Logistic curve fitting in epidemiology: Difference between revisions
Logistic curve fitting in epidemiology (view source)
Revision as of 14:41, 14 October 2021
, 2 years ago→{{header|jq}}
Line 564:
{{out}}
<pre>r = 0.1123021756975501, R0 = 3.8482792809167194</pre>
=={{header|jq}}==
{{trans|Wren}}
{{works with|jq}}
'''Works with gojq, the Go implementation of jq'''
<lang jq>def K: 7800000000; # approx world population
def n0: 27; # number of cases at day 0
def 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
];
def f:
. as $r
| reduce range(0; y|length) as $i (0;
(($r*$i)|exp) as $eri
| ((n0*$eri)/(1+n0*($eri-1)/K) - y[$i]) as $dst
| . + $dst * $dst);
def solve(f; $guess; $epsilon):
{ f0: ($guess|f),
delta: $guess,
factor: 2,
$guess}
# double until f0 best, then halve until delta <= epsilon
| until(.delta <= $epsilon;
.nf = ((.guess - .delta)|f)
| if .nf < .f0
then .f0 = .nf
| .guess = .guess - .delta
else .nf = ((.guess + .delta)|f)
| if .nf < .f0
then .f0 = .nf
| .guess = .guess + .delta
else .factor = 0.5
end
end
| .delta = .delta * .factor )
| .guess;
((solve(f; 0.5; 0) * 1e10)|round / 1e10 ) as $r
| ((((12 * $r)|exp) * 1e8)|round / 1e8) as $R0
| "r = \($r), R0 = \($R0)"</lang>
{{out}}
<pre>
r = 0.1123021757, R0 = 3.84827928
</pre>
=={{header|Julia}}==
| |||