比较两个单字符数组并返回差值？

Question

I need to be able to compare, and get the difference, between two arrays containing single character letters appearing multiple times in each array.

Example:

arr_a = ["E", "F", "X", "O", "U", "R", "T", "I", "D", "E", "N", "G", "Ø", "M", "F", "Æ", "A", "M", "Ø", "R", "Å", "N", "D", "E", "D", "R", "A", "_", "S", "E", "L", "U", "V"]

arr_b = ["E", "F", "X", "O", "U", "R", "T", "I", "D", "E", "N", "G", "Ø", "M", "F", "Æ", "A", "M", "Ø", "R", "Å", "N", "D", "E", "D", "R", "A", "G", "S", "E", "L", "U", "V", "O", "I", "A", "R", "E", "S", "E"]

I've tried quite a few found on google, but none return the expected result. The problem seems to be with algorithms that loop through each letter, and then manipulate the result. I need something that take into account the case that each letter can appear multiple times, and calculate the difference of how many times that actually happens.

Any hints?

Answer 1

You could check each character at the same position and return only the parts who are different.

 function getDifference(first, second) { var min = Math.min(first.length, second.length), i = 0, result = []; while (i < min) { if (first[i] !== second[i]) { result.push(first[i], second[i]); } ++i; } return result.concat(first.slice(min), second.slice(min)); } var arr_a = ["E", "F", "X", "O", "U", "R", "T", "I", "D", "E", "N", "G", "Ø", "M", "F", "Æ", "A", "M", "Ø", "R", "Å", "N", "D", "E", "D", "R", "A", "_", "S", "E", "L", "U", "V"], arr_b = ["E", "F", "X", "O", "U", "R", "T", "I", "D", "E", "N", "G", "Ø", "M", "F", "Æ", "A", "M", "Ø", "R", "Å", "N", "D", "E", "D", "R", "A", "G", "S", "E", "L", "U", "V", "O", "I", "A", "R", "E", "S", "E"]; console.log(getDifference(arr_a, arr_b)); 

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Answer 2

I have implemented a code just like you needed.

 arr_a = ["E", "F", "X", "O", "U", "R", "T", "I", "D", "E", "N", "G", "Ø", "M", "F", "Æ", "A", "M", "Ø", "R", "Å", "N", "D", "E", "D", "R", "A", "_", "S", "E", "L", "U", "V"] arr_b = ["E", "F", "X", "O", "U", "R", "T", "I", "D", "E", "N", "G", "Ø", "M", "F", "Æ", "A", "M", "Ø", "R", "Å", "N", "D", "E", "D", "R", "A", "G", "S", "E", "L", "U", "V", "O", "I", "A", "R", "E", "S", "E"] a_json = {}; b_json = {}; arr_a.forEach(function (element){ if(a_json.hasOwnProperty(element)){ a_json[element]++; } else{ a_json[element] = 1; } }); arr_b.forEach(function (element){ if(b_json.hasOwnProperty(element)){ b_json[element]++; } else{ b_json[element] = 1; } }); diff_json = {}; keys_a = Object.keys(a_json); keys_a.forEach(function(key_a){ if(b_json.hasOwnProperty(key_a)){ diff_json[key_a] = Math.abs(a_json[key_a] - b_json[key_a] ); } }); console.log(diff_json) 

Answer 3

What about transforming them into one (char -> number) object?

 var keys={};
 arr1.forEach(function(e){
   keys[e]=keys[e]||0;
   keys[e]++;
 }
 arr2.forEach(function(e){
   keys[e]=keys[e]||0;
   keys[e]--;
 }
console.log(keys);

Keys will contain the difference between the appearance of a key in the two arrays. Something like:

"A":0, // A appears equal in both
"B":1, //B appears one more time in arr1 then in arr2
"C":-2 //C appears 2 times more in arr2 then in arr1

Or if you want to find the differences in arr2 compared to arr1 as an array of chars, do:

var diff=arr2.map((e,i)=>e==arr1[i]?undefined:e);