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Assuming i have an array of arrays:
[[2012-12-20, 10], [2012-12-20, 25], [2012-12-19, 10], [2012-12-20, 5]]
How to reorganize this array by distinct DATE summing their respective values, so i have something like this:
[[2012-12-20, 40], [2012-12-19, 10]]
EDIT
I've tried something like this:
var array = [[2012-12-20, 10], [2012-12-20, 25], [2012-12-19, 10], [2012-12-20, 5]];
pastDate = '';
pastVal = '';
for(var data in array) {
var curDate = data[0];
var curVal = data[1];
// ??
// Can't figure out how to organize each value of array organized by distinct date.
// Sorry for any fail, I tottaly sux in arrays.
}
Someone?
Thanks in advance!
Try this :
http://jsbin.com/aNeQArek/2/edit
var g=[['2012-12-20', 10], ['2012-12-20', 25], ['2012-12-19', 10], ['2012-12-20', 5]];
var a={};
for( var i=0;i<g.length;i++)
{
a[g[i][0]] = (a[g[i][0]] || 0) + g[i][1] ;
}
console.log(a);
result : (it's object with your desired values).
{2012-12-20: 40, 2012-12-19: 10}
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I have an array like so:
const dynamicNumber = 4
const array = [10, 15, 20, 25, dynamicNumber]
So what i would like to do, is eliminate all the terms that are smaller than that dynamic number.
So in this case, I would like to create a copy of the array and remove the values which are greater:
const newArray = [4]
Check out array.filter:
const dynamicNumber = 4
const array = [10, 15, 1, 2, 20, 25, dynamicNumber]
const filteredArray = array.filter(number => number < dynamicNumber)
console.log(filteredArray)
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How do I compare elements in an array using Javascript?
I want to know if cas[1], cas[2] and cas[3] have the same value.
cas = ["0","1", "2","3","4","5","6","7","8","9"];
if(cas[1]== cas[2] && cas[2]==cas[3]){
console.log("yea");
}
The first of all it's better to use ===, because 0 == false is true, but 0 === false is false.
=== works without type casting.
You don't need full cycle loop here, you have to compare only the first 3 elements so if statement is ok here.
If you want to do that with every 3 elements, you can do something like this.
const getThreeArrayElements = (array, startIndex) => array.slice(startIndex, startIndex + 3);
getThreeArrayElements([1, 2, 3, 3, 3, 3], 0); // [1, 2, 3]
getThreeArrayElements([1, 2, 3, 3, 3, 3], 3); // [3, 3, 3]
So you can easily get an array with 3 required elements.
The another task is how to compare them.
const areArrayElementsEqual = array => array.every(item => item === array[0]);
areArrayElementsEqual([1, 2, 3]); // false
areArrayElementsEqual([3, 3, 3]); // true
if you're looking for an algorithm to check the entire array :
let cas = ["0","1", "2","3","4","5","6","7","8","9"];
let checkCas = (cur = 1)=>{return cas[cur-1]===cas[cur]?true:cur<cas.length?checkCas(++cur):false;};
console.log(checkCas());
cas = ["0","1", "2","3","4","5","5","7","8","9"];
console.log(checkCas());
as for ticktacktoe :
let casa = ["0","1", "2","3","4","5","6","7","8","9"];
let allEqual = (cas,cur = 1)=>{return cas[cur-1]===cas[cur]?true:cur<cas.length?allEqual(cas,++cur):false;};
for(let i = 0;i<Math.floor(casa.length/3);i++)
{
let tempLengArr = [casa[i],casa[i+3],casa[i+6]];
if(allEqual(casa.slice(i*3,(i*3)+3))||allEqual(tempLengArr))
console.log("won");
}
if(allEqual([casa[0],casa[4],casa[9]])||
allEqual([casa[3],casa[4],casa[2]])) console.log("won");
might have messed up a bit on my allEqual call but it's an idea
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I have started learning Javascript and is doing some simple exercises. Given an array of numbers, return uneven numbers. I Have used both the "Classical" way of doing it as well as an arrow function. However they are acting a bit weird. This is the arrow function:
const answer2 = [1, 2, 3, 4, 5, 6, 7, 8, 9].filter(value => {
return value % 2 != 0;
})
It looks alright to me and it seems to work. Then I did the classical one:
function filterArrayToOdd(inputArray) {
let outputArray = [];
for (let i = 0; i < inputArray.length; i++) {
if (i % 2 != 0) {
outputArray.push(inputArray[i]);
}
}
return outputArray;
}
This returns an array with all the even numbers! Changing the comparrison from != to == works, but why?!
It would be better if your test array was not in sequence, [1, 2, 3, 4, 5, 6, 7, 8, 9], but rather something like, [2, 5, 9, 11, 12, 3].
In sequence, you can produce false positives to this problem, because, as #Pointy is saying, in your second case, you are checking your array index, not the actual value.
// instead of
if (i % 2 != 0) {
// do this:
if (inputArray[i] % 2 != 0) {
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I have array of arrays in javascript that looks like: [[1, 23.34, -5.22], [1, 2.34, -52.22], [2, 0.34, -5.02], ...]. I need to group by by the first number in the arrays and produce 2 different ouputs:
Object (key: 2D array) that looks like this:
{1: [[23.34, -5.22], [2.34, -52.22]],
2: [[0.34, -5.02], ...],
...}
3D array that would just signify the grouping:
[[[23.34, -5.22], [2.34, -52.22]], [[0.34, -5.02], ...], ...]
I want to use ES6. Any suggestions would be greatly appreciated.
checkout this code
let array = [[1, 23.34, -5.22], [1, 2.34, -52.22], [2, 0.34, -5.02]];
let obj = {};
array.forEach(subArray =>
{
let key = subArray[0];
if(!obj[key])
obj[key] = [];
subArray.splice(0,1);
obj[key].push(subArray);
});
console.log(obj);
let keys = Object.keys(obj);
let array3D = [];
keys.forEach(key => {
array3D.push(obj[key])
});
console.log(array3D);
There's already a answer which uses map method but I find using forEach in this case more appropriate.
const a = [
[1, 23.34, -5.22], [1, 2.34, -52.22], [2, 0.34, -5.02]
];
let b = {};
let c = [];
a.forEach(x=>{
if(!b[x[0]]){
b[x[0]] = [];
}
b[x[0]].push(x.slice(1));
});
console.log(b);
Object.values(b).forEach(x => c.push(x));
console.log(c);
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Say you have a list of numbers:
var list = [4, -12, 18, 1, -3];
What is a more elegant solution to finding the value closest to zero without nesting a whole lot of if/else statements?
use reduce:
list.reduce((pre,cur) => Math.abs(pre) > Math.abs(cur) ? cur : pre)
This is what you could do with a single traversal of the array.
function findClosestToZero(arr) {
if (arr.length === 0) {
return;
}
var closeNumber = Math.abs(arr[0]);
arr.forEach(function(number) {
if (Math.abs(number) < closeNumber) {
closeNumber = Math.abs(number)
}
});
return closeNumber;
}
var arr = [-5, 98, -4, 7, 9, 213, 4, -2, 1];
console.log(findClosestToZero(arr));
use sort :
list.sort((a,b) => Math.abs(a)-Math.abs(b))[0];
get the absolute value of all numbers
Math.abs()
sort the number in ascending order
Array.sort()
The first number is the required answer.