計算價值多少次發生在陣列 - 的JavaScript - 角 - lodash

Question

我有兩個JavaScript數組

var openDates = ["6/14/2015", "6/15/2015", "6/16/2015", "6/17/2015", "6/18/2015", "6/19/2015", ...] 

var entries = ["6/16/2015", "6/16/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", ...] 

我怎麼算，有多少次openDates發生在項格式。 。

var entriesPerDay = [0, 0, 10, 2, 16, 18, 20, ...] 

我已經安裝lodash，但無法弄清楚。如果沒有匹配，我需要返回0值。

Answer 1

我現在所能想到的。

var openDates = ["6/14/2015", "6/15/2015", "6/16/2015", "6/17/2015", "6/18/2015", "6/19/2015"] 

var entries = ["6/16/2015", "6/16/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015"] 

var entriesPerDay = []; 

for(var i = 0; i < openDates.length; i++) { 
var currDate = openDates[i]; 
var temp = _.filter(entries, function(date) { 
    return date === currDate; 
}); 
entriesPerDay.push(temp.length); 
} 

對於上面的例子中entriesPerDay是這樣的：[0，0，2,0，42，0]

PS：我使用lodash，只要在問題說你有lodash。

PPS：如果它有效，請接受並注意。

Answer 2

下面的這個例子將幫助您開始並瞭解如何實現這個想法。

在此示例中，forEach正用於數組比較值。

var openDates = ["6/14/2015", "6/15/2015", "6/16/2015", "6/17/2015", "6/18/2015", "6/19/2015"] 
 
var entries = ["6/16/2015", "6/16/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015"] 
 

 
function countEntries(arr1, arr2) { 
 
    var count, total = []; 
 
    arr1.forEach(function(date) { 
 
     count = 0; 
 
     arr2.forEach(function(entry) { 
 
      count += date == entry ? 1 : 0; 
 
     }); 
 
     total.push(count); 
 
    }); 
 
    return total; 
 
} 
 

 
console.log(countEntries(openDates, entries));

在這裏，我們只運行一個foreach循環，並在每次迭代我們重新計數。請參閱下面的輸出：

輸出：[0, 0, 2, 0, 42, 0]

Answer 3

對於O（n）的解決方案，而不是（在^ 2）：

var openDates = ["6/14/2015", "6/15/2015", "6/16/2015", "6/17/2015", "6/18/2015", "6/19/2015"]; 
var entries = ["6/16/2015", "6/16/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015"]; 

function countEntries(open, existing) { 
    var openHash = {}; 
    var entriesHash = {}; 

    for(var i = 0; i < open.length; i ++){ 
    openHash[open[i]] = 0; 
    } 

    for(i = 0; i < existing.length; i ++){ 
    if(entriesHash.hasOwnProperty(entries[i])){ 
     entriesHash[entries[i]] ++; 
    }else{ 
     entriesHash[entries[i]] = 1; 
    }  
    } 

    var result = []; 
    for(var date in openHash){ 
    if(openHash.hasOwnProperty(date)){ 
     if(entriesHash[date]){ 
     openHash[date] = entriesHash[date]; 
     } 
     result.push(openHash[date]); 
    } 
    } 

    return result; 
} 

console.log(countEntries(openDates, entries)); 

http://jsbin.com/jinolapuza/edit?js,console

Answer 4

可以使用lodash的.countBy()做和.at()。您可以使用_.pick()而不是_.at（）來獲取包含找到的條目（至少1次出現）的對象作爲屬性名稱。

複雜性 - O（n）。

var openDates = ["6/14/2015", "6/15/2015", "6/16/2015", "6/17/2015", "6/18/2015", "6/19/2015"]; 
 

 
var entries = ["6/16/2015", "6/16/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015", "6/18/2015"]; 
 

 
var counts = _(entries) 
 
      .countBy(function(entry) { return entry }) // count the number of occurances of each entry 
 
      .at(openDates) // get the values of the keys that appear in openDates 
 
      .map(function(count) { return count || 0 }) // map undefined (not found) t0 0 
 
      .value(); // get the values 
 

 
var countsObject = _(entries) 
 
      .countBy(function(entry) { return entry }) // count the number of occurances of each entry 
 
      .pick(openDates) // get the values of the keys that appear in openDates 
 
      .value(); // get the values 
 

 
document.write('_.at(): ' + JSON.stringify(counts)); 
 
document.write('<br />'); 
 
document.write('_.pick(): ' + JSON.stringify(countsObject));

<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/3.10.1/lodash.js"></script>

Answer 5

使用map()和filter()：

_.map(openDates, function(item) { 
    return _.filter(entries, function(entry) { 
     return item === entry; 
    }).length; 
});