检查两个数组是否有具有相同属性的对象

Question

这是我的目标:比较两个对象并找出是否有1个或多个相同的项。如果有1个或多个共有项，则返回true，否则(无共有项)，返回false。

当前问题:我试图使用.some()方法与1对象从API和1对象是本地的，但有点困惑为什么它不工作…什么好主意吗?它应该返回true，因为John在两个对象中都有，但是它返回false🙃

代码示例:在这个示例中，它应该返回true，因为John是对象1 (result1)和对象2 (result 2)的名称。然而，它返回false。

有人能帮我理解我在这里做错了什么吗?

var result1 = [
{id:1, name:'Sandra', type:'user', username:'sandra'},
{id:2, name:'John', type:'admin', username:'johnny2'},
{id:3, name:'Peter', type:'user', username:'pete'},
{id:4, name:'Bobby', type:'user', username:'be_bob'}
];
var result2 = [
{id:2, name:'John', email:'[e-mail]'},
{id:4, name:'Bobby', email:'[e-mail]'}
];
const hasSimilarElement = result1.some((item) => item.name === result2.name); 
console.log(hasSimilarElement);

Answer 1

您可以创建一个函数，该函数接受一个函数f用于查找，然后使用两个数组xs&ys:

const checkBy = f => (xs, ys) => {
let [long, short] = xs.length > ys.length ? [xs, ys] : [ys, xs];
short = new Set(short.map(f));
return long.some(x => short.has(f(x)));
};

:

const checkByName = checkBy(x => x.name);
checkByName( [ {id: 1, name:'Sandra', type:  'user', username:  'sandra'}
, {id: 2, name:  'John', type: 'admin', username: 'johnny2'}
, {id: 3, name: 'Peter', type:  'user', username:    'pete'}
, {id: 4, name: 'Bobby', type:  'user', username:  'be_bob'}]
, [ {id: 2, name:  'John', email: '[e-mail]'}
, {id: 4, name: 'Bobby', email:  '[e-mail]'}]);
//=> true
checkByName( [ {id: 1, name:'Sandra', type:  'user', username:  'sandra'}
, {id: 2, name:  'John', type: 'admin', username: 'johnny2'}
, {id: 3, name: 'Peter', type:  'user', username:    'pete'}
, {id: 4, name: 'Bobby', type:  'user', username:  'be_bob'}]
, [ {id: 2, name:  'xxxx', email: '[e-mail]'}
, {id: 4, name: 'yyyyy', email:  '[e-mail]'}]);
//=> false

Answer 2

我将使用customcommander提出的技术，用于您的具体场景，以及您可以从对象中提取唯一键作为基本类型的任何地方。(这里的name是字符串，所以可以工作。)

如果你需要一些更通用的东西，那么你可以提供一个函数来判断两个元素是否相等，并像这样使用它:

const overlap = (equal) => (xs, ys) => 
xs .some (x => ys .some (y => equal (x, y)))
const result1 = [{id:1, name:'Sandra', type:'user', username:'sandra'}, {id:2, name:'John', type:'admin', username:'johnny2'}, {id:3, name:'Peter', type:'user', username:'pete'}, {id:4, name:'Bobby', type:'user', username:'be_bob'}]
const result2 = [{id:2, name:'John', email:'[e-mail]'}, {id:4, name:'Bobby', email:'[e-mail]'}]
console .log (overlap ((x, y) => x .name == y .name) (result1, result2))

在可以生成唯一键的情况下效率较低，但它更通用。您甚至可以将它用于两个数组，其中两个数组具有不同结构的对象，例如传递(x, y) => x .id == y .primaryKey。

Answer 3

Array.prototype.some()some()方法测试数组中是否至少有一个元素通过所提供的函数实现的测试。

在您的示例中，您正在尝试比较两个对象数组。如果result2.name未定义，则可能需要指定特定元素的索引，例如result2[0].name

如果你想比较两个数组与一些你需要迭代通过result2项目以及。你可以这样写:

var result1 = [
{id:1, name:'Sandra', type:'user', username:'sandra'},
{id:2, name:'John', type:'admin', username:'johnny2'},
{id:3, name:'Peter', type:'user', username:'pete'},
{id:4, name:'Bobby', type:'user', username:'be_bob'}
];
var result2 = [
{id:2, name:'John', email:'[e-mail]'},
{id:4, name:'Bobby', email:'[e-mail]'}
];    
const hasSimilarElement = result2.filter((item1) => !result1.some(item2 => item1.name === item2.name ));
console.log(hasSimilarElement);

Answer 4

代码失败的原因是您试图在result1 Array的元素中找到name与result2.name的匹配

然而，result2也是一个Array，数组没有一个.name恰好是所有的元素.names，以某种方式匹配你正在搜索的-这不是它的工作方式

你需要迭代两个数组来寻找匹配

var result1 = [
{id:1, name:'Sandra', type:'user', username:'sandra'},
{id:2, name:'John', type:'admin', username:'johnny2'},
{id:3, name:'Peter', type:'user', username:'pete'},
{id:4, name:'Bobby', type:'user', username:'be_bob'}
];
var result2 = [
{id:2, name:'John', email:'[e-mail]'},
{id:4, name:'Bobby', email:'[e-mail]'}
];
const hasSimilarElement = result1.some((item) => result2.some(item2 => item.name === item2.name)); 
console.log(hasSimilarElement);

.as-console-wrapper { min-height: 100%!important; top: 0; }

然而，对于非常大的结果集，这是低效的-我不是"大0"表示法专家，但我相信这将是O(n²) -您可能会比较result1.length*result2.length乘以

对于这个简单的例子，您可以从两个结果中提取.name，并创建一个Set…如果Set的大小小于合并后的结果.length，则表示存在重复

在这里，您将每个结果迭代一次-如果两个集合都有1000个元素，则迭代2000次，而使用上面的朴素方法则为1,000,000次

var result1 = [
{id:1, name:'Sandra', type:'user', username:'sandra'},
{id:2, name:'John', type:'admin', username:'johnny2'},
{id:3, name:'Peter', type:'user', username:'pete'},
{id:4, name:'Bobby', type:'user', username:'be_bob'}
];
var result2 = [
{id:2, name:'John', email:'[e-mail]'},
{id:4, name:'Bobby', email:'[e-mail]'}
];
const hasSimilarElement = new Set([...result1, ...result2].map(({name}) => name)).size < result1.length + result2.length
console.log(hasSimilarElement);

.as-console-wrapper { min-height: 100%!important; top: 0; }

然而，有一个缺陷如果在其中一个结果中有一个重复的名字，这将不能像预期的那样工作

这里每个集合迭代两次-但是，给定2 x 1,000个长结果，4,000次迭代仍然优于1,000,000

var result1 = [
{id:1, name:'Sandra', type:'user', username:'sandra'},
{id:2, name:'John', type:'admin', username:'johnny2'},
{id:3, name:'Peter', type:'user', username:'pete'},
{id:4, name:'Bobby', type:'user', username:'be_bob'}
];
var result2 = [
{id:2, name:'John', email:'[e-mail]'},
{id:4, name:'Bobby', email:'[e-mail]'}
];
const set1 = new Set(result1.map(({name}) => name));
const set2 = new Set(result2.map(({name}) => name));
const both = new Set([...set1, ...set2]);
const hasSimilarElement = both.size < set1.size + set2.size;
console.log(hasSimilarElement);

.as-console-wrapper { min-height: 100%!important; top: 0; }

Answer 5

从上面的注释…

" OP需要在result2中result1的item.name值相等的项目find，而不是直接比较result2中不存在的(例如undefined)name属性。">

或者正如3limin4t0or正确建议的那样，可以再次将find交换为另一个some。

const result1 = [
{ id: 1, name: 'Sandra', type: 'user', username: 'sandra' },
{ id: 2, name: 'John', type: 'admin', username: 'johnny2' },
{ id: 3, name: 'Peter', type: 'user', username: 'pete' },
{ id: 4, name: 'Bobby', type: 'user', username: 'be_bob' },
];
const result2 = [
{ id: 5, name: 'Lea', email: '[e-mail]' },
{ id: 2, name: 'John', email: '[e-mail]' },
];
const result3 = [
{ id: 5, name: 'Lea', email: '[e-mail]' },
];
const result4 = [];
console.log(
'similar elements within `result1` and `result2` ..?',
result1.some(itemA =>
result2.some(itemB => itemB.name === itemA.name)
)
);
console.log(
'similar elements within `result1` and `result3` ..?',
result1.some(itemA =>
result3.some(itemB => itemB.name === itemA.name)
)
);
console.log(
'similar elements within `result1` and `result4` ..?',
result1.some(itemA =>
result4.some(itemB => itemB.name === itemA.name)
)
);

.as-console-wrapper { min-height: 100%!important; top: 0; }

在下一次迭代中，其中一个原因可能会提出一个通用实现的函数，该函数允许自由选择被比较的属性名。

实现一个函数还提供了摆脱两个嵌套的some迭代的机会(尽管some因为成功的比较而提前退出)。

在这里，我们将从较短的数组中构建一个key(属性名)特定的值-索引(单个迭代任务)，其中，当迭代较长的数组时，我们将查找(不迭代)key特定的值来查找项目相似性。

function hasSimilarItemsByKey(key, arrA, arrB) {
const [
// the shorter array will be the base
// for a `key` specific value index.
lookupArray,
// the longer array will be iterated
// and have its item `key` specific 
// values looked up at a value index.
targetArray,
] = [
arrA,
arrB,
].sort((a, b) => a.length - b.length);
// create the object based value index
// (from the shorter array).
const valueIndex = lookupArray
.reduce((lookup, item) => {
const value = item[key];
lookup[value] ??= value;
return lookup;
}, Object.create(null));  // A `prototype`-less object. Thus,
// using the `in` operator is safe.
return targetArray.some(item => item[key] in valueIndex);
}
const result1 = [
{ id: 1, name: 'Sandra', type: 'user', username: 'sandra' },
{ id: 2, name: 'John', type: 'admin', username: 'johnny2' },
{ id: 3, name: 'Peter', type: 'user', username: 'pete' },
{ id: 4, name: 'Bobby', type: 'user', username: 'be_bob' },
];
const result2 = [
{ id: 5, name: 'Lea', email: '[e-mail]' },
{ id: 2, name: 'John', email: '[e-mail]' },
];
const result3 = [
// Attention,
// A matching `id` and a non matching `name`.
{ id: 1, name: 'Lea', email: '[e-mail]' },
];
const result4 = [];
console.log(
'similar elements within `result1` and `result2`n',
'...by `name`..?', hasSimilarItemsByKey('name', result1, result2),
'...by `id`..?', hasSimilarItemsByKey('id', result1, result2),
);
console.log(
'similar elements within `result1` and `result3`n',
'...by `name`..?', hasSimilarItemsByKey('name', result1, result3),
'...by `id`..?', hasSimilarItemsByKey('id', result1, result3),
);
console.log(
'similar elements within `result1` and `result4`n',
'...by `name`..?', hasSimilarItemsByKey('name', result1, result4),
'...by `id`..?', hasSimilarItemsByKey('id', result1, result4),
);

.as-console-wrapper { min-height: 100%!important; top: 0; }