 FAQ: Iterators - The .reduce() Method

This community-built FAQ covers the “The .reduce() Method” exercise from the lesson “Iterators”.

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Not unless we consider the dictionary meaning of the words, and no they are not built into JavaScript.

Parameters are locally defined variables (in function scope). We declare them arbitrarily, but in keeping with common sense use terms that describe what the variable references, or its purpose.

(accumulator, currentValue) => accumulator + currentValue

The above callback is kept going until the object array is reduced to a single value. The array itself is unchanged. Put in simple terms,

Array.reduce((a, b) => a + b))

In a sense we can think of a as being the first term in the array, and b as the last. The last term is popped off the array and added to the first term, and the new value replaces the first term. This process is repeated until there is only one term, which is the resulting value of the method.

The heavy lifting is done by the reduce method, which has a built-in iterator for stepping through the array object.

Consider this mockup of the possible code behind the method…

function reduce(array) {
var clone = array.slice();  // a copy so the original is untouched
while (clone.length > 1) {
clone += clone.pop();
}
return clone
}
array = [1,1,2,3,5,8,13,21,34,55]
console.log(reduce(array))    // 143
4 Likes

Does the first chart of results miss the first iteration? Shouldn’t row 1 be:
Iteration accumulator current val return val
first 1 1

Study the array, then fill in the blanks in the table. Compute the sum and store in column 4.

[1, 2, 4, 10]

We first load the accumulator with the leftmost element of the array (index 0).

First    1

The currentValue is the next element in the array (index 1) which gives,

First    1    2

Now we compute the sum and return it. This becomes the accumulator in the second iteration. The index pointer is incremented and the process repeated.

First    1    2    3

If we were to use the algo posted earlier, the table would look like,

First      1    10    11
second    11     4    15
Third     15     2    17

since we load the currentValue with the rightmost element in the array.

2 Likes

2 posts were split to a new topic: What is a real world application of .reduce()?

13 posts were split to a new topic: How does .reduce() actually work?

For step one (and only step one, not moving on) I write:

const newSum = newNumbers.reduce( => {

});

But this is incorrect. What please is correct line of code for step one?

The function lacks the required two formal parameters:

(a, b) => {
return a + b;
}

Thank you! That was a great explanation and now I feel like I actually understand what’s going on in the example they used for teaching the reduce() method.

1 Like

can someone puhleaze!!! edit the exercises and make it more understandable?

I’m confused about the ‘second argument’ idea. How does this work???

const numbers = [1, 2, 4, 10];

const summedNums = numbers.reduce((accumulator, currentValue) => {
return accumulator + currentValue
}, 100) // <- Second argument for .reduce()

console.log(summedNums); // Output: 117

Sorry to pop by again, but I just don’t get it…

THIS …

const newNumbers = [1, 3, 5, 7];

const newSum = newNumbers.reduce((accumulator, currentValue) => {
console.log('The value of accumulator: ', accumulator);
console.log('The value of currentValue: ', currentValue);
return accumulator + currentValue;
}, 10);

console.log(newSum);

RETURNS THIS…

The value of accumulator: 10
The value of currentValue: 1
The value of accumulator: 11
The value of currentValue: 3
The value of accumulator: 14
The value of currentValue: 5
The value of accumulator: 19
The value of currentValue: 7
26

BUT THE ACCUMULATOR WAS ONLY DECLARED WITHIN THE CALLBACK FUNCTION, so how can it be that 10, which was only used as an argument outside of the callback function, can somehow equate to the accumulator???

The reduce method permits us to specify an initial value to accumulate from. That is the second argument. If omitted, the value is defined by the type of data in the array. It could be an empty string or zero (for addition) or one (for multiplication), depending whether working with character data or numbers. The initial value is loaded into the accumulator and the process of winding down the array begins, one value at a time computed with the accumulator value.

join() is a form of reducing an array to form a string.

a = 'abcdefghijklmnopqrstuvwxyz'
"abcdefghijklmnopqrstuvwxyz"
a.length
26
b = a.split('')
(26) ["a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l", "m", "n", "o", "p", "q", "r", "s", "t", "u", "v", "w", "x", "y", "z"]
c = b.reduce((u,v) => u + v, "\$ ")
"\$ abcdefghijklmnopqrstuvwxyz"
c = b.reduce((u,v) => u + v)
"abcdefghijklmnopqrstuvwxyz"
a = '1234567'
b = a.split('').reduce((u, v) => +u + +v)
28
c = a.split('').reduce((u, v) => +u * +v)
5040

What happens if I try to pass only one argument into .reduce method?

const numbers = [1, 2, 4, 10];
const summedNums = numbers.reduce((accumulator) => {
return accumulator + accumulator
})
console.log(summedNums) // Output: 8

Why is the output 8?

As near as I can tell that outcome is,

2 ** (numbers.length -1)

As the array size is increased the outcome goes up in powers of two.

At any length, the reduce method functions correctly when we give the callback two parameters.

summedNumbers = numbers.reduce((a, b) => a + b)
1 Like

Here’s why the output is 8.

Since you do not provide an initial value (second argument) in your call to reduce, the first element (index 0) in the array is passed as the initial value of accumulator the first time reduce calls your callback. In addition, the second element (index 1) of the array is passed as the second argument to the callback the first time the callback is called.

After the first call to the callback, each subsequent call receives the value returned from the previous call as the first argument, and the next element in the array (continuing with the element at index 2) as the second argument.

So here are the calls that reduce performs on your callback, where I’ve given your callback the name callback as the name that the reduce method might use for it internally:

callback(1, 2) // returns 2 (1 + 1)
callback(2, 4) // returns 4 (2 + 2)
callback(4, 10) // returns 8 (4 + 4)

Because your callback ignores the second argument, there are only 2 things that affect the result: (1) the first element in the array, and (2) the length of the array.

If the array is empty, you’ll get an exception because you don’t supply a second argument to reduce. Otherwise, the result is numbers * 2 ** (numbers.length - 1)