Averages/Pythagorean means

Solutions

Solution 1 (Click to Show/Hide)
``````function pythagoreanMeans(rangeArr) {
// arithmeticMean :: [Number] -> Number
const arithmeticMean = xs =>
foldl((sum, n) => sum + n, 0, xs) / length(xs);

// geometricMean :: [Number] -> Number
const geometricMean = xs =>
raise(foldl((product, x) => product * x, 1, xs), 1 / length(xs));

// harmonicMean :: [Number] -> Number
const harmonicMean = xs =>
length(xs) / foldl((invSum, n) => invSum + (1 / n), 0, xs);

// GENERIC FUNCTIONS ------------------------------------------------------

// A list of functions applied to a list of arguments
// <*> :: [(a -> b)] -> [a] -> [b]
const ap = (fs, xs) => //
Array.prototype.concat(...fs.map(f => //
Array.prototype.concat(...xs.map(x => [f(x)]))));

// foldl :: (b -> a -> b) -> b -> [a] -> b
const foldl = (f, a, xs) => xs.reduce(f, a);

// length :: [a] -> Int
const length = xs => xs.length;

// mapFromList :: [(k, v)] -> Dictionary
const mapFromList = kvs =>
foldl((a, [k, v]) =>
(a[(typeof k === 'string' && k)] = v, a), {}, kvs);

// raise :: Num -> Int -> Num
const raise = (n, e) => Math.pow(n, e);
/*
// show :: a -> String
// show :: a -> Int -> String
const show = (...x) =>
JSON.stringify.apply(
null, x.length > 1 ? [x[0], null, x[1]] : x
);
*/
// zip :: [a] -> [b] -> [(a,b)]
const zip = (xs, ys) =>
xs.slice(0, Math.min(xs.length, ys.length))
.map((x, i) => [x, ys[i]]);

// TEST -------------------------------------------------------------------
// mean :: Dictionary
const mean = mapFromList(zip(
['Arithmetic', 'Geometric', 'Harmonic'],
ap([arithmeticMean, geometricMean, harmonicMean], [
rangeArr
])
));

return {
values: mean,
test: `is A >= G >= H ? \${mean.Arithmetic >= mean.Geometric &&
mean.Geometric >= mean.Harmonic ? 'yes' : 'no'}`
};
}
``````