Iterate an array as a pair (current, next) in JavaScript
In Ruby, this is called each_cons
(each consecutive):
(1..5).each_cons(2).to_a # => [[1, 2], [2, 3], [3, 4], [4, 5]]
It was proposed for Lodash, but rejected; however, there's an each-cons module on npm:
const eachCons = require('each-cons')
eachCons([1, 2, 3, 4, 5], 2) // [[1, 2], [2, 3], [3, 4], [4, 5]]
There's also an aperture
function in Ramda which does the same thing:
const R = require('ramda')
R.aperture(2, [1, 2, 3, 4, 5]) // [[1, 2], [2, 3], [3, 4], [4, 5]]
This answer is inspired by an answer I saw to a similar question but in Haskell: https://stackoverflow.com/a/4506000/5932012
We can use helpers from Lodash to write the following:
const zipAdjacent = function<T> (ts: T[]): [T, T][] {
return zip(dropRight(ts, 1), tail(ts));
};
zipAdjacent([1,2,3,4]); // => [[1,2], [2,3], [3,4]]
(Unlike the Haskell equivalent, we need dropRight
because Lodash's zip
behaves differently to Haskell's`: it will use the length of the longest array instead of the shortest.)
The same in Ramda:
const zipAdjacent = function<T> (ts: T[]): [T, T][] {
return R.zip(ts, R.tail(ts));
};
zipAdjacent([1,2,3,4]); // => [[1,2], [2,3], [3,4]]
Although Ramda already has a function that covers this called aperture. This is slightly more generic because it allows you to define how many consecutive elements you want, instead of defaulting to 2:
R.aperture(2, [1,2,3,4]); // => [[1,2], [2,3], [3,4]]
R.aperture(3, [1,2,3,4]); // => [[1,2,3],[2,3,4]]
Just make the "ugly" part into a function and then it looks nice:
arr = [1, 2, 3, 4];
function pairwise(arr, func){
for(var i=0; i < arr.length - 1; i++){
func(arr[i], arr[i + 1])
}
}
pairwise(arr, function(current, next){
console.log(current, next)
})
You can even slightly modify it to be able to make iterate all i, i+n pairs, not just the next one:
function pairwise(arr, func, skips){
skips = skips || 1;
for(var i=0; i < arr.length - skips; i++){
func(arr[i], arr[i + skips])
}
}
pairwise([1, 2, 3, 4, 5, 6, 7], function(current,next){
console.log(current, next) // displays (1, 3), (2, 4), (3, 5) , (4, 6), (5, 7)
}, 2)