Symmetric difference

Question

Symmetric wiki difference
We need to find the symmetric difference of the arrays.
I wrote a function for comparing two arrays, in which I combine arrays into one and look for duplicate values in a loop. When finding delete.

function sym() { var args = Array.prototype.slice.call(arguments); var result = compareTwoArray(arguments[0], arguments[1]); if (arguments.length > 2) { for (var i = 2; i < arguments.length; i++) { // console.log('result = ' + result); result = compareTwoArray(result, arguments[i]); } } return result; } function compareTwoArray() { var args = Array.prototype.slice.call(arguments); var result = []; var newArr = args.reduce(function(prev, curr) { return prev.concat(curr); }) for (var i = 0; i < newArr.length; i++) { var count = 0; for (var j = i+1; j < newArr.length; j++) { if (newArr[i] === newArr[j]) { count += 1; newArr.splice(j, 1); j -= 1; } } if (count === 0) { result.push(newArr[i]); } } return result; } console.log(sym([1, 2, 3], [3, 1, 5])); // [2, 5] console.log(sym([1, 1, 3], [4, 6])); // [1, 3, 4, 6] console.log(sym([1, 1, 2, 5], [2, 2, 3, 5])); // [1, 3] console.log(sym([1, 1, 2, 5], [2, 2, 3, 5], [3, 4, 5, 5])); // [1, 4, 5] console.log(sym([1, 2, 5], [2, 3, 5], [3, 4, 5])); // [1, 4, 5]

I do not know how to handle duplicate values.
In one case, it must be removed.

 [1, 2, 3] [3, 1, 5] [2, 5] - должен получиться [2, 5] - получается с моим кодом

In the other leave.

 [1, 1, 3] [4, 6] [1, 3, 4, 6] - должен получиться [3, 4, 6] - получается с моим кодом

In the other and leave (1) and delete (2).

 [1, 1, 2, 5] [2, 2, 3, 5] [1, 3] - должен получиться [3] - получается с моим кодом

for example, what answer is expected with: console.log(sym([1, 1, 2, 5], [2, 2, 3, 5], [3, 4, 5, 5]));
@Grundy in the text of the question there are after all the answers.
The symmetric difference reminds me of XOR in the algebra of logic.
Works: take the first two arrays, delete intersecting numbers [1, 1, 2, 5], [2, 2, 3, 5] => [1, 3] .
We get a new array and compare it with the following array [1, 3], [3, 4, 5, 5] => [1, 4, 5] .
I have two functions in the code compareTwoArray() for comparison and sym() for returning the answer, in the case of arrays> 2, it is called in the cycle compareTwoArray() , where result return from the previous value, and arguments[i] next array.
@Grundy repeating elements in one array should be reduced to one element

Visman Visman 16.2k 8 gold signs 21 silver marks 52 bronze marks · Accepted Answer · 2016-10-03T03:47:55

Solution for two arrays. Although 3 or more arrays can be processed in steps: first the first two, then the result of the first processing and the third array, and so on.

 function sum(arr1, arr2) { var tmp = arr1.concat(arr2), result = [], value, sum; for (var i = 0; i < tmp.length; i++) { var value = tmp[i]; if (result.indexOf(value) == -1) { sum = 0; if (arr1.indexOf(value) != -1) { sum++; } if (arr2.indexOf(value) != -1) { sum++; } if (sum == 1) { result.push(value); } } } return result; } console.log(sum([1, 2, 3], [3, 1, 5])); // [2, 5] console.log(sum([1, 1, 3], [4, 6])); // [1, 3, 4, 6] console.log(sum([1, 1, 2, 5], [2, 2, 3, 5])); // [1, 3]

Grundy grundy 58.6k 7 golden marks 54 silver marks 95 bronze marks · Answer 2 · 2016-10-03T10:16:52

In javascript, there is a class Set , which in this case can be used as a set, as a result, the record is obtained almost according to the formula

In the example below, the spread operator was used , and the rest parameters

 function sym(...arrs) { return arrs.reduce((acc, arr, i) => { var accSet = new Set(acc), // множество из элементов первого массива arrSet = new Set(arr); // множество из элементов второго массива return [...accSet].filter(a => !arrSet.has(a)) // элементы первого множества без элементо второго .concat( // объединение [...arrSet].filter(a => !accSet.has(a)) // элементы второго множества без элементов первого ); }, []); } console.log(sym([1, 2, 3], [3, 1, 5])); // [2, 5] console.log(sym([1, 1, 3], [4, 6])); // [1, 3, 4, 6] console.log(sym([1, 1, 2, 5], [2, 2, 3, 5])); // [1, 3] console.log(sym([1, 1, 2, 5], [2, 2, 3, 5], [3, 4, 5, 5])); // [1, 4, 5] console.log(sym([1, 2, 5], [2, 3, 5], [3, 4, 5])); // [1, 4, 5]

Gorgeous, as always :) Only if it is in the browser, it should not be very new (and if it’s completely old, my decision will not work)
@tutankhamun, did not quite understand the last sentence :-)
Well, all this syntactic beauty does not work in all browsers (and rest, as I have looked, generally only in Firefox)

Answer 3 · 2016-10-03T09:17:32

Alternatively, you can still do this:

 function sym() { var diff = []; // Допускаем, что симметричная разность пустого массива - пустой массив, // а симметричная разность одного массива - разность пустого массива // и этого массива равна самому этому массиву. // Также симметрическая разность равна объединению минус пересечение, // поэтому... [].forEach.call(arguments, function(arg) { var sum = diff.concat(arg); // ...объединяем массивы... diff = sum.filter(function(item) { // ...оставляем только те элементы, которые не найдены хотя бы // в одном из массивов... return (-1 === arg.indexOf(item) || -1 === diff.indexOf(item)); }); }); // ...возвращаем уникальные элементы // (иначе sym([1, 1, 2, 5], [2, 2, 3, 5]) будет [1, 1, 3]) return diff.filter(function(item, index, self) { return self.indexOf(item) === index; }); } console.log(sym()); console.log(sym([1, 2, 3])); console.log(sym([1, 2, 3], [3, 1, 5])); console.log(sym([1, 1, 3], [4, 6])); console.log(sym([1, 1, 2, 5], [2, 2, 3, 5])); console.log(sym([1, 1, 2, 5], [2, 2, 3, 5], [3, 4, 5, 5])); console.log(sym([1, 2, 5], [2, 3, 5], [3, 4, 5]));

For console.log(sym([1, 1, 2, 5], [2, 2, 3, 5], [3, 4, 5, 5]));
@Visman Well, sort of the author of this question, and sought
@Visman and, by the way, your version gives the same result if you execute console.log(sum(sum([1, 1, 2, 5], [2, 2, 3, 5]), [3, 4, 5, 5]));
My function is calculated for two arrays, how did you check it with 3 arrays?
@Visman as you indicated "the result of the first processing and 3 array" that is sum(sum([1, 1, 2, 5], [2, 2, 3, 5]), [3, 4, 5, 5]) :)

user207618 · Answer 4 · 2016-10-03T04:02:02

Based on the description:

Symmetric difference of two sets

Here is an example for two sets (I think you will guess how to screw more sets):

 function getDiff(a, b){ let result = [], pointer = -1; if(!Array.isArray(a) || !Array.isArray(b)) throw new Error(`Both arguments must be arrays!`); a.forEach(e => { if((pointer = b.indexOf(e)) !== -1) b.splice(pointer, 1); else if(result.indexOf(e) === -1) result.push(e); }); b.forEach(e => { if((pointer = a.indexOf(e)) !== -1) a.splice(pointer, 1); else if(result.indexOf(e) === -1) result.push(e); }); return result; } // Пример с вики console.info(getDiff([1, 2, 3, 4, 5], [3, 4, 5, 6, 7])); // Ваши примеры console.info(getDiff([1, 2, 3], [3, 1, 5])); console.info(getDiff([1, 1, 3], [4, 6])); console.info(getDiff([1, 1, 2, 5], [2, 2, 3, 5]));

Symmetric difference

4 answers 4

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