Logistic curve fitting in epidemiology: Difference between revisions

 

The least-squares method (see references below) in statistics is used to fit data

including the growth of epidemics, are curves akin to the logistic curve:

 

 

extrapolate from incomplete data, curves similar to the logistic curve have had

good fits in population studies, including modeling the growth of past epidemics.

 

 

<pre>

Daily totals:

</pre>

 

 

 

 

new individuals will become infected for every individual currently infected.

It is an important measure of how quickly an infectious disease may spread.

 

 

 

 

 

 

;Task:

* Demonstrate code that finds a least-squares fits of the curve to the data.

 

 

:;*[[https://en.wikipedia.org/wiki/Basic_reproduction_number Basic reproduction number]]

:;*[[https://en.wikipedia.org/wiki/Logistic_function Logistic functions]]

:;*[[https://ourworldindata.org/coronavirus#all-charts-preview World covid-19 case tallies]]

:;*[[https://www.zoology.ubc.ca/~bio310/Blank%20Page%204_files/DL%20R%20and%20r.htm Calculating r and R0]]

 

 

{{libheader|lm}}

This uses the Levenberg-Marquardt method.

 

import (

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))

 

{{out}}

R0 is then approximately equal to 3.8482793

</pre>

 

=={{header|Julia}}==

 

const K = 7_800_000_000 # approximate world population

confidence_interval(fit, 0.05))

println("Since R0 ≈ exp(G * r), and G ≈ 12 days, R0 ≈ ", exp(12r[1]))

<pre>

The logistic curve r for the world data is: [0.11230217572265622]

Since R0 ≈ exp(G * r), and G ≈ 12 days, R0 ≈ 3.8482792820761063

</pre>

 

=={{header|Perl}}==

{{trans|Phix}}

use warnings;

 

my $r = solve(\&logistic_func, 0.5, 0);

my $R0 = exp(12 * $r);

{{out}}

<pre>r = %(0.112), R0 = %(3.848)</pre>

=={{header|Phix}}==

Simplified, my interpretation of shift-cutting (from [[wp:Non-linear_least_squares]])

{{out}}

<pre>

=={{header|Python}}==

Uses NumPy/SciPy's optimize package.

import scipy.optimize as opt

 

", with covariance of", cov)

print("The calculated R0 is then", np.exp(12 * r))

<pre>

The r for the world Covid-19 data is: [0.11230218] , with covariance of [[2.46164331e-08]]

 

=={{header|R}}==

library(minpack.lm)

 

 

cat("Therefore, R0 is approximately: ", exp(12 * coef(nls.out)))

<pre>

The r for the model is: 0.1123022

Original task numbers of cases per day as of April 05 plus updated, as of today, May 11 numbers.

 

my $n0 = 27; # cases @ day 0

 

say "Reproductive rate: R0 = ", $R0.fmt('%08f');

}

{{out}}

<pre>For period: Dec 31 - Apr 5

Instantaneous rate of growth: r = 0.093328

Reproductive rate: R0 = 3.064672</pre>

 

=={{header|Wren}}==

{{trans|Phix}}

var n0 = 27 // number of cases at day 0

 

var sq = 0

for (i in 0...y.count) {

var dst = (n0*eri)/(1+n0*(eri-1)/K) - y[i]

sq = sq + dst * dst

 

var r = (solve.call(f, 0.5, 0) * 1e10).round / 1e10

 

{{out}}

<pre>

r = 0.1123021757, R0 = 3.84827928

</pre>