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"Friendly Number" ( pythonic version ) solution in Clear category for [old] Friendly Number by capback250
function friendlyNumber(number, {decimals = 0, suffix = '', base = 1000, powers = ['', 'k', 'M', 'G', 'T', 'P', 'E', 'Z', 'Y']}={} ){
let exponent = 0;
while (exponent + 1 < powers.length && Math.abs(number) >= Math.pow(base, (exponent + 1))) exponent++;
number /= Math.pow(base, exponent);
number = decimals ? number.toFixed(decimals) : number > 0 ? Math.floor(number) : Math.round(number);
return number + powers[exponent] + suffix
}
May 17, 2017
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