Logistic curve fitting in epidemiology: Difference between revisions

{{trans|C++}}

 

-V n0 = 27

-V actual = [

V r = solve(f)

V R0 = exp(12 * r)

 

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=={{header|Ada}}==

{{trans|C++}}

with Ada.Numerics.Generic_Elementary_Functions;

 

Put (", R0 = "); Real_Io.Put (R0, Exp => 0, Aft => 7);

New_Line;

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<pre>r = 0.1123022, R0 = 3.8482793</pre>

=={{header|C}}==

{{trans|C++}}

#include <stdio.h>

 

printf("r = %f, R0 = %f\n", r, R0);

return 0;

{{out}}

<pre>r = 0.112302, R0 = 3.848279</pre>

=={{header|C++}}==

{{trans|Phix}}

#include <functional>

#include <iostream>

std::cout << "r = " << r << ", R0 = " << R0 << '\n';

return 0;

 

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=={{header|D}}==

{{trans|C++}}

import std.stdio;

 

double R0 = exp(12 * r);

writeln("r = ", r, ", R0 = ", R0);

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<pre>r = 0.112302, R0 = 3.84828</pre>

=={{header|FreeBASIC}}==

{{trans|Phix}}

Const n0 = 27 ' number of cases at day 0

Dim Shared As Integer actual(1 To ...) => { _

Dim As Double R0 = Exp(12 * r)

Print Using "r = #.##########, R0 = #.########"; r; R0

{{out}}

<pre>r = 0.1123021757, R0 = 3.84827928</pre>

{{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))

 

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=={{header|Java}}==

{{trans|Kotlin}}

import java.util.function.Function;

 

System.out.printf("r = %.16f, R0 = %.16f\n", r, r0);

}

{{out}}

<pre>r = 0.1123021756975501, R0 = 3.8482792809167194</pre>

{{works with|jq}}

'''Works with gojq, the Go implementation of jq'''

def n0: 27; # number of cases at day 0

((solve(f; 0.5; 0) * 1e10)|round / 1e10 ) as $r

| ((((12 * $r)|exp) * 1e8)|round / 1e8) as $R0

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<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]

=={{header|Kotlin}}==

{{trans|D}}

 

const val K = 7.8e9

val r0 = exp(12.0 * r)

println("r = $r, R0 = $r0")

{{out}}

<pre>r = 0.11230217569755005, R0 = 3.8482792809167194</pre>

=={{header|Nim}}==

{{trans|Kotlin}}

 

const

let r = f.solve()

let r0 = exp(12 * r)

 

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=={{header|Perl}}==

{{trans|Phix}}

use warnings;

 

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

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

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

=={{header|Phix}}==

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

<span style="color: #000080;font-style:italic;">-- demo\rosetta\Curve_fit.exw</span>

<span style="color: #008080;">with</span> <span style="color: #008080;">javascript_semantics</span>

<span style="color: #000000;">R0</span> <span style="color: #0000FF;">=</span> <span style="color: #7060A8;">exp</span><span style="color: #0000FF;">(</span><span style="color: #000000;">12</span> <span style="color: #0000FF;">*</span> <span style="color: #000000;">r</span><span style="color: #0000FF;">)</span>

<span style="color: #7060A8;">printf</span><span style="color: #0000FF;">(</span><span style="color: #000000;">1</span><span style="color: #0000FF;">,</span><span style="color: #008000;">"r = %.10f, R0 = %.8f\n"</span><span style="color: #0000FF;">,{</span><span style="color: #000000;">r</span><span style="color: #0000FF;">,</span><span style="color: #000000;">R0</span><span style="color: #0000FF;">})</span>

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<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');

}

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<pre>For period: Dec 31 - Apr 5

=={{header|Ruby}}==

{{trans|C++}}

N0 = 27

ACTUAL = [

end

 

{{out}}

<pre>r = 0.11230215850810021, R0 = 3.8482784871191575</pre>

=={{header|Vlang}}==

{{trans|wren}}

const (

k = 7800000000 // approx world population

r0 := math.round(math.exp(12 * r) * 1e8) / 1e8

println("r = $r, R0 = $r0")

{{out}}

<pre>r = 0.1123021757, R0 = 3.84827928</pre>

=={{header|Wren}}==

{{trans|Phix}}

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

 

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

var R0 = ((12 * r).exp * 1e8).round / 1e8

 

{{out}}