Question

我有两个javascript数组

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", ...]

如何计算格式中条目中出现 openDates 的次数..

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

我安装了lodash，但无法弄明白。如果没有匹配，我需要返回0值。

Answer 1

到目前为止我能想到什么。

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);
}

对于上面的例子，entriesPerDay是这样的：[0,0,2,0,42,0]

PS：我正在使用lodash，正如你所说的那样，你有lodash。

PPS：如果有效，请接受并支持。

Answer 2

以下示例将帮助您开始并了解如何实现这一目标。

在此示例中，forEach正在数组上用于比较值。

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));

这里我们只是运行一个forEach循环并重置每次迭代的计数。请参阅以下输出：

输出：[0, 0, 2, 0, 42, 0]

Answer 3

对于O（n）解决方案而不是（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()来完成此操作。您可以使用_.pick()而不是_.at（）来获取具有找到的条目（至少出现1次）的对象作为属性名称。

复杂性 - 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;
});

计算一个值在数组中出现的次数 - javascript - angular - lodash

5 个答案: