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I am wondering how you would go about deleting arrays that contain the same elements in a 2 dimensional array.
For example:
let 2dArr = [ [1, 2, 3],
[3, 2, 1],
[2, 4, 5],
[4, 5, 2],
[4, 3, 1] ];
This array would delete the second and fourth elements, returning the 2d array:
returnedArr = [ [1, 2, 3],
[2, 4, 5],
[4, 3, 1] ];
How exactly could this be done, preserving the 2d array? I could only think to loop through elements, comparing elements via a sort, and deleting them as you go along, but this would result in an indexing error if an element is deleted.
1) You can easily achieve the result using reduce and Set as:
let twodArr = [
[1, 2, 3],
[3, 2, 1],
[2, 4, 5],
[4, 5, 2],
[4, 3, 1],
];
const set = new Set();
const result = twodArr.reduce((acc, curr) => {
const key = [...curr].sort((a, b) => a - b).join();
if (!set.has(key)) {
set.add(key);
acc.push(curr);
}
return acc;
}, []);
console.log(result);
2) You can also use filter as:
let twodArr = [
[1, 2, 3],
[3, 2, 1],
[2, 4, 5],
[4, 5, 2],
[4, 3, 1],
];
const set = new Set();
const result = twodArr.filter((curr) => {
const key = [...curr].sort((a, b) => a - b).join();
return !set.has(key) ? (set.add(key), true) : false;
});
console.log(result);
const seen = []
const res = array.filter((item) => {
let key = item.sort().join()
if(!seen.includes(key)){
seen.push(key)
return item
}
})
console.log(res)
You can use hash map
let arr = [ [1, 2, 3], [3, 2, 1],[2, 4, 5],[4, 5, 2],[4, 3, 1] ];
let obj = {}
let final = []
for(let i=0; i<arr.length; i++){
// create a key
let sorted = [...arr[i]].sort((a,b)=> a- b).join`,`
// check if this is not present in our hash map
// add value to final out and update hash map accordingly
if(!obj[sorted]){
obj[sorted] = true
final.push(arr[i])
}
}
console.log(final)
Using Array.prototype.filter() and a Set as thisArg
let arr = [ [1, 2, 3],
[3, 2, 1],
[2, 4, 5],
[4, 5, 2],
[4, 3, 1] ];
let res = arr.filter(function(e){
const sorted = [...e].sort((a,b) => a-b).join('|');
return this.has(sorted) ? false : this.add(sorted)
},new Set)
console.log(res)
I am trying to see if a particular pair of numbers exist in an array. I know that .some() works effectively when it's a simple array, but I can't seem to get it to work in nested arrays.
Thank you in advance.
const array = [[1, 0], [2, 0], [3, 1], [4, 3], [5, 2]];
// checks whether an element exists
const exists = (i) => i == [3, 1] || [1, 3];
console.log(array.some(exists));
// expected output: true
You need to check every value because of different arrays, you have different object references, and you misused the logical OR operator. This does not include a comparison with another value.
const array = [[1, 0], [2, 0], [3, 1], [4, 3], [5, 2]];
const exists = ([a, b]) => a === 3 && b === 1 || a === 1 && b === 3;
console.log(array.some(exists));
Here is a general solution.
const array = [[1, 0], [2, 0], [3, 1], [4, 3], [5, 2]];
const expected = [[1, 3], [3, 1]];
const exists = (i) => expected.some((j) => {
return i.every((k, n) => j[n] === k);
});
console.log(array.some(exists));
This could be wrapped in a function to make it reusable.
const array = [[1, 0], [2, 0], [3, 1], [4, 3], [5, 2]];
const expected = [[1, 3], [3, 1]];
const containsAny = (array, expected) => {
return array.some((i) => expected.some((j) => {
return i.every((k, n) => j[n] === k);
}))
};
console.log(containsAny(array, expected));
You could make a custom function out of it which makes it reusable.
This is a simple example. By flattening the array you can check if values are the same.
Got my inspiration on https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/some
const array = [[1, 0], [2, 0], [3, 1], [4, 3], [5, 2]];
const checkArrayFor = (needle, haystack) => {
const flattenNeedle = needle.join(':');
const flattenArray = haystack.map(item => item.join(':'));
return flattenArray.includes(flattenNeedle);
};
console.log("Check for [1, 0]", checkArrayFor([1, 0], array));
console.log("Check for [1, 1]", checkArrayFor([1, 1], array));
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I have multiple arrays in a main/parent array like this:
var arr = [
[1, 17],
[1, 17],
[1, 17],
[2, 12],
[5, 9],
[2, 12],
[6, 2],
[2, 12],
[2, 12]
];
I have the code to select the arrays that are repeated 3 or more times (> 3) and assign it to a variable.
The code is:
var arr = [[1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]]
arr.sort((a, b) => a[0] - b[0] || a[1] - b[1])
// define equal for array
const equal = (arr1, arr2) => arr1.every((n, j) => n === arr2[j])
let GROUP_SIZE = 3
first = 0, last = 1, res = []
while(last < arr.length){
if (equal(arr[first], arr[last])) last++
else {
if (last - first >= GROUP_SIZE) res.push(arr[first])
first = last
}
}
if (last - first >= GROUP_SIZE) res.push(arr[first])
console.log(res)
So the final result is:
console.log(repeatedArrays);
>>> [[1, 17], [2, 12]]
My problem: But the problem is, I have an array like this {from: [12, 0], to: [14, 30]}.
var arr = [
[1, 17],
[1, 17],
[1, 17],
[2, 12],
[5, 9],
[2, 12],
[6, 2],
{from: [12, 0], to: [14, 5]},
{from: [12, 0], to: [14, 5]},
{from: [4, 30], to: [8, 20]},
{from: [12, 0], to: [14, 5]},
{from: [4, 30], to: [8, 20]},
[2, 12],
[2, 12]
];
When I try to use the above code, it doesn't work. The error message is:
Uncaught TypeError: arr1.every is not a function
The final result should be:
console.log(repeatedArrays);
>>> [[1, 17], [2, 12], {from: [12, 0], to: [14, 5]}]
How can I make that code above work?
If you introduce a non array into the mix, you need to handle it differently.
Yours already work with array so I'm adding object style check for both sort and equal.
var arr = [
[1, 17],
[1, 17],
[1, 17],
[2, 12],
[5, 9],
[2, 12],
[6, 2],
{ from: [4, 30], to: [8, 21] },
{ from: [12, 0], to: [14, 5] },
{ from: [12, 0], to: [14, 5] },
{ from: [4, 30], to: [8, 20] },
{ from: [12, 0], to: [14, 5] },
{ from: [4, 30], to: [8, 20] },
[2, 12],
[2, 12]
];
arr.sort((a, b) => {
if (a instanceof Array && b instanceof Array) {
return a[0] - b[0] || a[1] - b[1]
} else if (a instanceof Array || b instanceof Array) {
return a instanceof Array ? -1 : 1
} else {
return a.from[0] - b.from[0] || a.from[1] - b.from[1] || a.to[0] - b.to[0] || a.to[1] - b.to[1]
}
});
// define equal for array
const equal = (arr1, arr2) => {
if (arr1 instanceof Array) {
return arr1.every((n, j) => n === arr2[j]);
} else {
if (arr2 instanceof Array) return false;
for (let k in arr1) {
if (!arr1[k].every((n, j) => n === arr2[k][j])) {
return false
}
}
return true;
}
};
let GROUP_SIZE = 3;
(first = 0), (last = 1), (res = []);
while (last < arr.length) {
if (equal(arr[first], arr[last])) last++;
else {
if (last - first >= GROUP_SIZE) res.push(arr[first]);
first = last;
}
}
if (last - first >= GROUP_SIZE) res.push(arr[first]);
console.log(res);
You can use the function reduce for grouping and counting the objects and then execute the function filter for getting the object with count >= 3.
var array = [ [1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12], [2, 12] ];
let result = Object.values(array.reduce((a, [c, b]) => {
let key = `${c}|${b}`;
(a[key] || (a[key] = {count: 0, value: [c, b]})).count++;
return a;
}, {})).filter(o => {
if (o.count >= 3) {
delete o.count;
return true;
}
return false;
}).map(({value}) => value);
console.log(result);
.as-console-wrapper { min-height: 100%; }
Really simple - filter it all, then remove duplicates with Set and JSON methods (because it's nested arrays not objects):
var array = [
[1, 17],
[1, 17],
[1, 17],
[2, 12],
[5, 9],
[2, 12],
[6, 2],
[2, 12],
[2, 12]
];
var repeatedArrays = [...new Set(array.filter(e => array.filter(f => JSON.stringify(e.sort()) == JSON.stringify(f.sort()))).map(JSON.stringify))].map(JSON.parse);
console.log(repeatedArrays);
I'm looking for a way to make comparisons between arrays in arrays.
let small = [[1, 3], [2, 2], [2, 3], [3, 0]];
let large = [[1, 0], [1, 1], [2, 0], [2, 2], [2, 5], [3, 0], [3, 2]];
For example, I would like to be able to find out how many of the arrays in small are found in large. Some function that, given the arrays above as arguments, would return 2, since [2, 2] and [3, 0] from small are found in large.
How would you go about doing that?
You can convert one of the arrays into a Set of hashes, and than filter the 2nd array using the set:
const small = [[1, 3], [2, 2], [2, 3], [3, 0]];
const large = [[1, 0], [1, 1], [2, 0], [2, 2], [2, 5], [3, 0], [3, 2]];
const containsCount = (arr1, arr2, hashFn) => {
const arr1Hash = new Set(arr1.map(hashFn));
return arr2.filter(s => arr1Hash.has(hashFn(s))).length;
}
const result = containsCount(small, large, ([a, b]) => `${a}-${b}`);
console.log(result);
You can try something like:
let small = [[1, 3], [2, 2], [2, 3], [3, 0]];
let large = [[1, 0], [1, 1], [2, 0], [2, 2], [2, 5], [3, 0], [3, 2]];
let z = zeta(small, large);
console.log(z);
function zeta(a, b) {
let join = m => m.join();
let x = a.map(join);
let y = b.map(join);
return x.reduce((n, m) => (y.indexOf(m)>0) ? ++n : n, 0);
}
I hope this helps!
Use every and some to compare the arrays with each other.
If you want to get an array containing the subarrays that match, use filter:
let result = small.filter(arr =>
large.some(otherArr =>
otherArr.length === arr.length && otherArr.every((item, i) => item === arr[i])
)
);
Which filters the subarray from small that some subarray from large has the same length and the same elements/items.
Demo:
let small = [[1, 3], [2, 2], [2, 3], [3, 0]];
let large = [[1, 0], [1, 1], [2, 0], [2, 2], [2, 5], [3, 0], [3, 2]];
let result = small.filter(arr =>
large.some(otherArr =>
otherArr.length === arr.length && otherArr.every((item, i) => item === arr[i])
)
);
console.log(result);
And if you want just a count, then use reduce instead of filter to count the mathched items (this makes use of the fact that the numeric value of true is 1 and that of false is 0):
let count = small.reduce((counter, arr) =>
counter + large.some(otherArr =>
otherArr.length === arr.length && otherArr.every((item, i) => item === arr[i])
)
, 0);
Demo:
let small = [[1, 3], [2, 2], [2, 3], [3, 0]];
let large = [[1, 0], [1, 1], [2, 0], [2, 2], [2, 5], [3, 0], [3, 2]];
let count = small.reduce((counter, arr) =>
counter + large.some(otherArr =>
otherArr.length === arr.length && otherArr.every((item, i) => item === arr[i])
)
, 0);
console.log(count);
Note: If the subarrays contain only numbers, the code could be simplified to use Array#toString instead of every and length comparaison:
let result = small.filter(arr => large.some(otherArr => "" + otherArr === "" + arr));
Which casts both arrays into strings and compares the two strings instead. This can be used with the reduce as well.
Create two nesting map() function oute for small and inner for large then use JSON.stringify()
let small = [[1, 3], [2, 2], [2, 3], [3, 0]];
let large = [[1, 0], [1, 1], [2, 0], [2, 2], [2, 5], [3, 0], [3, 2]];
var same=[];
small.map(function(element){
large.map(function(element2){
if(JSON.stringify(element)==JSON.stringify(element2)){
same.push(element);
}
});
});
console.log(same);
let small = [[1, 3], [2, 2], [2, 3], [3, 0]];
let large = [[1, 0], [1, 1], [2, 0], [2, 2], [2, 5], [3, 0], [3, 2]];
let largeArrayStringForm = large.map(item => item.toString())
let matchingItems = small.filter(item => largeArrayStringForm.includes(item.toString()))
console.log(`matchCount: ${matchingItems.length}`)
Definition：Creates an array of grouped elements, the first of which contains the first elements of the given arrays, the second of which contains the second elements of the given arrays, and so on.
Current Solution：
const zip = (...arr) => {
let maxLength = 0
let res = []
for (let el of arr) {
maxLength = Math.max(maxLength, el.length)
}
for (let j = 0; j < maxLength; j++) {
const foo = []
for (let n of arr) {
foo.push(n[j])
}
res.push(foo)
}
return res
}
Test Case:
test(('zip', () => {
expect(zip([1, 2], [4, 5], [9, 1])).toEqual([[1, 4, 9], [2, 5, 1]])
}
test('zip', () => {
expect(zip([1, 2, 3], [4, 5, 6])).toEqual([[1, 4], [2, 5], [3, 6]])
})
test('zip', () => {
expect(zip([1, 2], [], [3, 4, 5])).toEqual([
[1, undefined, 3],
[2, undefined, 4],
[undefined, undefined, 5],
])
})
I want to get a better way to achieve zip, current solution is ugly
See Destructuring Assignment and Array.prototype.map for more info.
// Proof.
const zip = (...args) => [...new Array(Math.max(...args.map(arr => arr.length)))].map((x, i) => args.map((y) => y[i]))
// Proof.
console.log(zip([1, 2], [4, 5], [9, 1])) // [[1, 4, 9], [2, 5, 1]]
console.log(zip([1, 2, 3], [4, 5, 6])) // [[1, 4], [2, 5], [3, 6]]
console.log(zip([1, 2], [], [3, 4, 5])) // [[1, undefined, 3], [2, undefined, 4], [undefined, undefined, 5]]