JAVASCRIPT

javascript reduce

var array = [36, 25, 6, 15];

array.reduce(function(accumulator, currentValue) {
  return accumulator + currentValue;
}, 0); // 36 + 25 + 6 + 15 = 82

js reduce

// syntax: array.reduce(function, accumulator-initial-value)
let array = [3, 7, 2, 9, 5]
const result = array.reduce((accumulator, currentValue, currentIndex, arr) => {
  // code
}, initialValue)

// accumulator = will store return value of the function
// currentValue = iterate through each array element
// currentIndex = index of currentValue
// arr = original array
// initialValue = set accumulator initial value

reduce javascript

let array = [36, 25, 6, 15];
 
array.reduce((acc, curr) => acc + curr, 0)
// 36 + 25 + 6 + 15 = 82

A.reduce

const arr = [3, 5, 4, 11, 7, 22];

arr.reduce((previousValue, currentValue) => {
    return previousValue + currentValue;  // <--- 3 + 5 + 4 + 11 + 7 + 22 = 52
}, 0) 

reduce javascript

var depthArray = [1, 2, [3, 4], 5, 6, [7, 8, 9]];

depthArray.reduce(function(flatOutput, depthItem) {
    return flatOutput.concat(depthItem);
}, []);

=> [1, 2, 3, 4, 5, 6, 7, 8, 9];


// --------------
const prices = [100, 200, 300, 400, 500];
const total = prices.reduce((total, cur) => {return total + cur} )
//  init value = 0 => total = 1500;


// --------------
const prices = [100, 200, 300, 400, 500];
const total = prices.reduce(function (total, cur) => {
                   return total + cur
                }, 1 )
// init value = 1 => so when plus all numnber will be result => 1501;

// --------------
const prices = [100, 200, 300, 400, 500];
const total = prices.reduce((total, cur) => total + cur )
// 1500;

reduce()

const array1 = [1, 2, 3, 4];
const reducer = (accumulator, currentValue) => accumulator + currentValue;

// 1 + 2 + 3 + 4
console.log(array1.reduce(reducer));
// expected output: 10

// 5 + 1 + 2 + 3 + 4
console.log(array1.reduce(reducer, 5));
// expected output: 15

THE REDUCE() METHOD IN JAVASCRIPT

/*
This method takes a callback that takes two parameters,
one that represents the element from the last iteration and the other is the current
element of the iteration
*/

let nums = [2,4,6,8,3,5]

let result = nums.reduce((prev,curr)=>prev+curr)
console.log(result); // 28

Using Array.reduce() method

// define a reusable function
const calculateSum = (arr) => {
    return arr.reduce((total, current) => {
        return total + current;
    }, 0);
}

// try it
console.log(calculateSum([1, 2, 3, 4, 5]));

reduce

// Array.prototype.reduce()

const array1 = [1, 2, 3, 4];
const reducer = (accumulator, currentValue) => accumulator + currentValue;

// 1 + 2 + 3 + 4
console.log(array1.reduce(reducer));
// expected output: 10

// 5 + 1 + 2 + 3 + 4
console.log(array1.reduce(reducer, 5));
// expected output: 15

javascript reduce

const sum = array.reduce((accumulator, element) => {
  return accumulator + element;
}, 0);

JavaScript Array Methods .reduce()

const list = [1, 2, 3, 4, 5];
console.log(list.reduce((number, nextNumber) => number + nextNumber));
// --> 15

js .reducer method

const arrayOfNumbers = [1, 2, 3, 4];
 
const sum = arrayOfNumbers.reduce((accumulator, currentValue) => {  
  return accumulator + currentValue;
});
 
console.log(sum); // 10

js reduce

Sum array elements:
let myArr = [1,2,3,-1,-34,0]
return myArr.reduce((a,b) => a + b,0)

javascript reduce

[3, 2.1, 5, 8].reduce((total, number) => total + number, 0)

// loop 1: 0 + 3
// loop 2: 3 + 2.1
// loop 3: 5.1 + 5
// loop 4: 10.1 + 8
// returns 18.1

reduce javascript

 let sum = arr.reduce((curr, index) => curr + index);

JS Reduce example

const grades = [60, 55, 80, 90, 99, 92, 75, 72];
const total = grades.reduce(sum);
function sum(acc, grade) {  
return acc + grade;
}
const count = grades.length;
const letterGradeCount = grades.reduce(groupByGrade, {});
function groupByGrade(acc, grade) {  
const { a = 0, b = 0, c = 0, d = 0, f = 0 } = acc;  
if (grade >= 90) {    
return { ...acc, a: a + 1 };  
} else if (grade >= 80) {    
return { ...acc, b: b + 1 };  
} else if (grade >= 70) {    return { ...acc, c: c + 1 };  
} else if (grade >= 60) {    return { ...acc, d: d + 1 };  
} else {    return { ...acc, f: f + 1 };  }}
console.log(total, total / count, letterGradeCount);

reduce in js

let x = [1, 2, 3]; // 2 % 2 == 0
let v = x.reduce(e => e % 2 == 0);
console.log(v) // true 
//reduce return true or false

syntax of reduce in js

[1,2,3,4,5].reduce((acc, current)=>acc+current, 0)

.reduce javascript

reduce function sample solution

js array reduce

// Arrow function
reduce((previousValue, currentValue) => { /* ... */ } )
reduce((previousValue, currentValue, currentIndex) => { /* ... */ } )
reduce((previousValue, currentValue, currentIndex, array) => { /* ... */ } )
reduce((previousValue, currentValue, currentIndex, array) => { /* ... */ }, initialValue)

// Callback function
reduce(callbackFn)
reduce(callbackFn, initialValue)

// Inline callback function
reduce(function(previousValue, currentValue) { /* ... */ })
reduce(function(previousValue, currentValue, currentIndex) { /* ... */ })
reduce(function(previousValue, currentValue, currentIndex, array) { /* ... */ })
reduce(function(previousValue, currentValue, currentIndex, array) { /* ... */ }, initialValue)

javascript reduce function array

let numbers = [1, 2, 3];
let sum = numbers.reduce(function (previousValue, currentValue) {
    return previousValue + currentValue;
});

reduce method in javascript

const months = ['april','may','june','may','may','june'];

const countDuplicates = months.reduce((obj,month)=>{
    if(obj[month] == undefined){
        obj[month] = 1;
        return obj;
    }else{
        obj[month]++;
        return obj;
    }
},{});
console.log(countDuplicates);//output:{april: 1, may: 3, june: 2}

array reduce

arr.reduce(callback( accumulator, currentValue[, index[, array]] ) {
  // return result from executing something for accumulator or currentValue
}[, initialValue]);

js reduce

// Arrow function
reduce((previousValue, currentValue) => { ... } )
reduce((previousValue, currentValue, currentIndex) => { ... } )
reduce((previousValue, currentValue, currentIndex, array) => { ... } )
reduce((previousValue, currentValue, currentIndex, array) => { ... }, initialValue)

// Callback function
reduce(callbackFn)
reduce(callbackFn, initialValue)

// Inline callback function
reduce(function(previousValue, currentValue) { ... })
reduce(function(previousValue, currentValue, currentIndex) { ... })
reduce(function(previousValue, currentValue, currentIndex, array) { ... })
reduce(function(previousValue, currentValue, currentIndex, array) { ... }, initialValue)

reduce method

function reduce(array, combine, start) {
  let current = start;
  for (let element of array) {
    current = combine(current, element);
  }
  return current;
}

console.log(reduce([1, 2, 3, 4], (a, b) => a + b, 0));
// → 10

reduce

var sum = _.reduce([1, 2, 3], function(memo, num){ return memo + num; }, 0);
=> 6

.reduce mdn

arr.reduce(callback( accumulator, currentValue[, index[, array]] )[, initialValue])

javascript reduce

You can return whatever you want from reduce method, reduce array method good for computational problems
const arr = [36, 25, 6, 15];
array.reduce(function(accumulator, currentValue) {
  return accumulator + currentValue;
}, 0); // this will return Number
array.reduce(function(accumulator, currentValue) {
  return accumulator + currentValue;
}, []); // this will return array
let {total} = array.reduce(function(accumulator, currentValue) {
  let {currentVal} = currentValue
  return accumulator + currentVal;
}, {
total: 0
}); // this will return object
And one additional rule is always always retrun the first param, which do arithmetic operations.

reduce method

The reduce() method executes a reducer function (that you provide) on each element of the array, resulting in single output value.

The reducer function takes four arguments:
Accumulator (acc)
Current Value (cur)
Current Index (idx)
Source Array (src)

//syntax
arr.reduce(callback( accumulator, currentValue, [, index[, array]] )[, initialValue])
//example flatten an array

let flattened = [[0, 1], [2, 3], [4, 5]].reduce(
  ( accumulator, currentValue ) => accumulator.concat(currentValue),
  []
)

reduce in javascript

// Reduce in javascript advance
// output: { '22': 2, '24': 2, '26': 1 }
const user = [
  {name:"Abhishek",age:24},
  {name:"Dhruval",age:22},
  {name:"Anish",age:26},
  {name:"Aakash",age:22},
  {name:"Darshil",age:24},
]
const newUser = user.reduce((acc,cur)=>{
  if(acc[cur.age]){
    acc[cur.age] = acc[cur.age] +1 
  }
  else{
    acc[cur.age] = 1 
  }
  return acc;
},{})

console.log(newUser)

reduce in js

var totalYears = pilots.reduce(function (accumulator, pilot) {
  return accumulator + pilot.years;
}, 0);

reduce javascript

//note idx and sourceArray are optional
const sum = array.reduce((accumulator, element[, idx[, sourceArray]]) => {
	//arbitrary example of why idx might be needed
	return accumulator + idx * 2 + element 
}, 0);

reduce method

[0, 1, 2, 3, 4].reduce(
  (accumulateur, valeurCourante) => accumulateur + valeurCourante;
);

array reduce javascript

const numbers = [260, 25, 35];

console.log(numbers.reduce(myFunc))

function myFunc(total, num) {
  return total - num;
}
/*console.log will show 200 since the first element minus 35 and minus 25 is 260-25-35=200*/
/*while the name is "reduce", you do NOT have to subtract. Reduce() is more about "take the first element as total, and do something(whatever the function says) to it for each of the remaining elements; the remaining elements will be identified as the second parameter num for the function myFunc, you do whatever once for all remaining elements*/

reduce javascript

// use of reduce on an array, returns an array into a single value

// define the array
let numbers = [25, 25, 20]

// using the .reduce() to ADD all the numbers into a single value
// we are storing the reduce method into a single variable called sum
let sum = numbers.reduce(function(total, current) {
	return total + current // adds the current value (10) and the total value (0) 
}, 0); // the }, 0) is important because we want our total value to be 0.

JavaScript Reducer

const reviews = [4.5, 4.0, 5.0, 2.0, 1.0, 5.0, 3.0, 4.0, 1.0, 5.0, 4.5, 3.0, 2.5, 2.0];
const reviewsLetter = ['c', 'd', 'f', 'a', 'g', 'a', 'd', 'f', 'e', 'b'];
// 1. Using the reduce function, create an object that// has properties for each review value, where the value// of the property is the number of reviews with that score.// for example, the answer should be shaped like this:// { 4.5: 1, 4: 2 ...}
const countGroupedByReview = reviewsLetter.reduce(groupBy, {});
function groupBy (acc, review){  
  const count = acc[review] || 0;  
  return {...acc, [review]: count + 1}}
console.log(countGroupedByReview);

Reduce use

//Array Reduce
let arr=[12,14,15,28]
let numarr=arr.reduce((h1,h2)=>{
  return h1+h2
})
console.log(numarr)

reduce javascript

let fil = res.map(x=>Object.keys(x).filter(x=>x!='process' && x!='relay_type')
  .reduce((a,b)=>{
    a[b] = x[b]
    return a
  },{})).map(x=>Object.assign(x,{total:'wew'}))

javascript reduce

let arr = [1,2,3]

/*
reduce takes in a callback function, which takes in an accumulator
and a currentValue.

accumulator = return value of the previous iteration
currentValue = current value in the array

*/

// So for example, this would return the total sum:

const totalSum = arr.reduce(function(accumulator, currentVal) {
	return accumulator + currentVal
}, 0);

> totalSum == 0 + 1 + 2 + 3 == 6

/* 
The '0' after the callback is the starting value of whatever is being 
accumulated, if omitted it will default to whatever the first value in 
the array is, i.e: 

> totalSum == 1 + 2 + 3 == 6

if the starting value was set to -1 for ex, the totalSum would be:
> totalSum == -1 + 1 + 2 + 3 == 5
*/


// arr.reduceRight does the same thing, just from right to left

javascript reducers

// function that takes in the current state, and an action to update it 
// you just pass whatever the action you want to perform on the current state
// and it will go ahead and update it 

const arr = [2, 10, 40];
const initValue = 0;

const sumWithInitValue = arr.reduce((prevValue, currentValue) => {
  		return prevValue + currentValue 
	}, initValue);


console.log(`sum is ${sumWithInitValue}`)	// > "sum is 52"

reduce javascript

let arr = [{
name:"MAERON",
target:1,
},
{
name:"MAERON1",
target:1,
}]

let final = arr.map(x=> Object.keys(x).filter(x=> ['target'].includes(x)).reduce((a,b)=> {
a[b] = x[b]
return a
},{}))
console.log(final)

reduce method

const doubledAndSummed = [1, 2, 3].reduce(function(accumulator, element){ return element * 2 + accumulator}, 0)
// => 12

reduce function javascript

array.reduce(function(acumulador, elementoAtual, indexAtual, arrayOriginal), valorInicial)

js reduce

let newArray = array.reduce( (acc, element, index, originalArray) => {
    // return acc += element
}, initialValue)

JavaScript Array reduce()

const numbers = [45, 4, 9, 16, 25];
let sum = numbers.reduce(myFunction);

function myFunction(total, value, index, array) {
  return total + value;
}

reduce

const arr = [5, 7, 1, 8, 4];const sum = arr.reduce(function(accumulator, currentValue) {  return accumulator + currentValue;});// prints 25console.log(sum);

reduce function in javascript

reduce method does not changed original array

reduce function in javascript

reduce function