計算值在數組中出現的次數-Javascript-angular-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", ...]

如何計算格式中的條目中openDates出現的次數。

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

我安裝了lodash，但無法弄清楚。 如果沒有匹配項，我需要返回0值。

Answer 1

到目前為止我能想到的。

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);
}

對於以上示例，entriesPerDay是這樣的：[0，0，2，0，42，0]

PS：我正在使用lodash，正如您所說的，您有lodash。

PPS：請接受並支持（如果可行）。

Answer 2

下面的示例將幫助您入門，並獲得有關如何完成此操作的想法。

在此示例中，在數組上使用了forEach來比較值。

 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)); 

在這里，我們只是運行一個forEach循環並在每次迭代中重置計數。 參見下面的輸出：

輸出： [0, 0, 2, 0, 42, 0]

Answer 3

對於O（n）解決方案而不是（On ^ 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

您可以使用lodash的.countBy（） 和 .at（） 。 您可以使用_.pick（）而不是_.at（）來獲得一個對象，該對象具有找到的條目（至少出現1次）作為屬性名稱。

復雜度-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;
});