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I want to filter a large array list into multiple arrays for every 5 items in a certain way so that [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] would be [[1, 2, [3, 4, 5]], [6, 7, [8, 9, 10]]] or [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15] would be [[1, 2, [3, 4, 5]], [6, 7, [8, 9, 10]], [11, 12, [13, 14, 15]]]. (All arrays will be a multiple of 5 in my program.)
How would I do this?
Right now I'm doing this
for (var i = 1; i < (stoneTextureUnfiltered.length+1)/1.01; i++) {
stoneTexture.push([stoneTextureUnfiltered[i], stoneTextureUnfiltered[i+1], stoneTextureUnfiltered[i+2], [stoneTextureUnfiltered[i+3], stoneTextureUnfiltered[i+4], stoneTextureUnfiltered[i+5]]]);
}
but it doesn't seem to be working.
Thanks,
-Voxel
Assuming you've chunked the array already into parts of 5 with these answers and it's stored in a variable named chunks, to wrap the last 3 in each chunk you can use map:
const final = chunks.map((chunk) => [chunk[0], chunk[1], chunk.slice(2)]);
You add the first and second elements to the new list, then add the rest of the chunk as a whole.
Demo below:
// using second answer
var perChunk = 5 // items per chunk
var inputArray = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15]
var chunks = inputArray.reduce((resultArray, item, index) => {
const chunkIndex = Math.floor(index/perChunk)
if(!resultArray[chunkIndex]) {
resultArray[chunkIndex] = [] // start a new chunk
}
resultArray[chunkIndex].push(item)
return resultArray
}, [])
// answer below
const final = chunks.map((chunk) => [chunk[0], chunk[1], chunk.slice(2)]);
console.log(final);
As you can see, it works nicely!
what's up? I hope you going well.
So, my question is, I have a array with number that I have to compare with another arrays (like 1 to X), what is the best way to:
1º compare the arrays and retrieve the numbers that are equals.
2º the numbers of elements that are equal (without using .length on the array with numbers are equals).
Example:
Array 1 = [1, 3, 4, 5, 6, 7, 8, 9, 11, 13, 16]
Array 2 = [1, 2, 3, 7, 9, 12, 16, 17]
That way, the total numbers is 5
And the numbers are: [1, 3, 7, 9, 16]
My method is using forEach and compare each item and using .length on the array with the numbers that are equals, there's another way or best way to do this?
Another example using more arrays:
Arr1 = [1, 2, 3, 5, 6, 7, 8, 10, 15]
Arr2 = [
[1, 3, 5, 7, 8, 9, 10, 11, 12],
[2, 5, 6, 7, 9, 10],
[1, 3, 5, 7, 10, 11, 13, 14, 15]
]
// Output
6, [1, 3, 5, 7, 8, 10]
5, [2, 5, 6, 7, 10]
7, [1, 3, 5, 6, 7, 10, 15]
Thanks for the answer.
I like using Set for this purpose. You can create a Set from your first array and then any lookup in that Set (using Set.has) is O(1) efficiency.
const arr1 = [1, 3, 4, 5, 6, 7, 8, 9, 11, 13, 16];
const arr2 = [1, 2, 3, 7, 9, 12, 16, 17];
const arr1Items = new Set(arr1);
const matched = arr2.filter(el => arr1Items.has(el));
console.log(matched.length, matched);
Arr2 is an array, not an object, your code would change accordingly
Arr1 = [1, 2, 3, 5, 6, 7, 8, 10, 15]
Arr2 = [
[1, 3, 5, 7, 8, 9, 10, 11, 12,],
[2, 5, 6, 7, 9, 10],
[1, 3, 5, 7, 10, 11, 13, 14, 15,]
]
Given that, the solution is a oneliner:
res = Arr2.map(a=>a.filter(x=>Arr1.indexOf(x)!=-1).length)
It should be straightforward but, just in case:
The [].indexOf(el) method give you the position of the parameter in the array, if that element is not present it will return -1. Therefore, the function
x => Arr1.indexOf(x)!=-1
returns true or false if x is present or not in the Arr1 array
The [].filter(fn) method use the fn function to evaluate every array element and give as result an array with the evaluated true elements.
a.filter(x => Arr1.indexOf(x)!=-1)
Means give me all the elements of array a presents in Arr1
Now we just have to count the lenght of that array
a.filter(x => Arr1.indexOf(x)!=-1).length
and pass this count to the [].map(fn) method to have the result we need.
The function I wrote below will give the results you want, but remember the function returns an array of arrays, even if the second parameter had only one array or was an array of elements instead of array of arrays (Works for both).
Arr1 = [1, 2, 3, 5, 6, 7, 8, 10, 15]
Arr2 = [
[1, 3, 5, 7, 8, 9, 10, 11, 12],
[2, 5, 6, 7, 9, 10],
[1, 3, 5, 7, 10, 11, 13, 14, 15]
]
Arr3 = [1, 5, 6, 9, 12, 15, 17]
function check(base_array,search_values)
{
if(base_array.length===0 || search_values.length===0)
{
return [];
}
else if(Array.isArray(search_values[0]))// Check if second parameter is an array of arrays.
{
var result=[];
search_values.forEach(search=>{
var result_sub=[];
search.forEach(key=>{
if(base_array.includes(key))
{
result_sub.push(key);
}
});
result.push(result_sub);
});
return result;
}
else
{
var result=[];
search_values.forEach(key=>{
if(base_array.includes(key))
{
result.push(key);
}
});
return [result];
}
}
console.log("Array of Arrays");
console.log(check(Arr1,Arr2));
console.log("Array of Elements");
console.log(check(Arr1,Arr3));
From the returned result you can loop through the value to get the elements and the number of elements by checking length of array.
result.forEach(element=>{
console.log(result.length, result);// number of elements doesn't have to be passed
});
What the Function does is it checks if any array is empty , then returns empty array [], if the second array is an array of arrays it loops through each array and then to each element in the sub array and checks if it exists in the first array, else if the array was array of elements, then it just loops through the elements and checks if it exists in the first array. And returns the result stored
I have two arrays.
var allAuth = [1, 2, 3, 4, 5, 6, 7, 8]
let thomasauth = [2, 3, 4, 5, 6, 7, 8]
let res = allAuth.filter(f => !thomasauth.includes(f));
I am returning the missing matching value from allAuth array which is 1.
I want to remove 1 from allAuth array.
I currently have
let filteredArr = allAuth.filter(e => e !== res)
but this just gives me back the allAuth array without anything being removed.
my desired output should be
let filteredArr = [2,3,4,5,6,7,8]
Can anyone help? Thanks.
Change !thomasauth.includes(f) to thomasauth.includes(f) so to filter only the included items (1 will be excepted in that case.)
var allAuth = [1, 2, 3, 4, 5, 6, 7, 8]
let thomasauth = [2, 3, 4, 5, 6, 7, 8]
let res = allAuth.filter(f => thomasauth.includes(f));
console.log(res);
Have you tried out your code? You will see that res is actually an array containing the values you want to remove ([1]) and not a single scalar value.
So I think you intended to do this:
let filteredArr = allAuth.filter(e => !res.includes(e));
var allAuth = [1, 2, 3, 4, 5, 6, 7, 8];
let thomasauth = [2, 3, 4, 5, 6, 7, 8];
let result = allAuth.filter(val => thomasauth.indexOf(val) !== -1);
console.log(result);
I don't understand this method of pop and unshift in this array
let nums = [1, 2, 3, 6, 9, 8, 7, 4];
const ids = [1, 2, 3, 6, 9, 8, 7, 4];
let btn5 = document.getElementById("btn5");
btn5.onclick = function() {
nums.unshift(nums.pop());
for (i = 0; i <= 7; i++) {
document.getElementById("btn" + ids[i]).innerHTML = nums[i];
}
}
nums.unshift(nums.pop()); is:
// Remove the last entry from the array
const tmp = nums.pop();
// Insert it at the beginning of the array
nums.unshift(tmp);
So for instance, the first time that runs, nums starts with:
[1, 2, 3, 6, 9, 8, 7, 4]
so pop removes the 4 from the end, and inserts it at the beginning:
[4, 1, 2, 3, 6, 9, 8, 7]
Live Example:
const nums = [1, 2, 3, 6, 9, 8, 7, 4];
console.log("before:", JSON.stringify(nums));
nums.unshift(nums.pop());
console.log("after: ", JSON.stringify(nums));
Details on MDN: pop, unshift.
Array pop method removes and return last element of an array. If you write something like,
let nums = [1, 2, 3, 6, 9, 8, 7, 4];
const ids = [1, 2, 3, 6, 9, 8, 7, 4];
const poppedValue = nums.pop(); //poppedValue = 4 and nums = [1, 2, 3, 6, 9, 8, 7]
And unshift method push the item at the beginning of the array.
nums.unshift(poppedValue); // nums = [4, 1, 2, 3, 6, 9, 8, 7];
I have an array of arrays
[
[1,3,5,7,8,8],
[1,3,5,7,8,8],
[1,3,5,7,8,8],
[1,3,5,7,8,8],
[1,3,5,7,8,8]
]
I am trying to insert a value between each item. So I have this:
let reelList = [
[1, 3, 5, 7, 8, 8],
[1, 3, 5, 7, 8, 8],
[1, 3, 5, 7, 8, 8],
[1, 3, 5, 7, 8, 8],
[1, 3, 5, 7, 8, 8]
]
reelList.map(reel => {
// Adds the separator (works)
let v = separate(reel, '-')
console.log(v)
return v
})
function separate(arr, value) {
return arr.reduce((result, element, index, array) => {
result.push(element)
index < array.length - 1 && result.push(value)
return result
}, []);
}
// Logs the new list to the console (doesn't work)
console.log(reelList)
When I log the values after I run the separate function they are separated, however, when I display reelList they are not separated. Why is that?
The map() function returns a new array. It doesn't modify the existing one.
You would need to set the results of reelList.map() to something else.
let reelList = [
[1, 3, 5, 7, 8, 8],
[1, 3, 5, 7, 8, 8],
[1, 3, 5, 7, 8, 8],
[1, 3, 5, 7, 8, 8],
[1, 3, 5, 7, 8, 8]
]
const finalResult = reelList.map(reel => {
// Adds the separator (works)
let v = separate(reel, '-')
console.log(v)
return v
})
function separate(arr, value) {
return arr.reduce((result, element, index, array) => {
result.push(element)
index < array.length - 1 && result.push(value)
return result
}, []);
}
// Logs the new list to the console (doesn't work)
console.log(finalResult);
If you want to edit the array in place, instead of using map(), use a forEach() with a callback that has a second and third parameter, which are index and array. Then you can update the array with the new values as you go.
let reelList = [
[1, 3, 5, 7, 8, 8],
[1, 3, 5, 7, 8, 8],
[1, 3, 5, 7, 8, 8],
[1, 3, 5, 7, 8, 8],
[1, 3, 5, 7, 8, 8]
]
reelList.forEach((reel, index, arr) => {
// Adds the separator (works)
let v = separate(reel, '-')
arr[index] = v;
console.log(v)
})
function separate(arr, value) {
return arr.reduce((result, element, index, array) => {
result.push(element)
index < array.length - 1 && result.push(value)
return result
}, []);
}
// Logs the new list to the console (doesn't work)
console.log(reelList);
You could map a new array with a calculated lenght and take the result of the calculated index ot the dash.
var array = [[1, 3, 5, 7, 8, 8], [1, 3, 5, 7, 8, 8], [1, 3, 5, 7, 8, 8], [1, 3, 5, 7, 8, 8], [1, 3, 5, 7, 8, 8]],
result = array.map(a => Array.from({ length: a.length * 2 - 1 }, (_, i) => a[i / 2] || '-'));
console.log(result);