比较两个数组并计数事件
[英]Compare two arrays and count incidents
问题描述
I have two array
我有两个数组
$array1=('18753933','18753933','18771982')
$array2=('18753933','18771982')
I'm going through each one to compare the values that are the same in each array我正在逐个比较每个数组中相同的值
var contarArticulosCargados=0;
for(var $i=0;$i<$array1.length;$i++) {
for(var $j=0;$j<$array2.length;$j++) {
if($array1[$i]===$array[$j]) {
countArticulosCargados++;
}
}
}
console.log(countArticlesLoaded);
This example that I have put in my code works well, that is to say, I am doing the comparison and it works without problem.我放在代码中的这个例子运行良好，也就是说，我正在做比较并且它没有问题。
What do I want to do?我想做什么？ (It's what I have no idea how to do) （这是我不知道该怎么做）
As you can see in $array1 is the value 18753933 and in $array2 is also the value 18753933 .正如您在$array1看到的值 18753933 而在$array2中也是值18753933 。
What I want is that when the value is repeated this is counted, in the $array1 the value 18753933 is 2 times and the value 18771982 is 1 time.我想要的是，当值重复时，这被计算在内，在$array1 ，值18753933是2次，值18771982是1次。
I want to see something like this:我想看到这样的事情：
18753933 repeats: 2
18771982 is repeated: 1
could you give me an idea of what I'm missing?你能告诉我我错过了什么吗？ I always find it hard to make comparisons between arrays.我总是发现很难在数组之间进行比较。 Thank you.谢谢你。
3 个解决方案
解决方案1
1 2019-04-13 11:19:54
Simply map over array2 and take the length of the filtered array1.只需映射 array2 并获取过滤后的 array1 的长度。
const a1 = ['18753933','18753933','18771982']; const a2 = ['18753933','18771982']; const res = a2.map(e => ({[e]: a1.filter(k => k === e).length})); console.log(res);
To remove elements that were not found, you could filter first要删除未找到的元素，您可以先过滤
a2.filter(e => a1.includes(e)).map(e => ({[e]: a1.filter(k => k === e).length}))
or after或之后
a2.map(e => ({[e]: a1.filter(k => k === e).length})).filter(e => Object.values(e).every(x => x > 0));
Or use reduce或使用减少
const a1 = ['18753933','18753933','18771982']; const a2 = ['18753933','18771982']; const res = a2.reduce((a, b) => { const l = a1.filter(e => e === b).length; if (l) a[b] = l; return a; }, {}); console.log(res);
解决方案2
0 2019-04-13 12:36:47
What it looks like you'll need to do is create an Array of Objects.看起来您需要做的是创建一个对象数组。 Something like:就像是：
<!doctype html>
<html>
<head>
<meta charset="utf-8">
</head>
<body>
<h2>JavaScript Array of Objects</h2>
<p id="demo"></p>
<script>
var array1 = ['18753933', '18753933', '18771982'];
var array2 = ['18753933', '18771982'];
var contarArticulosCargados = [];
// for each of the numbers in array2
for (var i = 0; i < array2.length; i++) {
// create an object
contarArticulosCargados.push({
articulos: array2[i],
contar: 0
});
// for each of the numbers in array2
for (var j = 0; j < array1.length; j++) {
if (array2[i] === array1[j]) {
// increment counter for the object
var foo = contarArticulosCargados.length - 1;
contarArticulosCargados[foo].contar++;
}
}
}
// create string for result
var result = "";
for (var k = 0; k < contarArticulosCargados.length; k++) {
result += contarArticulosCargados[k].articulos + " repeats: " + contarArticulosCargados[k].contar + ". ";
}
// send result to html
document.getElementById("demo").innerHTML = result;
// or console as string
console.log(result);
// or console as array
console.log(contarArticulosCargados);
</script>
</body>
</html>
You can copy & paste the html to test, to see the results.您可以复制并粘贴 html 进行测试，以查看结果。
You might want to switch array1 & array2, but I think I have them the right way around (from your example).您可能想切换 array1 和 array2，但我认为我有正确的方法（从您的示例中）。
Change array1 to something else might help visualisation将 array1 更改为其他内容可能有助于可视化
var array1 = ['18753933', '18753933', '18771982', '18771982', '18771982', '18771982', '18771982', '18771982', '18771982'];
If array1 was as above, the output is:如果 array1 如上所述，则输出为：
18753933 repeats: 2. 18771982 repeats: 7.
Or if array1 is as originally stated in the question, the output is:或者，如果 array1 与问题中最初陈述的一样，则输出为：
18753933 repeats: 2. 18771982 repeats: 1.
But, of course the question tags are JavaScript & jQuery but the code is PHP .但是，当然问题标签是JavaScript和jQuery但代码是PHP 。 So I'm wrong either way.所以我错了。
解决方案3
0 2019-04-13 15:11:41
const a1 = ['18753933','18753933','18771982']; const a2 = ['18753933','18771982']; const r = {}; a2.forEach(_ => r[_] = 0) a1.forEach(_ => { if (r[_] === 0 || r[_]) { r[_] += 1 } }) console.log(r)
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