// 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
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;
arr.reduce(callback(accumulator, currentValue[, index[, array]]){// return result from executing something for accumulator or currentValue}[, initialValue]);
functionreduce(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
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 arraylet{total}= array.reduce(function(accumulator, currentValue){let{currentVal}= currentValue
return accumulator + currentVal;},{total:0});// this will return objectAnd one additional rule is always always retrun the first param, which do arithmetic operations.
Thereduce() 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)CurrentValue(cur)CurrentIndex(idx)SourceArray(src)//syntax
arr.reduce(callback( accumulator, currentValue,[, index[, array]])[, initialValue])//example flatten an arraylet flattened =[[0,1],[2,3],[4,5]].reduce((accumulator, currentValue)=> accumulator.concat(currentValue),[])
//note idx and sourceArray are optionalconst sum = array.reduce((accumulator, element[, idx[, sourceArray]])=>{//arbitrary example of why idx might be neededreturn accumulator + idx *2+ element
},0);
// use of reduce on an array, returns an array into a single value// define the arraylet 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 sumlet 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.
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,{});functiongroupBy(acc, review){const count = acc[review]||0;return{...acc,[review]: count +1}}console.log(countGroupedByReview);
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'}))
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
// 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"
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)