Given two 2D arrays, how can I assert whether or not they are equal?
For example:
array1 = [[1,0],[2,1],[3,0]]
array2 = [[1,0],[2,1],[3,1]]
What is an efficient way to check whether array1 == array2?
If by equality you mean the array contents have same elements in the same order, then the shortest (though not the fastest) way will be:
JSON.stringify(array1) === JSON.stringify(array2)
This will work with arrays of any dimensions.
UPDATE: If you need a really fast algorithm then simple iteration will work better. However it is less fool-proof, hence to make it really safe and secure you'll need to spend more development time. Here is one possible solution for modern browsers:
function equal(array1, array2) {
if (!Array.isArray(array1) && !Array.isArray(array2)) {
return array1 === array2;
}
if (array1.length !== array2.length) {
return false;
}
for (var i = 0, len = array1.length; i < len; i++) {
if (!equal(array1[i], array2[i])) {
return false;
}
}
return true;
}
The following JSPerf speed test shows the supremacy of this algorithm over the short JSON approach: http://jsperf.com/2d-array-comparion.
The most efficient way would always depend on the size of the array/s and on your application and usage for them. You can check for the lengths of the 2 arrays for an early termination in case of a non match, but this might be considered as an extra step if that case rarely happens.
boolean areEqual(array array1, array array2){
if array1.length != array2.length
return false;
for (int i=0;i<array1.length; i++)
if(!areEqual(array1[i], array2[i])
return false;
return true;
}
boolean areEqual(int first, int second){
return first == second;
}
The way to go would be to iterate all values in a nested loop, compare value by value. Use a boolean which you set to false at the first mismatch, and assert that this boolean is true in your test.
Related
What is a quick way to determine if an array is only composed of 0's and 1's?
I'm vaguely familiar with a boolean solution but am not away how to implement it. In order to implement is you would have to assign values to the boolean
true = 0
and
false = 1
But how would you implement this if you are given an array such that
array = [1,1,1,1,0,0,0,0]
isArrayBool (){
}
Very very naive solution:
function isArrayBool(array) {
for (var i of array) {
if (i !== 0 && i !== 1) return false;
}
return true;
}
console.log(isArrayBool([1,0,0,0,1])); // true
console.log(isArrayBool([1])); // true
console.log(isArrayBool([2,3,0])); // false
So I threw a few functions together in jsperf to test. The fastest one I found so far is the one below (which is much faster than the for of version):
function isBoolFor(arr) {
for(var i=arr.length-1; i>=0; --i){
if(arr[i] !== 0 && arr[i] !== 1) return false
}
return true
}
Comparison is here
EDIT: I played around with another version that turned out to be quicker:
function isBoolFor(arr) {
for(var i=0; arr[i] === 0 || arr[i] === 1; i++){}
return i === arr.length
}
The key here is that because most of the array you are checking will consist of zeros and ones, you can avoid doing two boolean checks every iteration by using the || instead of the && negative check. It is only a subtle improvement, and could be argued to be no better than the one above in practicality.
UPDATE: So the difference between all the for and while variants is too subtle to declare an overall winner. So in that case I would go for readability. If you want binary use typed arrays!
FINAL UPDATE:
I was curious and wanted to find an even faster version, and it can be done if the array is known to be only numbers. If the array contains only numbers, then you can use a bitwise operator check for either of the values in question. For example:
function isBoolFor(arr) {
const test = ~0x01
for(var i=arr.length-1; i>=0; --i){
if(test & arr[i]){ return false }
}
return true
}
https://jsperf.com/bool-array-test-2/1
As we can see here, the fastest way to do this can be seen here... With a simple for loop, as suggested by user Jaromanda in a comment. This for loop blows other solutions out of the water in terms of speed.
var someArray = [1,0,1,1,1,1,0,0,1,0,1,0,0,1,0,1,0,1,1,1,1,0,1,1,0,1,0];
function simpleForLoop(array){
var numArgs = someArray.length;
for(var loop = 0; loop < numArgs; loop++){
var arg = someArray[loop];
if(arg !== 0 && arg !== 1){ return false; }
}
return true;
}
var isbool = simpleForLoop(someArray);
function ipsBetween(start, end){
var count = 0;
for(var i = 0; i < 4; i++) {
if(start.split('.').slice(i, i + 1) != end.split('.').slice(i, i + 1)) {
count++;
}
}
return count;
}
I am trying to find all possible IP's between a range. The above code is just a starting. I was trying to split the IP in pieces and check if they are equal or not. While I was doing so, interesingly even if the values are equal it evaluates the if statement as true and increases the count. Here is my test case:
ipsBetween("10.0.0.0", "10.0.0.50")
This test case returns 4, whereas it should return 1. I don't know why this is happening. I implicity looked the values of start.split('.').slice(i, i + 1) and end.split('.').slice(i, i + 1) and the first three element is seem to be equal.
There's really no need to use .slice() here. (That's what's causing the problem: .slice() returns an array, and two different arrays will never be equal to each other.) Split the strings first and then just use array indexing:
var count = 0;
start = start.split("."); end = end.split(".");
for (var i = 0; i < start.length; ++i)
if (start[i] != end[i])
count++;
return count;
The reason is that operator != when comparing two list objects will return true if they're not the very same object: split returns a list of strings but slice(i, i+1) will return a list of length 1.
This means that you're comparing ["10"] with another ["10"] and they're two different list objects, so != will return true.
If you just compare the contents using x.slice(".")[i] instead of using slice then the result is what you were expecting.
PS: The operator != of Javascript is terrible and you should not use it and prefer instead !==. It would be the same in this case, but it's much nicer to work with because it doesn't do crazy things when the two types are different.
PS2: Seems a good idea to split the strings at each iteration?
You are comparing arrays not strings you want to compare the string values try this instead:
function ipsBetween(start, end){
var count = 0;
for(var i = 0; i < 4; i++) {
if(start.split('.').slice(i, i + 1)[0] != end.split('.').slice(i, i + 1)[0]) {
count++;
}
}
return count;
}
console.log(ipsBetween("10.0.0.0", "10.0.0.50"));
The problem is the array objects returned won't equal each other because they are not the same array ie. they are not located at the same spot in memory...
Primordial topic, I know, but can anyone please comment on the following algorithm for comparing two ES5 arrrays:
function equal_arrays(a, b) {
"use strict";
var c = 0;
return a.every(function (e) {
return e === b[c++];
});
}
I think this is very clean and obvious use of [].every(), to quickly compare two arrays?
I have that nagging "surely it can not be that simple ?" feeling about it ?
NOTE: two arrays are equal when they contain all elements in the same position strictly equal to each other. So both the element index and element value have to be exactly equal. Values of different types are considered not equal. Sparse arrays are also compared.
TEST CASES:
equal_arrays([],[]) ; // => true
equal_arrays([1,2],[1,2]) ; // => true
equal_arrays([1,,2],[1,,2]) ; // => true
equal_arrays([,],[,]); // => true
equal_arrays([1,,3,,,],[1,,3,,,]); // => true
Uses cases yielding => false, one can imagine herself. Comparing non-arrays is a syntax error.
Many thanks to the well meaning and helpful contributors. It appears that the "best" (there is never such a thing) implementation is this:
function has(element, index)
{
return this[index] === element;
}
function equal_arrays(a, b)
{
return (a.length === b.length) && a.every(has, b) && b.every(has, a);
}
#tom's implementation of the two-way every() which is necessary sp that test cases like this one work:
equal_arrays([1,,3],[1,2,3]); //=> false
Once again thanks ...
No, it is not valid. From the MDN docs:
callback is invoked only for indexes of the array which have assigned values; it is not invoked for indexes which have been deleted or which have never been assigned values.
So if the first array has 'gaps' they will be skipped over.
equal_arrays([1, 2, , 4], [1, 2, 4]); // true
equal_arrays([1, 2, 4], [1, 2, , 4]); // false
Here is a better implementation:
function equal_arrays(a, b) {
if (a.length != b.length) return false;
var i;
for (i = 0; i < a.length; i++) {
if (a[i] !== b[i]) return false;
}
return true;
}
Here is an elegant implementation of #RobG's two-way every() idea:
function has(element, index)
{
return this[index] === element;
}
function equal_arrays(a, b)
{
return (a.length === b.length) && a.every(has, b) && b.every(has, a);
}
I have that nagging "surely it can not be that simple ?" feeling about it ?
No, it isn't. Using every on one array will test it for every member of that array but not the other. You have to test both ways.
Also, it depends on your criteria for "equal". You might want every member of the same index to have the same value, but you might not be concerned about order so is [1,2] equal to [2,1]? and is [,,,2] equal to [2]?
Also, should [1,,2] equal [1,undefined,2], i.e. should a member that doesn't exist be "equal" to one that does but has the value undefined?
An array comparison function
The following may do the job. It compares both ways, checking own properties and values. I think it covers all cases but am willing to be convinced otherwise. Might be better as a separate function, but adding to Array.prototype is convenient.
// Test that own properties of two arrays have the same values
Array.prototype.isEqualTo = function(a) {
var visited = {};
if (this.length != a.length) return false;
// Test every own property of this, remember tested properties
for (var p in this) {
if (this.hasOwnProperty(p)) {
visited[p] = p;
if (!(a.hasOwnProperty(p)) || this[p] !== a[p]) {
return false;
}
}
}
// Reverse test that comparison array only has tested properties
for (var q in a) {
if (a.hasOwnProperty(q) && !(q in this)) {
return false;
}
}
return true;
}
console.log([1,,2].isEqualTo([1,undefined,2])); // false, not defined != undefined
console.log([1,,2].isEqualTo([1,2])); // false, different length
console.log([1,2].isEqualTo([1,2])); // true
Note that inherited properties should be ignored as if arrays from different windows are compared (e.g. one is from a frame) then inherited properties won't be equal.
As an alternative, if you just want to check if both arrays are exactly the same you could just do this:
var a = [1,2,3,4];
var b = [1,2,3,4];
JSON.stringify(a) == JSON.stringify(b); //= true
That should work with arrays of numbers and strings.
As CrazyTrain said, the increment value is useless and you need to check for length first:
function equalArrays( a, b ) {
"use strict";
if (a.length !== b.length) return false;
return a.filter(function(el){ return el;}).every(function(e, i) {
return e === b[i];
});
}
Is a valid algorithm to compare arrays by value.
I have this array (below) and I'm trying to check if it has specific values.
var a = [ true, "lipsum" ];
What I need to do, is to check if a[0] is true and if a[1] is "lipsum"
I could check both values separately:
a[0] === true && a[1] === 'lipsum' // true
...to shorten the code a bit, I tried to do this:
a === [ true, 'lipsum'] // false
Why is this code example above false and is there another way to achieve what I'm trying to do?
I could do this:
a.join() === 'true,lipsum' // true
though I can't help but feel that there is a better way..?
jsfiddle
For only two elements to check the straightforward way seems best, but I assume you want to do this for maintenance reasons because eventually you may have several conditions to check (not just two). If so, you can do something like the following, which seems verbose for only two conditions, but as you start adding more it would be more reasonable, so here's an example with 5 conditions to check:
// set a constant somewhere for your truth condition
var COND = [1, 'a', 5, 'b', 0];
// check `a` against the constant array using `every` (Thanks Bergi)
if (a.every(function(v, i){ return COND[i] === v; })) {
// all array elements the same
}
Each array is a separate object, so the equality operator cannot be used to compare them. Assuming that you have a strict comparison of known arguments to do, the first method you use is the best.
If you have another array of arguments that the original array must contain, you must use a loop, although you could abstract it:
Array.prototype.contains = function (array) {
for (var x = 0; x < array.length; x++) {
if (this.length < x || this[x] !== array[x]) {
return false;
}
}
return true;
}
http://jsfiddle.net/q5DvG/1/
Possible Duplicate:
Easiest way to find duplicate values in a javascript array
How do I check if an array has duplicate values?
If some elements in the array are the same, then return true. Otherwise, return false.
['hello','goodbye','hey'] //return false because no duplicates exist
['hello','goodbye','hello'] // return true because duplicates exist
Notice I don't care about finding the duplication, only want Boolean result whether arrays contains duplications.
If you have an ES2015 environment (as of this writing: io.js, IE11, Chrome, Firefox, WebKit nightly), then the following will work, and will be fast (viz. O(n)):
function hasDuplicates(array) {
return (new Set(array)).size !== array.length;
}
If you only need string values in the array, the following will work:
function hasDuplicates(array) {
var valuesSoFar = Object.create(null);
for (var i = 0; i < array.length; ++i) {
var value = array[i];
if (value in valuesSoFar) {
return true;
}
valuesSoFar[value] = true;
}
return false;
}
We use a "hash table" valuesSoFar whose keys are the values we've seen in the array so far. We do a lookup using in to see if that value has been spotted already; if so, we bail out of the loop and return true.
If you need a function that works for more than just string values, the following will work, but isn't as performant; it's O(n2) instead of O(n).
function hasDuplicates(array) {
var valuesSoFar = [];
for (var i = 0; i < array.length; ++i) {
var value = array[i];
if (valuesSoFar.indexOf(value) !== -1) {
return true;
}
valuesSoFar.push(value);
}
return false;
}
The difference is simply that we use an array instead of a hash table for valuesSoFar, since JavaScript "hash tables" (i.e. objects) only have string keys. This means we lose the O(1) lookup time of in, instead getting an O(n) lookup time of indexOf.
You could use SET to remove duplicates and compare, If you copy the array into a set it will remove any duplicates. Then simply compare the length of the array to the size of the set.
function hasDuplicates(a) {
const noDups = new Set(a);
return a.length !== noDups.size;
}
One line solutions with ES6
const arr1 = ['hello','goodbye','hey']
const arr2 = ['hello','goodbye','hello']
const hasDuplicates = (arr) => arr.length !== new Set(arr).size;
console.log(hasDuplicates(arr1)) //return false because no duplicates exist
console.log(hasDuplicates(arr2)) //return true because duplicates exist
const s1 = ['hello','goodbye','hey'].some((e, i, arr) => arr.indexOf(e) !== i)
const s2 = ['hello','goodbye','hello'].some((e, i, arr) => arr.indexOf(e) !== i);
console.log(s1) //return false because no duplicates exist
console.log(s2) //return true because duplicates exist
Another approach (also for object/array elements within the array1) could be2:
function chkDuplicates(arr,justCheck){
var len = arr.length, tmp = {}, arrtmp = arr.slice(), dupes = [];
arrtmp.sort();
while(len--){
var val = arrtmp[len];
if (/nul|nan|infini/i.test(String(val))){
val = String(val);
}
if (tmp[JSON.stringify(val)]){
if (justCheck) {return true;}
dupes.push(val);
}
tmp[JSON.stringify(val)] = true;
}
return justCheck ? false : dupes.length ? dupes : null;
}
//usages
chkDuplicates([1,2,3,4,5],true); //=> false
chkDuplicates([1,2,3,4,5,9,10,5,1,2],true); //=> true
chkDuplicates([{a:1,b:2},1,2,3,4,{a:1,b:2},[1,2,3]],true); //=> true
chkDuplicates([null,1,2,3,4,{a:1,b:2},NaN],true); //=> false
chkDuplicates([1,2,3,4,5,1,2]); //=> [1,2]
chkDuplicates([1,2,3,4,5]); //=> null
See also...
1 needs a browser that supports JSON, or a JSON library if not.
2 edit: function can now be used for simple check or to return an array of duplicate values
You can take benefit of indexOf and lastIndexOf. if both indexes are not same, you have duplicate.
function containsDuplicates(a) {
for (let i = 0; i < a.length; i++) {
if (a.indexOf(a[i]) !== a.lastIndexOf(a[i])) {
return true
}
}
return false
}
If you are dealing with simple values, you can use array.some() and indexOf()
for example let's say vals is ["b", "a", "a", "c"]
const allUnique = !vals.some((v, i) => vals.indexOf(v) < i);
some() will return true if any expression returns true. Here we'll iterate values (from the index 0) and call the indexOf() that will return the index of the first occurrence of given item (or -1 if not in the array). If its id is smaller that the current one, there must be at least one same value before it. thus iteration 3 will return true as "a" (at index 2) is first found at index 1.
is just simple, you can use the Array.prototype.every function
function isUnique(arr) {
const isAllUniqueItems = input.every((value, index, arr) => {
return arr.indexOf(value) === index; //check if any duplicate value is in other index
});
return isAllUniqueItems;
}
One nice thing about solutions that use Set is O(1) performance on looking up existing items in a list, rather than having to loop back over it.
One nice thing about solutions that use Some is short-circuiting when the duplicate is found early, so you don't have to continue evaluating the rest of the array when the condition is already met.
One solution that combines both is to incrementally build a set, early terminate if the current element exists in the set, otherwise add it and move on to the next element.
const hasDuplicates = (arr) => {
let set = new Set()
return arr.some(el => {
if (set.has(el)) return true
set.add(el)
})
}
hasDuplicates(["a","b","b"]) // true
hasDuplicates(["a","b","c"]) // false
According to JSBench.me, should preform pretty well for the varried use cases. The set size approach is fastest with no dupes, and checking some + indexOf is fatest with a very early dupe, but this solution performs well in both scenarios, making it a good all-around implementation.
function hasAllUniqueChars( s ){
for(let c=0; c<s.length; c++){
for(let d=c+1; d<s.length; d++){
if((s[c]==s[d])){
return false;
}
}
}
return true;
}