7 ms·
Idiomatic underscore.js solution: for (var i = 1 ; i <= 100; i++) { console.log( _.chain(["Fizz", "Buzz"]) .zip(_.map([3, 5], function(n)
by adam-a 12y ago
Idiomatic underscore.js solution:
for (var i = 1 ; i <= 100; i++) {
console.log(
_.chain(["Fizz", "Buzz"])
.zip(_.map([3, 5], function(n) { return i%n;}))
.where({"1": 0})
.pluck(0)
.value().join("")
|| i);
}
- erikwiffin 12y agoA for loop seems like a weird choice for "idiomatic underscore.js". _.each(_.range(1, 100), function (i) { console.log( _.chain(["Fizz", "Buzz"]) .zip(_.map([3, 5], function(n) { return i%n;})) .where({"1": 0}) .pluck(0) .value().join("") || i); });