blocks|key|1032886|text|我将使用键/值计数方法，其中日期是键，值是它出现的次数，如下所示：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1032887|var+counters+=+{};
$.each(dates,+function(i,+date)+{
+++counters[date]+=+counters[date]+?+counters[date]+%2B+1+:+1;
});|code-block|syntax|javascript|1032888|当然，这种方法假设日期都遵循相同的格式。|1032889|编辑：|offset|length|style|BOLD|1032890|http://jsfiddle.net/wMR9x/2/|1032891|然后，您可以像这样循环它，并简单地将结果连接到另一个数组中：|1032892|var+finalCounts+=+[];
var+i+=+0;
for(var+key+in+counters)
++++finalCounts[i%2B%2B]+=+counters[key];|1032893|entityMap|0|LINK|mutability|MUTABLE|url^0|0|0|0|0|3|0|0|S|0|0|0|0^^$0|@$1|2|3|4|5|6|7|11|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|12|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|13|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|14|8|@$K|15|L|16|M|N]]|9|@]|A|$]]|$1|O|3|P|5|6|7|17|8|@]|9|@$K|18|L|19|1|1A]]|A|$]]|$1|Q|3|R|5|6|7|1B|8|@]|9|@]|A|$]]|$1|S|3|T|5|D|7|1C|8|@]|9|@]|A|$E|F]]|$1|U|3|-4|5|6|7|1D|8|@]|9|@]|A|$]]]|V|$W|$5|X|Y|Z|A|$10|P]]]]

I would use a key/value counting approach where the date is the key and the value is the number of times it appears like so:

<pre><code>var counters = {};
$.each(dates, function(i, date) {
 counters[date] = counters[date] ? counters[date] + 1 : 1;
});
</code></pre>

This approach assumes the dates will all be following identical formats of course.

Edit: 

<a href="http://jsfiddle.net/wMR9x/2/" rel="nofollow">http://jsfiddle.net/wMR9x/2/</a>

Then you can loop over it like so and simply join the results into another array:

<pre><code>var finalCounts = [];
var i = 0;
for(var key in counters)
 finalCounts[i++] = counters[key];
</code></pre>

blocks|key|1032899|text|试一试|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1032900|var+src+=+['2013/03',+'2013/03',+'2012/01',+'2012/11',+'2012/09',+'2012/09',+'2012/09'];

var+cnts+=+[];
for(var+i+=+0;+i+<+src.length;+i%2B%2B){
++++var+val+=+src[i];
++++var+count+=+1;
++++while(+val+==+src[i+%2B+1]){
++++++++count%2B%2B;
++++++++i%2B%2B;
++++}
++++cnts.push(count)
}|code-block|syntax|javascript|1032901|演示：Fiddle|offset|length|1032902|entityMap|0|LINK|mutability|MUTABLE|url|http://jsfiddle.net/arunpjohny/pk7rY/1/^0|0|0|3|6|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@$I|V|J|W|1|X]]|A|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

Try

<pre><code>var src = ['2013/03', '2013/03', '2012/01', '2012/11', '2012/09', '2012/09', '2012/09'];

var cnts = [];
for(var i = 0; i &lt; src.length; i++){
 var val = src[i];
 var count = 1;
 while( val == src[i + 1]){
 count++;
 i++;
 }
 cnts.push(count)
}
</code></pre>

Demo: <a href="http://jsfiddle.net/arunpjohny/pk7rY/1/" rel="nofollow">Fiddle</a>

blocks|key|3159796|text|您可以通过使用映射来跟踪您已经看到的值，从而有效地计算递归值：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3159797|var+data+=+['2013/03',+'2013/03',+'2012/01',+'2012/11',+'2012/09',+'2012/09',+'2012/09'];

var+map+=+{};
var+sequence+=+[];
for+(var+i+=+0;+i+<+data.length;+i%2B%2B)+{
++++var+item+=+data[i];
++++if+(map[item])+{
++++++++map[item]%2B%2B;
++++}+else+{
++++++++map[item]+=+1;
++++++++sequence.push(item);
++++}
}

var+output+=+[];
for+(i+=+0;+i+<+sequence.length;+i%2B%2B)+{
++++output.push(map[sequence[i]]);
}

//+output+contains+[2,1,1,3]|code-block|syntax|javascript|3159798|工作演示：http://jsfiddle.net/jfriend00/rQ7KB/|offset|length|3159799|entityMap|0|LINK|mutability|MUTABLE|url|http://jsfiddle.net/jfriend00/rQ7KB/^0|0|0|5|10|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@$I|V|J|W|1|X]]|A|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

You can count the recurring values efficiently by using a map to keep track of the values you've already seen like this:

<pre><code>var data = ['2013/03', '2013/03', '2012/01', '2012/11', '2012/09', '2012/09', '2012/09'];

var map = {};
var sequence = [];
for (var i = 0; i &lt; data.length; i++) {
 var item = data[i];
 if (map[item]) {
 map[item]++;
 } else {
 map[item] = 1;
 sequence.push(item);
 }
}

var output = [];
for (i = 0; i &lt; sequence.length; i++) {
 output.push(map[sequence[i]]);
}

// output contains [2,1,1,3]
</code></pre>

Working demo: <a href="http://jsfiddle.net/jfriend00/rQ7KB/" rel="nofollow">http://jsfiddle.net/jfriend00/rQ7KB/</a>

blocks|key|3159844|text|您可以使用javascript数组的push方法来实现此目的|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|3159845|var+arr+=+['2013/03',+'2013/03',+'2012/01',+'2012/11',+'2012/09',+'2012/09',+'2012/09'];
var+x+=+[],y+=+[],+prev;
for+(var+i+=+0;+i+<+arr.length;+i%2B%2B)+{
++++if+(arr[i]+!==+prev)+{
++++++++x.push(arr[i]);
++++++++y.push(1);
++++}+else+{
++++++++y[y.length+-+1]%2B%2B;
++++}
++++prev+=+arr[i];
}
console.log(y);|code-block|syntax|javascript|3159846|JSFiddle|3159847|entityMap|0|LINK|mutability|MUTABLE|url|http://jsfiddle.net/FNxtb/1/^0|I|4|0|0|0|8|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|Y|8|@]|D|@$9|Z|A|10|1|11]]|E|$]]|$1|M|3|-4|5|6|7|12|8|@]|D|@]|E|$]]]|N|$O|$5|P|Q|R|E|$S|T]]]]

You can use push method of the javascript array to achieve this

<pre><code>var arr = ['2013/03', '2013/03', '2012/01', '2012/11', '2012/09', '2012/09', '2012/09'];
var x = [],y = [], prev;
for (var i = 0; i &lt; arr.length; i++) {
 if (arr[i] !== prev) {
 x.push(arr[i]);
 y.push(1);
 } else {
 y[y.length - 1]++;
 }
 prev = arr[i];
}
console.log(y);
</code></pre>

<a href="http://jsfiddle.net/FNxtb/1/" rel="nofollow">JSFiddle</a>

blocks|key|1033041|text|var+dates=['2013/03',+'2013/03',+'2012/01',+'2012/11',+'2012/09',+'2012/09',+'2012/09'];
var+count={};
var+count_array=[];
for+(i+in+dates){
++++count[dates[i]]+=count[dates[i]]?count[dates[i]]%2B1:1;
}

for+(i+in+count){
+++count_array.push(count[i]);
}
console.log(count_array);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1033042|jsfiddle|unstyled|offset|length|1033043|entityMap|0|LINK|mutability|MUTABLE|url|http://jsfiddle.net/UKDUp/^0|0|0|8|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|R|8|@]|9|@$G|S|H|T|1|U]]|A|$]]|$1|I|3|-4|5|F|7|V|8|@]|9|@]|A|$]]]|J|$K|$5|L|M|N|A|$O|P]]]]

<pre><code>var dates=['2013/03', '2013/03', '2012/01', '2012/11', '2012/09', '2012/09', '2012/09'];
var count={};
var count_array=[];
for (i in dates){
 count[dates[i]] =count[dates[i]]?count[dates[i]]+1:1;
}

for (i in count){
 count_array.push(count[i]);
}
console.log(count_array);
</code></pre>

<a href="http://jsfiddle.net/UKDUp/" rel="nofollow">jsfiddle</a>

blocks|key|1339336|text|在underscore.js中尝试countBy函数：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1339337|_.countBy(dates,_.identity);|code-block|syntax|javascript|1339338|，它将返回表单的对象|1339339|{'2013/03':2,+...}|1339340|如果只想检索该对象的计数，请使用_.values函数。|1339341|entityMap^0|H|7|0|0|0|0|G|8|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|W|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|Y|8|@$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|Q|3|-4|5|6|7|11|8|@]|D|@]|E|$]]]|R|$]]

Try the <code>countBy</code> function in underscore.js:

<pre><code>_.countBy(dates,_.identity);
</code></pre>

That will return an object of the form

<pre><code>{'2013/03':2, ...}
</code></pre>

If you want to retrieve just the counts from that object, use the <code>_.values</code> function.

I'm looking to create an array of counts based on the occurrences of the same object within a different array. So far, my code looks like this:

<pre><code>var dates = [];
$.each(items, function (i, item) {
 dates.push(item.date);
});
</code></pre>

which returns:

<pre><code>['2013/03', '2013/03', '2012/01', '2012/11', '2012/09', '2012/09', '2012/09']
</code></pre>

After that, I'd like to end up with an array that looks like this:

<pre><code>[2,1,1,3]
</code></pre>

Any help would be much appreciated!

Count occurrences of array objects

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我希望根据同一对象在不同数组中的出现情况创建一个计数数组。到目前为止，我的代码如下所示：var dates = [];$.each(items, function (i, item) {    dates.push(item.date);});它返回：['2013/03', '2013/03', '2012/01', ...

问计数数组对象的出现次数
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问计数数组对象的出现次数EN