Reputation: 989
Why does a return of the push
method cause
Uncaught TypeError: acc.push is not a function
But a return concat
results in the correct solution?
[1, 2, 3, 4].reduce(function name(acc, curr) {
if (even(curr)) {
return acc.push(curr);
}
return acc;
}, []);
function even(number) {
if (number % 2 === 0) {
return true;
}
return false;
}
[1, 2, 3, 4].reduce(function name(acc, curr) {
if (even(curr)) {
return acc.concat(curr);
}
return acc;
}, []);
function even(number) {
if (number % 2 === 0) {
return true;
}
return false;
}
Upvotes: 77
Views: 82858
Reputation: 22769
concat()
creates a new array by merging the array the method is called on with all arguments flattened (1 level deep) into an array.
push()
just adds elements to an existing array.
Since concat()
creates a new array avoid it if you use a repetitive task like filling an array with values from another array, it's extremely slow:
` Chrome/117
-----------------------------------------------------------------------
create new array by push 1.0x | x10000 510 518 525 541 546
create new array by concat 109.8x | x100 560 565 572 582 590
-----------------------------------------------------------------------
https://github.com/silentmantra/benchmark `
let people = [
{ name: "John", age: [30, 40], city: "New York" },
{ name: "Alice", age: [30, 40], city: "Los Angeles" },
{ name: "Bob", age: [30, 40], city: "Chicago" },
{ name: "Eva", age: [30, 60], city: "San Francisco" },
{ name: "David", age: [30, 50], city: "Miami" }
];
const COUNT = 3000;
// @benchmark create new array by push
{
let count = COUNT, result = [];
while(count--) result.push(...people);
result;
}
// @benchmark create new array by concat
{
let count = COUNT, result = [];
while(count--) result = result.concat(people);
}
/*@end*/eval(atob('e2xldCBlPWRvY3VtZW50LmJvZHkucXVlcnlTZWxlY3Rvcigic2NyaXB0Iik7aWYoIWUubWF0Y2hlcygiW2JlbmNobWFya10iKSl7bGV0IHQ9ZG9jdW1lbnQuY3JlYXRlRWxlbWVudCgic2NyaXB0Iik7dC5zcmM9Imh0dHBzOi8vY2RuLmpzZGVsaXZyLm5ldC9naC9zaWxlbnRtYW50cmEvYmVuY2htYXJrL2xvYWRlci5qcyIsdC5kZWZlcj0hMCxkb2N1bWVudC5oZWFkLmFwcGVuZENoaWxkKHQpfX0='));
Upvotes: 0
Reputation: 14208
According to the MDN document say that:
push()
method: adds one or more elements to the end of an array and returns the new length of the array.const count = ['pigs', 'goats'].push('cows');
console.log(count); // expected output: 3
concat()
method is used to merge two or more arrays. This method does not change the existing arrays but instead returns a new arrayconsole.log(['a'].concat(['b']));// expected output: Array ["a", "b"]
And combined with the final Array#reduce
's parameter is the array initialize [])
, which means that you want to return an array result.
==> So that's the reason why in case that you use concat
working well.
Array#reduce
and Array#push
const even = (number) => number%2 === 0;
const result = [1, 2, 3, 4].reduce(function name(acc, curr) {
if(even(curr)) acc.push(curr); // Just add one more item instead of return
return acc;
}, []);
console.log(result);
Array#filter
const even = (number) => number%2 === 0;
console.log([1, 2, 3, 4].filter(even));
Upvotes: 14
Reputation: 791
https://dev.to/uilicious/javascript-array-push-is-945x-faster-than-array-concat-1oki Concat is 945x slower than push only because it has to create a new array.
Upvotes: 8
Reputation: 2217
acc
should not be an array. Look at the documentation. It can be one, but..
It makes no sense at all to reduce
an array to an array. What you want is filter
. I mean, reduce
using an array as the accumulator and concat
ing each element to it technically does work, but it is just not the right approach.
var res = [1, 2, 3, 4].filter(even);
console.log(res);
function even(number) {
return (number % 2 === 0);
}
Upvotes: 7
Reputation: 34168
The push()
adds elements to the end of an array and returns the new length of the array. Thus your return here is invalid.
The concat()
method is used to merge arrays. Concat does not change the existing arrays, but instead returns a new array.
Better to filter, if you want a NEW array like so:
var arr = [1, 2, 3, 4];
var filtered = arr.filter(function(element, index, array) {
return (index % 2 === 0);
});
Note that assumes the array arr is complete with no gaps - all even indexed values. If you need each individual, use the element
instead of index
var arr = [1, 2, 3, 4];
var filtered = arr.filter(function(element, index, array) {
return (element% 2 === 0);
});
Upvotes: 108