Here's an example of an array:
const arr = [{
a: 12,
b: "A"
c: 17
},
{
a: 12,
b: "B"
c: 17
},
{
a: 12,
b: "C"
c: 17
}
];
What is the most efficient way to calculate the sum of all objects in the array? The expected result is [29,29,29]
I attempted to use
arr.map(a => Object.values(a).reduce((p,c) => p+c))To simplify the reduction process, filter values by type before performing the operation
const arr = [{a: 12,b: "A",c: 17},{a: 12,b: "B",c: 17},{a: 12,b: "C",c: 17}];
let final = arr.map(a => Object.values(a).filter(v => typeof v === 'number').reduce((p, c) => p + c))
console.log(final)You can also achieve the same result within the reduce method itself by adding the current value based on its type, adding 0 if the type is not number
const arr = [{a: 12,b: "A",c: 17},{a: 12,b: "B",c: 17},{a: 12,b: "C",c: 17}];
let final = arr.map(a => Object.values(a).reduce((p, c) => p + (typeof c === 'number' ? c : 0)))
console.log(final)To decrease the value directly, you can use a finiteness check.
const
array = [{ a: 12, b: "A", c: 17 }, { a: 12, b: "B", c: 17 }, { a: 12, b: "C", c: 17 }],
result = array.map(o => Object
.values(o)
.reduce((s, v) => s + (isFinite(v) && v), 0)
);
console.log(result);Whether you consider the possible construction {a: "12"} as a number or a string will determine the result you receive. Using the isFinite function in this case will lead to an incorrect result.
If you view {a: "12"} as a number, you can utilize the following code:
const
array = [{ a: "12", b: "A", c: 17 }, { a: 12, b: "B", c: 17 }, { a: 12, b: "C", c: 17 }],
result = array.map(o => Object
.values(o)
.reduce((s, v) => s += +v || 0, 0)
);
console.log(result);If you perceive {a: "12"} as a string, you can implement the following code:
const
array2 = [{ a: "12", b: "A", c: 17 }, { a: 12, b: "B", c: 17 }, { a: 12, b: "C", c: 17 }],
result2 = array2.map(o => Object
.values(o)
.reduce((s, v) => s += (typeof v === 'number') ? v : 0, 0)
);
console.log(result2);Here's an alternative way to approach the problem:
data = [{x: 10, y: "A", z: 20},{x: 15, y: "B", z: 25},{x: 20, y: "C", z: 30}];
var output=Object.values(data).reduce((accumulator, element)=>{
let entries=Object.entries(element), sum=0;
entries.forEach(([key, value])=>{if(!isNaN(value))sum+=value;});
accumulator.push(sum);
return accumulator;
},[]);
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