javascript reduce
var array = [36, 25, 6, 15];
array.reduce(function(accumulator, currentValue) {
return accumulator + currentValue;
}, 0); // 36 + 25 + 6 + 15 = 82javascript reduce
var array = [36, 25, 6, 15];
array.reduce(function(accumulator, currentValue) {
return accumulator + currentValue;
}, 0); // 36 + 25 + 6 + 15 = 82reduce 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;javascript reduce
/* reduce method: combine an array's values together, left to right */
let myArray = [1, 2, 3];
let a = myArray.reduce((initial, current) => initial + current, 0);
// [0 + 1 + 2 + 3] → 6
let b = myArray.reduce((initial, current) => initial + (current*10), 7);
// [7 + (1*10) + (2*10) + (3*10)] → 67
// Combine right to left with .reduceRight
let c = myArray.reduceRight((initial, current) => initial / current, 30);
// [30 / 3 / 2 / 1] → 5
/* Syntax: array.reduce((accumulator, currentValue) =>
combineFunction, accumulatorStartValue); */js .reducer method
const arrayOfNumbers = [1, 2, 3, 4];
const sum = arrayOfNumbers.reduce((accumulator, currentValue) => {
return accumulator + currentValue;
});
console.log(sum); // 10javascript 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 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);Copyright © 2021 Codeinu
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