Question

我无法使功能与下面的示例相同我只返回3分以上的返回值

过滤质量将根据以下公式确定：

q=((TP+TN)/(TP+TN+10*FP+FN))

正面案例（P）是过滤器分类为垃圾邮件的电子邮件，否定案例（N）是过滤器分类为普通电子邮件的电子邮件。因此，FP是错误标记为垃圾邮件的电子邮件数量，FN是未经标记的垃圾邮件数量。

值得注意的是，TP, FP, TN, FN是频率（案例数），而不是百分比。

您的过滤器将在3个不同的数据集上进行评估。对于每个数据集，您的过滤器可能会得到0到3个点：

   q       pts
<0, 0.3)    0
<0.3, 0.5)  1
<0.5, 0.7)  2
<0.7, 0.9)  2.5
<0.9, 1>    3

我的代码到目前为止：

def quality_score(TP, TN, FP, FN):
        q=(TP+TN//TP+TN+10*FP+FN)

        return q

Answer 1

将质量得分表放入具有范围起始值的列表中。首先，根据公式计算“q”值。然后，在表格中查找q值以获得最终质量点。

<link rel="stylesheet" type="text/css" href="style.css">
<div class="container">
<div class="hook">
<video onended="playNext()" width="640" height="480" id="myVideo"     controls autoplay>

    </video>
</div>

<div>
  <ul>
    <li><a id="vid1" href="#"><img src="http://placehold.it/200x100"></a>  </li>
    <li><a id="vid2" href="#"><img src="http://placehold.it/200x100"></a></li>
    <li><a id="vid3" href="#"><img src="http://placehold.it/200x100"></a></li>
  </ul>
</div>
<!-- /.hook -->
</div>
<!-- /.container -->

<script type="text/javascript"  src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.0.0-alpha1/jquery.min.js">  </script>

<script type="text/javascript">

var conUrlCount = 0;
var myVideo = document.getElementById("myVideo");

//parsedData should be declared outside the function scope since we want it     to be accessible from outside
var parsedData;
var callback = function (text) {
parsedData = JSON.parse(text);
console.log('parsed',parsedData)
conUrlCount = 0;
//after load play the first video
playNext();
};

//when page is loaded, data in json file is parsed and returned
function loadFunction() {
//returning json data
//ajax.get("clips.json", callback);

//simulate
getAjax(callback);
};


function playNext() {
if (!parsedData) {
    return
}
var myVideo = document.getElementById("myVideo");
myVideo.src = parsedData[conUrlCount]['content-url'];
myVideo.play();
conUrlCount+1;
}


//to simulate
function getAjax(callback) {
setTimeout(function () {
    callback(JSON.stringify([{
        "id": "vid1",
            "name": "#",
            "description": "#",
            "content-url": "video1.mp4",
            "thumb-url": "#"
    }, {
        "id": "vid2",
            "name": "#",
            "description": "#",
            "content-url": "video2.mp4",
            "thumb-url": "#"
    }, {
        "id": "vid3",
            "name": "#",
            "description": "#",
            "content-url": "video3.mp4",
            "thumb-url": "#"
    },{
        "id": "vid4",
            "name": "#",
            "description": "#",
            "content-url": "video4.mp4",
            "thumb-url": "#"
    }  ]))
}, 100)
}

loadFunction();

document.getElementById("vid1").addEventListener("click", loadFunction);
document.getElementById("vid2").addEventListener("click", loadFunction);
document.getElementById("vid3").addEventListener("click", loadFunction);

</script>

此代码还检查q值是否超出范围[0,1]。

用法：

pointsTable = [
    (0, 0),
    (0.3, 1),
    (0.5, 2),
    (0.7, 2.5),
    (0.9, 3)
    ]

def calculate_quality_points(tp, tn, fp, fn):
    q = get_quality_score(tp, tn, fp, fn)
    return get_quality_points(q)

def get_quality_score(tp, tn, fp, fn):
    return float(tp + tn) / (tp + tn + 10 * fp + fn)

def get_quality_points(q):
    i = 0
    if q < 0 or q > 1:
        return -1
    for i in range(len(pointsTable) - 1):
        if q >= pointsTable[i][0] and q < pointsTable[i + 1][0]:
            return pointsTable[i][1]
    return pointsTable[len(pointsTable) - 1][1]

我怎样才能获得过滤质量分数？

1 个答案: