比较两个数组并返回一个新数组，其中包含仅在原始数组之一中找到的任何项目
Posted
技术标签:
【中文标题】比较两个数组并返回一个新数组，其中包含仅在原始数组之一中找到的任何项目【英文标题】：Compare two arrays and return a new array with any items only found in one of the original arrays 【发布时间】：2016-04-22 15:02:41 【问题描述】：["diorite", "andesite", "grass", "dirt", "pink wool", "dead shrub"],
["diorite", "andesite", "grass", "dirt", "dead shrub"]
应该返回
["pink wool"]。
因为“粉色羊毛不存在于第一个数组中，即 arr1。但它返回一个空数组。此代码仅适用于数字数组。但当数组仅包含字符串或带有数字的字符串时，代码不起作用。
function diff(arr1, arr2)
var newArray = arr2.concat(arr1); //first joininng both arrays inn one and storing it in newArray
var newestArray = [];
for (var i=0 ; i<newArray.length ; i++) //NOW COMPARING EACH ELEMENT OF newArray WITH ARR1 AD ARR2 AND PUSHING NOT SAME VALUES TO newestArray
if (arr1.indexOf(newArray[i]) == -1)
newestArray.push(newArray[i]);
if (arr2.indexOf(newArray[i]) == -1)
newestArray.push(newArray[i]);
return newestArray.filter(Boolean); //It is returning an empty arrray but it should return "pink wool"
diff(["diorite", "andesite", "grass", "dirt", "pink wool", "dead shrub"], ["diorite", "andesite", "grass", "dirt", "dead shrub"]);
【问题讨论】：
它对我来说很好。 你可以参考这个链接。这可能会对您有所帮助。 http://***.com/questions/1187518/javascript-array-difference 缩进/格式化你的代码。 @HamletHakobyan 是的，它会起作用我在下一个 if 语句之后放置大括号的错误。注意到了吗？没有人在这里阅读代码来找出如此愚蠢的错误，他们只是给出了自己的解决方案。 Finding matches between multiple JavaScript Arrays的可能重复 【参考方案1】：此解决方案具有用于计数的对象的线性方法。
var array1 = ["diorite", "andesite", "grass", "dirt", "pink wool", "dead shrub"],
array2 = ["diorite", "andesite", "grass", "dirt", "dead shrub"];
function symmetricDifference(setA, setB)
var o = , result = [];
function count(i, o)
return function (a)
o[a] = o[a] || count: 0, value: a ;
o[a].count += i;
;
setA.forEach(count(1, o));
setB.forEach(count(-1, o));
Object.keys(o).forEach(function (k)
if (o[k].count)
o[k].count = Math.abs(o[k].count);
while (o[k].count--)
result.push(o[k].value);
);
return result;
document.write('<pre>' + JSON.stringify(symmetricDifference(array1, array2), 0, 4) + '</pre>');【讨论】：
【参考方案2】：对于 ES6，Set 如下。
function diff(arr1, arr2)
var s1 = new Set(arr1);
var s2 = new Set(arr2);
for (let item of s1)
if (s2.has(item))
s2.delete(item);
s1.delete(item);
return Array.from(s1).concat( Array.from(s2) );
//return [...s1].concat([...s2]);
【讨论】：
【参考方案3】：非常感谢你们的帮助，但是当有人问像我这样的问题时，我们并不是在为我们的问题寻求全新的解决方案。那将是清晰的复制，我将从中学到什么？我一直在解决我的问题怎么办。我的解决方案可以被纠正，我需要解决那个问题，这样我就不会重复这样的错误，并且可以知道我错在哪里。
我发现只使用大括号有一个非常愚蠢的错误，这解决了我的整个问题。
function diff(arr1, arr2)
var newArray = arr2.concat(arr1); //first joininng both arrays inn one and storing it in newArray
var newestArray = [];
for (var i=0 ; i<newArray.length ; i++) //NOW COMPARING EACH ELEMENT OF newArray WITH ARR1 AD ARR2 AND PUSHING NOT SAME VALUES TO newestArray
if (arr1.indexOf(newArray[i])===-1)
newestArray.push(newArray[i]);
//Solution to my problem,I put this braces after the next if, because of that next if was not running.
if (arr2.indexOf(newArray[i])===-1)
newestArray.push(newArray[i]);
return newestArray; //It is returning an empty arrray but it should return "pink wool"
diff(["diorite", "andesite", "grass", "dirt", "pink wool", "dead shrub"], ["diorite", "andesite", "grass", "dirt", "dead shrub"]);
【讨论】：
正确嵌套/缩进/格式化您的代码可以帮助避免此类问题。【参考方案4】：您可以使用Array.forEach 循环和Array.indexOf 来检查数组。
我们将最大的数组与最短的数组循环，然后确保您也获得每个数组的唯一值，您可以索引找到的匹配项，然后添加在最短的数组中未找到的项数组。
'use strict';
var arr1 = ["diorite", "andesite", "grass", "dirt", "pink wool", "dead shrub", "alpha"],
arr2 = ["diorite", "andesite", "grass", "dirt", "dead shrub", "beta"];
function compare(left, right)
if (!left || !left.length)
return right;
if (!right || !right.length)
return left;
var i, len, source, target, value, result = [],
indexes = ;
// swap to make sure we iterate the longest array
if (left.length > right.length)
source = left;
target = right;
else
target = left;
source = right;
source.forEach(function(item)
var index = target.indexOf(item);
if (index >= 0)
indexes[index] = true;
return;
result.push(item);
);
for (i = 0, len = target.length; i < len; i++)
if (!indexes[i])
result.push(target[i]);
return result;
console.log(compare(arr1, arr2));
console.log(compare(arr2, arr1));【讨论】：
【参考方案5】：只是你需要找到两个数组之间的差异：
let diff = (a, b) => a.filter(x => b.indexOf(x) === -1);
let fullDiff = (a, b) => diff(a, b).concat(diff(b, a));
/*
var a = ["diorite", "andesite", "grass", "dirt", "pink wool", "dead shrub"]
var b = ["diorite", "andesite", "grass", "dirt", "dead shrub"]
fullDiff(a,b) // ["pink wool"]
*/
或者在 ES5 中：
var diff = function(a, b)
return a.filter(function(value) return b.indexOf(value) === -1; );
,
fullDiff = function(a, b)
return diff(a, b).concat(diff(b, a));
;
附：如果数组真的很大或者它在系统的性能关键部分，最好使用不太复杂的方法（就 big-O 而言）。
【讨论】：
【参考方案6】：function diffArray(arr1, arr2)
return arr1.concat(arr2).filter(
item => !arr1.includes(item) || !arr2.includes(item)
)
diffArray(["df","sds","sdsd",], ["as","as","as"]);
【讨论】：
@RazvanDumitru 是的，我认为是 Razvan【参考方案7】：这是一个简单的示例，将重复的值替换为“x”并将它们过滤掉：
function diffArray(arr1, arr2)
var newArr = [];
var result = [];
var array1 = arr1;
var array2 = arr2;
//a nested loop to replace duplicate values with 'x'
for (var i = 0; i < arr1.length; i++)
for (var j = 0; j < arr2.length; j++)
if (array1[i] == array2[j])
array1.splice(i, 1, 'x');
array2.splice(j, 1, 'x');
newArr = array1.concat(array2);
//remove the 'x's
for (var k = 0; k < newArr.length; k++)
if (newArr[k] != 'x')
result.push(newArr[k]);
return result;
diffArray([1, 2, 3, 5], [1, 2, 3, 4, 5, 6, 7, 7]);
【讨论】：
【参考方案8】：对我来说更容易！
function diffArray(arr1, arr2)
var newArr = [];
function isIn(value)
if (arr2.indexOf(value) === -1)
return true;
arr2.splice(arr2.indexOf(value), 1);
newArr = arr1.filter(isIn);
return newArr.concat(arr2);
filter 和 indexOf 完成了大部分工作，而 splice 为我们提供了其余未匹配的元素，因此无需检查数组是否比其他数组大！祝你好运！
【讨论】：
【参考方案9】：我觉得这样比较容易理解
function diffArray(arr1, arr2)
var newArray = [];
var condition1 = arr1.forEach(function(x)
if(!arr2[arr2.indexOf(x)])
newArray.push(x);
);
var condition2 = arr2.forEach(function(y)
if(!arr1[arr1.indexOf(y)])
newArray.push(y);
);
var compare = arr1.length > arr2.length ? condition1 : condition2;
return newArray;
【讨论】：
谢谢@fernandosavio【参考方案10】：https://lodash.com/docs/4.17.4#difference 可以使用 lodash
使用 _.difference(array, [values]) 找出两个数组值之间的差异
_.difference([2, 1], [2, 3]); // => [1]
如果你想检查更多参数的差异，你可以使用 differenceWith 或 DifferenceBy。
_.differenceWith(数组, [值], [比较器]) https://lodash.com/docs/4.17.4#differenceWith
.differenceBy(array, [values], [iteratee=.identity]) https://lodash.com/docs/4.17.4#differenceBy
【讨论】：
【参考方案11】：这是解决这个问题的好方法：
function diffArray(arr1, arr2)
var newArray = arr1.concat(arr2);
function find(item)
if (arr1.indexOf(item) === -1 || arr2.indexOf(item) === -1)
return item;
return newArray.filter(find);
diffArray([1, 2, 3, 5], [1, 2, 3, 4, 5, 6, 7, 7]);
【讨论】：
【参考方案12】：答案较长，但逻辑很混乱。
function diffArray(arr1, arr2)
var newArrUn;
// Same, same; but different.
if (arr2.length >= arr1.length)
var newArr = [];
var newArrY = [];
var UnusualElement = [];
var UnusualElementY = [];
for (var i = 0; i < arr2.length; i++)
newArr[i] = arr1.indexOf(arr2[i]);
for (var t = 0; t < arr1.length; t++)
newArrY[t] = arr2.indexOf(arr1[t]);
for (var j = 0; j < newArr.length; j++)
if (newArr[j] === -1)
UnusualElement[j] = arr2[j];
for (var e = 0; e < newArrY.length; e++)
if (newArrY[e] === -1)
UnusualElementY[e] = arr1[e];
return (UnusualElement.filter(Boolean)).concat(UnusualElementY.filter(Boolean));
else
if (arr1.length >= arr2.length)
var newArrX = [];
var newArrXX = [];
var UnusualElementX = [];
var UnusualElementXX = [];
for (var b = 0; b < arr1.length; b++)
newArrX[b] = arr2.indexOf(arr1[b]);
for (var u = 0; u < arr2.length; u++)
newArrXX[u] = arr1.indexOf(arr2[u]);
for (var x = 0; x < newArrX.length; x++)
if (newArrX[x] === -1)
UnusualElementX[x] = arr1[x];
for (var z = 0; z < newArrXX.length; z++)
if (newArrXX[z] === -1)
UnusualElementXX[z] = arr2[z];
return (UnusualElementX.filter(Boolean)).concat(UnusualElementXX.filter(Boolean));
【讨论】：
【参考方案13】：function diffArray(arr1, arr2)
var newArr = []; // Same, same; but different.
for(let i = 0; i< arr1.length;i++)
if(arr2.indexOf(arr1[i])==-1)
newArr.push(arr1[i]);
for(let i = 0; i< arr2.length;i++)
if(arr1.indexOf(arr2[i])==-1)
newArr.push(arr2[i]);
return newArr;
diffArray([1, 2, 3, 5], [1, 2, 3, 4, 5]);
【讨论】：
我怀疑这是否有帮助或根本有效。为了说服别人，请解释你的代码是如何工作的以及它为什么有用。以上是关于比较两个数组并返回一个新数组，其中包含仅在原始数组之一中找到的任何项目的主要内容，如果未能解决你的问题，请参考以下文章