i) dst[i] = (n0*math.Exp(p[0]*t))/(1+n0*(math.Exp(p[0]*t)-1)/K) - y[i] }
}
func main() {
j := lm.NumJac{Func: f} prob := lm.LMProblem{ Dim: 1, Size: len(y), Func: f, Jac: j.Jac, InitParams: []float64{0.5}, Tau: 1e-6, Eps1: 1e-8, Eps2: 1e-8, } res, err := lm.LM(prob, &lm.Settings{Iterations: 100, ObjectiveTol: 1e-16}) if err != nil { log.Fatal(err) } r := res.X[0] fmt.Printf("The logistic curve r for the world data is %.8f\n", r) fmt.Printf("R0 is then approximately equal to %.7f\n", math.Exp(12*r))
}</lang>
- Output:
The logistic curve r for the world data is 0.11230218 R0 is then approximately equal to 3.8482793
Julia
<lang julia>using LsqFit
const K = 7_800_000_000 # approximate world population const n0 = 27 # starting at day 0 with 27 Chinese cases
""" The model for logistic regression with a given r """ @. model(t, r) = (n0 * exp(r * t)) / (( 1 + n0 * (exp(r * t) - 1) / K))
- Daily world totals of covid cases, all countries
ydata = [ 27, 27, 27, 44, 44, 59, 59, 59, 59, 59, 59, 59, 59, 60, 60, 61, 61, 66, 83, 219, 239, 392, 534, 631, 897, 1350, 2023, 2820, 4587, 6067, 7823, 9826, 11946, 14554, 17372, 20615, 24522, 28273, 31491, 34933, 37552, 40540, 43105, 45177, 60328, 64543, 67103, 69265, 71332, 73327, 75191, 75723, 76719, 77804, 78812, 79339, 80132, 80995, 82101, 83365, 85203, 87024, 89068, 90664, 93077, 95316, 98172, 102133, 105824, 109695, 114232, 118610, 125497, 133852, 143227, 151367, 167418, 180096, 194836, 213150, 242364, 271106, 305117, 338133, 377918, 416845, 468049, 527767, 591704, 656866, 715353, 777796, 851308, 928436, 1000249, 1082054, 1174652, ] tdata = collect(LinRange(0.0, 96, 97))
- starting approximation for r of 1/2
rparam = [0.5]
fit = curve_fit(model, tdata, ydata, rparam)
- Our answer for r given the world data and simplistic model
r = fit.param println("The logistic curve r for the world data is: ", r) println("The confidence interval at 5% significance is: ",
confidence_interval(fit, 0.05))
println("Since R0 ≈ exp(G * r), and G ≈ 12 days, R0 ≈ ", exp(12r[1]))
</lang>
- Output:
The logistic curve r for the world data is: [0.11230217572265622] The confidence interval at 5% significance is: [(0.11199074156706985, 0.11261360987824258)] Since R0 ≈ exp(G * r), and G ≈ 12 days, R0 ≈ 3.8482792820761063
Perl
<lang perl>use strict; use warnings;
my $K = 7_800_000_000; # population my $n0 = 27; # cases @ day 0
my @y = (
27, 27, 27, 44, 44, 59, 59, 59, 59, 59, 59, 59, 59, 60, 60, 61, 61, 66, 83, 219, 239, 392, 534, 631, 897, 1350, 2023, 2820, 4587, 6067, 7823, 9826, 11946, 14554, 17372, 20615, 24522, 28273, 31491, 34933, 37552, 40540, 43105, 45177, 60328, 64543, 67103, 69265, 71332, 73327, 75191, 75723, 76719, 77804, 78812, 79339, 80132, 80995, 82101, 83365, 85203, 87024, 89068, 90664, 93077, 95316, 98172, 102133, 105824, 109695, 114232, 118610, 125497, 133852, 143227, 151367, 167418, 180096,
194836, 213150, 242364, 271106, 305117, 338133, 377918, 416845, 468049, 527767, 591704, 656866, 715353, 777796, 851308, 928436,1000249,1082054,1174652 );
sub logistic_func {
my($r) = @_; my $sq = 0; for my $i (0 .. @y-1) { my $eri = exp($r * $i); my $dst = ($n0 * $eri) / (1 + $n0 * ($eri-1) / $K) - $y[$i]; $sq = $sq + $dst**2; } $sq
}
sub solve {
my($fn, $guess, $epsilon) = @_; my($nfm,$nfp); my $f0 = &$fn($guess); my $delta = $guess; my $factor = 2; while ($delta > $epsilon) { ($nfm = &$fn($guess - $delta)) < $f0 ? ($f0 = $nfm, $guess -= $delta, $delta *= $factor) : ($nfp = &$fn($guess + $delta)) < $f0 ? ($f0 = $nfp, $guess += $delta, $delta *= $factor) : $delta /= $factor } $guess
}
my $r = solve(\&logistic_func, 0.5, 0); my $R0 = exp(12 * $r); printf "r = %%(%.3f), R0 = %%(%.3f)\n", $r, $R0;</lang>
- Output:
r = %(0.112), R0 = %(3.848)
Phix
Simplified, my interpretation of shift-cutting (from wp:Non-linear_least_squares) <lang Phix>-- demo\rosetta\Curve_fit.exw constant K = 7_800_000_000, -- approx world population
n0 = 27 -- number of cases at day 0
sequence y = {
27, 27, 27, 44, 44, 59, 59, 59, 59, 59, 59, 59, 59, 60, 60, 61, 61, 66, 83, 219, 239, 392, 534, 631, 897, 1350, 2023, 2820, 4587, 6067, 7823, 9826, 11946, 14554, 17372, 20615, 24522, 28273, 31491, 34933, 37552, 40540, 43105, 45177, 60328, 64543, 67103, 69265, 71332, 73327, 75191, 75723, 76719, 77804, 78812, 79339, 80132, 80995, 82101, 83365, 85203, 87024, 89068, 90664, 93077, 95316, 98172, 102133, 105824, 109695, 114232, 118610, 125497, 133852, 143227, 151367, 167418, 180096, 194836, 213150, 242364, 271106, 305117, 338133, 377918, 416845, 468049, 527767, 591704, 656866, 715353, 777796, 851308, 928436, 1000249, 1082054, 1174652,
}
function f(atom r)
atom sq = 0 for i=1 to length(y) do atom eri = exp(r*(i-1)) sq += power((n0*eri)/(1+n0*(eri-1)/K) - y[i],2) end for return sq
end function
function solve(integer f, atom guess=0.5, epsilon=0)
atom f0 = f(guess), delta = guess, factor = 2 -- double until f0 best, then -- halve until delta<=epsilon while delta>epsilon do atom nf = f(guess-delta) if nf<f0 then f0 = nf guess -= delta else nf = f(guess+delta) if nf<f0 then f0 = nf guess += delta else factor = 0.5 end if end if delta *= factor end while return guess
end function
atom r = solve(f),
R0 = exp(12 * r)
printf(1,"r = %.10f, R0 = %.8f\n",{r,R0})</lang>
- Output:
r = 0.1123021757, R0 = 3.84827928
Python
Uses NumPy/SciPy's optimize package. <lang python>import numpy as np import scipy.optimize as opt
n0, K = 27, 7_800_000_000
def f(t, r):
return (n0 * np.exp(r * t)) / (( 1 + n0 * (np.exp(r * t) - 1) / K))
y = [ 27, 27, 27, 44, 44, 59, 59, 59, 59, 59, 59, 59, 59, 60, 60, 61, 61, 66, 83, 219, 239, 392, 534, 631, 897, 1350, 2023, 2820, 4587, 6067, 7823, 9826, 11946, 14554, 17372, 20615, 24522, 28273, 31491, 34933, 37552, 40540, 43105, 45177, 60328, 64543, 67103, 69265, 71332, 73327, 75191, 75723, 76719, 77804, 78812, 79339, 80132, 80995, 82101, 83365, 85203, 87024, 89068, 90664, 93077, 95316, 98172, 102133, 105824, 109695, 114232, 118610, 125497, 133852, 143227, 151367, 167418, 180096, 194836, 213150, 242364, 271106, 305117, 338133, 377918, 416845, 468049, 527767, 591704, 656866, 715353, 777796, 851308, 928436, 1000249, 1082054, 1174652, ] x = np.linspace(0.0, 96, 97)
r, cov = opt.curve_fit(f, x, y, [0.5])
- Our answer for r given the world data and simplistic model
print("The r for the world Covid-19 data is:", r,
", with covariance of", cov)
print("The calculated R0 is then", np.exp(12 * r))
</lang>
- Output:
The r for the world Covid-19 data is: [0.11230218] , with covariance of [[2.46164331e-08]] The calculated R0 is then [3.8482793]
Wren
<lang ecmascript>var K = 7800000000 // approx world population var n0 = 27 // number of cases at day 0
var y = [
27, 27, 27, 44, 44, 59, 59, 59, 59, 59, 59, 59, 59, 60, 60, 61, 61, 66, 83, 219, 239, 392, 534, 631, 897, 1350, 2023, 2820, 4587, 6067, 7823, 9826, 11946, 14554, 17372, 20615, 24522, 28273, 31491, 34933, 37552, 40540, 43105, 45177, 60328, 64543, 67103, 69265, 71332, 73327, 75191, 75723, 76719, 77804, 78812, 79339, 80132, 80995, 82101, 83365, 85203, 87024, 89068, 90664, 93077, 95316, 98172, 102133, 105824, 109695, 114232, 118610, 125497, 133852, 143227, 151367, 167418, 180096, 194836, 213150, 242364, 271106, 305117, 338133, 377918, 416845, 468049, 527767, 591704, 656866, 715353, 777796, 851308, 928436, 1000249, 1082054, 1174652
]
var exp = Fn.new { |x|
var e = 2.718281828459045 return e.pow(x)
}
var f = Fn.new { |r|
var sq = 0 for (i in 0...y.count) { var eri = exp.call(r*i) var dst = (n0*eri)/(1+n0*(eri-1)/K) - y[i] sq = sq + dst * dst } return sq
}
var solve = Fn.new { |f, guess, epsilon|
var f0 = f.call(guess) var delta = guess var factor = 2 // double until f0 best then halve until delta <= epsilon while (delta > epsilon) { var nf = f.call(guess - delta) if (nf < f0) { f0 = nf guess = guess - delta } else { nf = f.call(guess + delta) if (nf < f0) { f0 = nf guess = guess + delta } else { factor = 0.5 } } delta = delta * factor } return guess
}
var r = (solve.call(f, 0.5, 0) * 1e10).round / 1e10 var R0 = (exp.call(12 * r) * 1e8).round / 1e8 System.print("r = %(r), R0 = %(R0)")</lang>
- Output:
r = 0.1123021757, R0 = 3.84827928