Grouping consecutive elements together using Javascript
You can use a counter variable which has to be incremented and the difference between the index and the consecutive elements are the same, group them in a temporary array. If the difference is varies for two consecutive array elements, the temporary element has to be moved to the result
and the temporary array has to be assigned a new array object.
var array = [2, 3, 4, 5, 8, 9, 12, 13, 14, 15, 16, 17, 20];
var result = [], temp = [], difference;
for (var i = 0; i < array.length; i += 1) {
if (difference !== (array[i] - i)) {
if (difference !== undefined) {
result.push(temp);
temp = [];
}
difference = array[i] - i;
}
temp.push(array[i]);
}
if (temp.length) {
result.push(temp);
}
console.log(result);
# [ [ 2, 3, 4, 5 ], [ 8, 9 ], [ 12, 13, 14, 15, 16, 17 ], [ 20 ] ]
Given :
var data = [ undefined, undefined, 2, 3, 4, 5,
undefined,undefined, 8, 9,
undefined, undefined, 12, 13, 14, 15, 16, 17,
undefined, undefined, 20];
(or the almost equivalent array where the undefined
elements don't exist at all, but where the defined elements have the same indices as above) this reduce
call will return a two-dimensional array where each top level element is the contents of the original array, grouped by contiguously defined entries:
var r = data.reduce(function(a, b, i, v) {
if (b !== undefined) { // ignore undefined entries
if (v[i - 1] === undefined) { // if this is the start of a new run
a.push([]); // then create a new subarray
}
a[a.length - 1].push(b); // append current value to subarray
}
return a; // return state for next iteration
}, []); // initial top-level array
i.e. [[ 2, 3, 4, 5], [8, 9], [12, 13, 14, 15, 16, 17], [20]]
NB: this could also be written using a .forEach
call, but I like .reduce
because it requires no temporary variables - all state is encapsulated in the function parameters.