var objs = [
{name: "Peter", age: 35},
{name: "John", age: 27},
{name: "Jake", age: 28}
];
objs.reduce(function(accumulator, currentValue) {
return accumulator + currentValue.age;
}, 0); // 35 + 27 + 28 = 90let array = [36, 25, 6, 15];
array.reduce((acc, curr) => acc + curr, 0)
// 36 + 25 + 6 + 15 = 82function getArraySum(a){
var total=0;
for(var i in a) {
total += a[i];
}
return total;
}
var payChecks = [123,155,134, 205, 105];
var weeklyPay= getArraySum(payChecks); //sums up to 722var array = [36, 25, 6, 15];
array.reduce(function(accumulator, currentValue) {
return accumulator + currentValue;
}, 0); // 36 + 25 + 6 + 15 = 82You 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.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