基於部分重複連接行

Question

我正在尋找一種更有效的方式來連接基於2個字段中的重複項的數組成員，理想情況下不需要在循環內部使用循環。

的陣列已經被name和email升序 - 如果兩行具有相同的name和email，我想「合併」這些行連成一排與drink場從第一行成爲drink_1第二行的drink字段變成drink_2。

輸入：

[ 
    {name: 'bob', email: 'bob@bob.com', drink: 'beer'}, 
    {name: 'bobs_alias', email: 'bob@bob.com', drink: 'beer'}, 
    {name: 'john', email: 'john@john.com', drink: 'beer'}, 
    {name: 'john', email: 'john@john.com', drink: 'cider'}, 
    {name: 'mike', email: 'mike@mike.com', drink: 'wine'}, 
    {name: 'mike', email: 'mike@mike.com', drink: 'water'} 
] 

所需的輸出：

[ 
    { name: 'bob', email: 'bob@bob.com', drink: 'beer'}, 
    { name: 'bobs_alias', email: 'bob@bob.com', drink: 'beer'}, 
    { name: 'john', email: 'john@john.com', drink_1: 'beer', drink_2: 'cider'}, 
    { name: 'mike', email: 'mike@mike.com', drink_1: 'wine', drink_2: 'water'} 
] 

Answer 1

你可以嘗試這樣的事情：

var d = [{name: "bob", email: "bob@bob.com", drink: "beer"}, 
 
{name: "bobs_alias", email: "bob@bob.com", drink: "beer"}, 
 
{name: "john", email: "john@john.com", drink: "beer"}, 
 
{name: "john", email: "john@john.com", drink: "cider"}, 
 
{name: "john", email: "john@john.com", drink: "wine"}, 
 
{name: "mike", email: "mike@mike.com", drink: "wine"}, 
 
{name: "mike", email: "mike@mike.com", drink: "water"}]; 
 

 
var r = []; 
 
d.sort(function(a, b) { 
 
    return a.name > b.name ? 1 : a.name < b.name ? -1 : 0; 
 
}).reduce(function(p, c) { 
 
    if (p.name === c.name && p.email === c.email) { 
 
    var index = Object.keys(p).reduce(function(i, k) { 
 
     var str = k.replace('drink_', ''); 
 
     if (!isNaN(str)) i = Math.max(+str, i) + 1; 
 
     return i; 
 
    }, 1); 
 
    if (p.drink) { 
 
     p['drink_' + index++] = p.drink; 
 
    } 
 
    p['drink_' + index] = c.drink; 
 
    delete p.drink; 
 
    return p; 
 
    } else { 
 
    r.push(c) 
 
    return c; 
 
    } 
 
}) 
 

 
console.log(r)

Answer 2

使用Array.prototype.reduce和hash table以獲得所需的結果 - 因爲它使用散列表，因此即使沒有排序的輸入，該提議也可以運行。

見下面的演示：

var array=[{name:'bob',email:'bob@bob.com',drink:'beer'},{name:'bobs_alias',email:'bob@bob.com',drink:'beer'},{name:'john',email:'john@john.com',drink:'beer'},{name:'john',email:'john@john.com',drink:'cider'},{name:'mike',email:'mike@mike.com',drink:'wine'},{name:'mike',email:'mike@mike.com',drink:'water'}]; 
 

 
var result = array.reduce(function(hash) { 
 
    return function(p, c) { 
 
    let key = `${c.name}_${c.email}`; 
 
    if (hash[key]) { 
 
     if (hash[key].drink) { 
 
     hash[key]['drink_' + ++hash[key + '_num']] = hash[key].drink; 
 
     delete hash[key].drink; 
 
     } 
 
     hash[key]['drink_' + ++hash[key + '_num']] = c.drink; 
 
    } else { 
 
     hash[key] = c; 
 
     hash[key + '_num'] = 0; 
 
     p.push(hash[key]); 
 
    } 
 
    return p; 
 
    }; 
 
}(Object.create(null)), []); 
 

 
console.log(result);

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Answer 3

使用map()和filter()我們可以實現這個任務的組合。 試試這個：

console.clear(); 
 
var ps = [ 
 
    {name: 'bob', email: 'bob@bob.com', drink: 'beer'}, 
 
    {name: 'bobs_alias', email: 'bob@bob.com', drink: 'beer'}, 
 
    {name: 'john', email: 'john@john.com', drink: 'beer'}, 
 
    {name: 'john', email: 'john@john.com', drink: 'cider'}, 
 
    {name: 'mike', email: 'mike@mike.com', drink: 'wine'}, 
 
    {name: 'mike', email: 'mike@mike.com', drink: 'water'} 
 
] 
 

 
var temp = [], n = 1, obj = {}; 
 
var x = ps.map(function(el, i) { 
 
    var t = el.name + el.email; 
 
    if(t !== temp[0]) { 
 
    temp[0] = t; 
 
    n = 1; 
 
    obj = null; 
 
    return el; 
 
    } else {  
 
    obj = ps[i-1]; 
 
    obj['drink_' + n] = obj['drink']; 
 
    delete obj['drink']; 
 
    obj['drink_' + (n + 1)] = el['drink'];  
 
    n++; 
 
    return null; 
 
    } 
 
}) 
 

 
//filter null value from array element 
 
var y = x.filter(function(el,i){ 
 
    return el !== null; 
 
}) 
 

 
//print in console 
 
y.forEach(function(e) {console.log(e);})

Answer 4

通常情況下，我使用某種散列的粉絲，但是因爲數組已經排序，它可以簡單地向前看，每個條目：

var arr=[{name:'bob',email:'bob@bob.com',drink:'beer'},{name:'bobs_alias',email:'bob@bob.com',drink:'beer'},{name:'john',email:'john@john.com',drink:'beer'},{name:'john',email:'john@john.com',drink:'cider'},{name:'mike',email:'mike@mike.com',drink:'wine'},{name:'mike',email:'mike@mike.com',drink:'water'}]; 
 

 

 
for(let i=0; i < arr.length; i++) 
 
\t for(let j =i +1, ind = 2; j< arr.length && arr[i].name === arr[j].name && arr[i].email === arr[j].email; j++){ \t 
 
    \t if(arr[i].drink){ 
 
     \t arr[i].drink_1 = arr[i].drink; 
 
      delete arr[i].drink; 
 
     } 
 
    \t arr[i]['drink_' + ind++] = arr.splice(j--,1)[0].drink; 
 
    } 
 
    
 
console.log(arr);