Logo

Programming-Idioms

# 208 Formula with arrays
Given the arrays a,b,c,d of equal length and the scalar e, calculate a = e*(a+b*c+cos(d)).
Store the results in a.
New implementation

Be concise.

Be useful.

All contributions dictatorially edited by webmasters to match personal tastes.

Please do not paste any copyright violating material.

Please try to avoid dependencies to third-party libraries and frameworks.

Other implementations 
#include <math.h>
  for (int i=0; i<n; i++)
    a[i] = e*(a[i]+b[i]*c[i]+cos(d[i]));
#include <valarray>
using namespace std;
valarray a {1, 2}, b {3, 4},
         c {5, 6}, d {7, 8};
a = e * (a + b * c + cos(d));
for (int i = 0; i < a.Length; i++)
  a[i] = e * (a[i] + b[i] * c[i] + Math.Cos(d[i]));
a = e*(a+b*c+cos(d))
import "math"
func applyFormula(a, b, c, d []float64, e float64) {
	for i, v := range a {
		a[i] = e * (v + b[i] + c[i] + math.Cos(d[i]))
	}
}
a.forEach((aa, i) => a[i] = e * (aa + b[i] * c[i] + Math.cos(d[i])))
import static java.lang.Math.cos;
import static java.util.stream.IntStream.range;
range(0, a.length)
    .forEach(i -> {
        a[i] = e * (a[0] + (b[1] * c[2]) + cos(d[3]));
    });
import static java.lang.Math.cos;
int i, n = a.length;
for (i = 0; i < n; ++i)
    a[i] = e * (a[0] + (b[1] * c[2]) + cos(d[3]));
for i := 0 to High(a) do
    a[i] = e*(a[i]+b[i]*c[i]+cos(d[i]);
$a[$_] = $e * ($a[$_] + $b[$_] * $c[$_] + cos $d[$_]) for 0 .. $#a;

import math
a = [e*(a[i] + b[i] + c[i] + math.cos(d[i])) for i in range(len(a))]
from math import cos
f = lambda a, b, c, d: \
    e * (a + (b * c) + cos(d))
a = list(map(f, a, b, c, d))
from math import cos
def f(a, b, c, d):
    return e * (a + (b * c) + cos(d))
a = list(map(f, a, b, c, d))
import math
for i in xrange(len(a)):
	a[i] = e*(a[i] + b[i] + c[i] + math.cos(a[i]))

a = a.zip(b,c,d).map{|i,j,k,l| e*(i+j*k+Math::cos(l)) }
for i in 0..a.len() {
    a[i] = e * (a[i] + b[i] * c[i] + d[i].cos());
}

Preview