JavaScript Array Reduce 

Just came from MDN 'cause I didnt understant a sheep!

Exactly.

You are very right! MDN didnt help at all...

Ikr he is awesome

So much clarity in 7 minutes video !

understand the elegant way and use the readable way if you want, after some months you will like to use the elegant way

I also think the "old way" is better. The "elegant way" is limited to iteration of type y_n = f(y_{n-1}, x_n) where f is the callback function. There is this new movement in programming where people think old fashion for-loops are ugly code.

If you were using an array, you could return an array in the reduce, and spread the accumulator. Then add any new value after. Example: nums = [1,2,3,4,5] const doubledNums = nums.reduce( (acc, num) => [...acc, num*2], [ ]) console.log(doubledNums) // [2,4,6,8,10] Obviosuly a .map could do the same thing but there are more complex examples where it is useful.

@@liamwelsh5565 Hi, I'm very new to coding, may I know why did you add the "..." before acc? I'm not really sure what it does here

@@Xetron1978 … is the spread operator. If you have an array, let’s call x and want to make a new array, let’s call y, that has all the values of x, you can spread x into y. Example: Const x = [1,2,3] Const y = […x] Console.log(y) // [1,2,3] You can also spread objects into objects

jajaj mosh es muy bien!

ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-a_Bfu1XWGmI.html

it is the implementation of the function reduce(). A implementation could be like that: const reduce = function(iterable, reduceFn, accumulator){ for(let i of iterable){ accumulator = reduceFn(accumulator, i) } return accumulator }

? each time we add to the accumulator. the current value starts at 0 mosh didn't explain this part but you can change the starting value with a last argument after teh callback.. in the exmple below the starting value and starting accumulator value would be 50. const final=numbers.reduce((accumulator, current val)=>{ return accumulator+current val }, 50); we start by adding the first number in the array to 50(or the first current value we are working with) then if that result was 52... then the accumulator is 52, then the current value moves on to the next index number in the array and keeps going until it adds each value in the array together onto the accumulator basically the current value is the current value in the array we are currently working with. each array method does whatver the callback says to do on EACH value in the array. the current value in moshes example becomes 0 at one point because we add -1 to 1... which is 0 hope this helped?

ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-a_Bfu1XWGmI.html

ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-a_Bfu1XWGmI.html

ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-a_Bfu1XWGmI.html

as in time complexity

@@rogerh2694 yes

ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-a_Bfu1XWGmI.html

ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-a_Bfu1XWGmI.html