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连续行数(Number of successive rows)

 考虑一个data.table ，它包含多年来id1 ， id2之间的匹配。 
 

        id1 year      id2
1: 51557094 2003 65122111
2: 51557094 2004 65122111
3: 51557094 2005 65122111
4: 51557094 2007 65122111
5: 51557094 2008 65122111
6: 51557093 2006 65122111

 

 对于这些比赛中的任何一个，我想找出持续时间以及比赛开始的年份。 如果特定年份没有数据，则匹配结束（如果再次有数据，则在下一年开始新匹配）。 
 

 因此，对于上面的样本数据，预期的输出将是 
 

        id1 year      id2 length
1: 51557094 2003 65122111      3
2: 51557094 2007 65122111      2
3: 51557093 2006 65122111      1

 

 我接受了其中一个答案，因为它给我带来了足够的意义，但注意到它不正确。 虽然它适用于样本数据，但不适用于以下情况 
 

> dtId
        id1 year      id2
1: 51557094 2003 65122111
2: 51557094 2004 65122111
3: 51557094 2005 65122111
4: 51557094 2007 65122111
5: 51557094 2008 65122111
6: 51557094 2006 65122112

> setkey(dtId, id1, id2, year)
> dtId[,grp := cumsum(c(1,diff(year)) > 1),by=id1]
> dtId[,list(year=year[1],length=length(year)),by=list(id1,id2,grp)]
        id1      id2 grp year length
1: 51557094 65122111   0 2003      5
2: 51557094 65122112   0 2006      1

 

 相反，在id1, id2创建匹配变量grp可以解决问题： 
 

> dtId[,grp := cumsum(c(1,diff(year)) > 1),by=list(id1, id2)]
> dtId[,list(year=year[1],length=length(year)),by=list(id1,id2,grp)]
        id1      id2 grp year length
1: 51557094 65122111   0 2003      3
2: 51557094 65122112   0 2006      1
3: 51557094 65122111   1 2007      2


Consider a data.table that contains matches between id1, id2 for several years.
 

        id1 year      id2
1: 51557094 2003 65122111
2: 51557094 2004 65122111
3: 51557094 2005 65122111
4: 51557094 2007 65122111
5: 51557094 2008 65122111
6: 51557093 2006 65122111

 

For any of these matches, I want to find out the duration, together with the year the match started. If there was no data for a specific year, a match ends (and the next year, if there is data again, a new match starts).
 

Hence, for the sample data above, the expected output would be
 

        id1 year      id2 length
1: 51557094 2003 65122111      3
2: 51557094 2007 65122111      2
3: 51557093 2006 65122111      1

 

I accepted one of the answers because it brought me far enough, but notice that it was not correct. While it worked for the sample data, it will not for the following
 

> dtId
        id1 year      id2
1: 51557094 2003 65122111
2: 51557094 2004 65122111
3: 51557094 2005 65122111
4: 51557094 2007 65122111
5: 51557094 2008 65122111
6: 51557094 2006 65122112

> setkey(dtId, id1, id2, year)
> dtId[,grp := cumsum(c(1,diff(year)) > 1),by=id1]
> dtId[,list(year=year[1],length=length(year)),by=list(id1,id2,grp)]
        id1      id2 grp year length
1: 51557094 65122111   0 2003      5
2: 51557094 65122112   0 2006      1

 

Instead, creating the match variable grp over both id1, id2 solves the issue:
 

> dtId[,grp := cumsum(c(1,diff(year)) > 1),by=list(id1, id2)]
> dtId[,list(year=year[1],length=length(year)),by=list(id1,id2,grp)]
        id1      id2 grp year length
1: 51557094 65122111   0 2003      3
2: 51557094 65122112   0 2006      1
3: 51557094 65122111   1 2007      2


原文：https://stackoverflow.com/questions/27283589
更新时间：2022-09-17 08:09

最满意答案

 在那里你可以看到这里的例子我必须展示如何编码ajax响应所以我在我的域名中添加了示例 
 

 你的PHP文件看起来像下面这样，我省略了SQL部分，并且只添加了关于如何在数组中使用json_encode的逻辑，希望你发现它有帮助，你可以在你的本地机器上使用这段代码来看看事情是如何工作的 
 

<?php 
$response   =   array('success'=>false,'likes'=>0);


if(isset($_POST['count'])){
    $counter =   $_POST['count'];
    $response['likes']=$counter+1;
    $response['success']=true;
}

echo json_encode($response);
?>

 

 您的HTML页面如下 
 

<html>

<head>
    <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
    <link href="https://maxcdn.bootstrapcdn.com/font-awesome/4.7.0/css/font-awesome.min.css" rel="stylesheet" />

    <script src="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/js/bootstrap.min.js"></script>
    <link href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/css/bootstrap.min.css" rel="stylesheet" />
    <style>
        .feed {
            width: 95%;
            height: auto;
        }

        i.fa {
            cursor: pointer;
        }
    </style>
    <script type="text/javascript">
        $(document).ready(function () {
            $(".voteup").click(function () {
                var curElement = $(this);
console.log(curElement.parent().find('.likes').text());
                $.ajax({
                    url: 'test.php',
                    dataType: 'json',
                    data: 'count=' + curElement.parent().find(".likes").text(),
                    method: 'POST'
                }).done(function (data) {
                    if (data.success) {
                        curElement.parent().find(".likes").html(data.likes);
                    } else {
                        alert('Some Error occoured at the server while liking the feed');
                    }
                });
            });
        });
    </script>
</head>

<body>
    <div class="panel panel-default">
        <div class="panel-heading">Panel Heading</div>
        <div class="panel-body">
            <div class="feed">
                <p>This is my feed can someone like it</p>
                <i class="fa fa-thumbs-up voteup" aria-hidden="true" ></i>
                <span class="likes">0</span>
                <i class="fa fa-thumbs-down votedown" aria-hidden="true" ></i>
                <span class="dlikes">0</span>
            </div>
            <div class="feed">
                <p>Another feed item</p>
                <i class="fa fa-thumbs-up voteup" aria-hidden="true" ></i>
                <span class="likes">0</span>
                <i class="fa fa-thumbs-down votedown" aria-hidden="true" ></i>
                <span class="dlikes">0</span>
            </div>
            <div class="feed">
                <p>This is my feed can someone like it</p>
                <i class="fa fa-thumbs-up voteup" aria-hidden="true" ></i>
                <span class="likes">0</span>
                <i class="fa fa-thumbs-down votedown" aria-hidden="true" ></i>
                <span class="dlikes">0</span>
            </div>
            <div class="feed">
                <p>This is my feed can someone like it</p>
                <i class="fa fa-thumbs-up voteup" aria-hidden="true" ></i>
                <span class="likes">0</span>
                <i class="fa fa-thumbs-down votedown" aria-hidden="true" ></i>
                <span class="dlikes">0</span>
            </div>
            <div class="feed">
                <p>This is my feed can someone like it</p>
                <i class="fa fa-thumbs-up voteup" aria-hidden="true" ></i>
                <span class="likes">0</span>
                <i class="fa fa-thumbs-down votedown" aria-hidden="true" ></i>
                <span class="dlikes">0</span>
            </div>
            <div class="feed">
                <p>This is my feed can someone like it</p>
                <i class="fa fa-thumbs-up voteup" aria-hidden="true" ></i>
                <span class="likes">0</span>
                <i class="fa fa-thumbs-down votedown" aria-hidden="true" ></i>
                <span class="dlikes">0</span>
            </div>
        </div>
    </div>


</body>


</html>

 

 编辑： 
 

 基本上，我只是增加发布的变量count你不必做，只需要在数据库中更新喜欢就可以了，一旦你发送了ajax调用，然后用SQL查询计数并以我用过的相同格式显示输出。关于$.parseJSON()你会注意到这里使用的ajax调用将dataType设置为JSON如果你设置了不需要解析响应的dataType ，否则你应该使用var myData=$.parseJSON(data); 然后像myData.likes myData.success一样使用 

there you go you can see the example here I had to show how the ajax response should be encoded so I added the example on my domain 
 

your PHP file will look like the following, I have omitted the SQL part and added only the logic on how to use json_encode with the arrays hope you find it helpful you can use this code on your local machine to look into how things are working 
 

<?php 
$response   =   array('success'=>false,'likes'=>0);


if(isset($_POST['count'])){
    $counter =   $_POST['count'];
    $response['likes']=$counter+1;
    $response['success']=true;
}

echo json_encode($response);
?>

 

your HTML page is below
 

<html>

<head>
    <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
    <link href="https://maxcdn.bootstrapcdn.com/font-awesome/4.7.0/css/font-awesome.min.css" rel="stylesheet" />

    <script src="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/js/bootstrap.min.js"></script>
    <link href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/css/bootstrap.min.css" rel="stylesheet" />
    <style>
        .feed {
            width: 95%;
            height: auto;
        }

        i.fa {
            cursor: pointer;
        }
    </style>
    <script type="text/javascript">
        $(document).ready(function () {
            $(".voteup").click(function () {
                var curElement = $(this);
console.log(curElement.parent().find('.likes').text());
                $.ajax({
                    url: 'test.php',
                    dataType: 'json',
                    data: 'count=' + curElement.parent().find(".likes").text(),
                    method: 'POST'
                }).done(function (data) {
                    if (data.success) {
                        curElement.parent().find(".likes").html(data.likes);
                    } else {
                        alert('Some Error occoured at the server while liking the feed');
                    }
                });
            });
        });
    </script>
</head>

<body>
    <div class="panel panel-default">
        <div class="panel-heading">Panel Heading</div>
        <div class="panel-body">
            <div class="feed">
                <p>This is my feed can someone like it</p>
                <i class="fa fa-thumbs-up voteup" aria-hidden="true" ></i>
                <span class="likes">0</span>
                <i class="fa fa-thumbs-down votedown" aria-hidden="true" ></i>
                <span class="dlikes">0</span>
            </div>
            <div class="feed">
                <p>Another feed item</p>
                <i class="fa fa-thumbs-up voteup" aria-hidden="true" ></i>
                <span class="likes">0</span>
                <i class="fa fa-thumbs-down votedown" aria-hidden="true" ></i>
                <span class="dlikes">0</span>
            </div>
            <div class="feed">
                <p>This is my feed can someone like it</p>
                <i class="fa fa-thumbs-up voteup" aria-hidden="true" ></i>
                <span class="likes">0</span>
                <i class="fa fa-thumbs-down votedown" aria-hidden="true" ></i>
                <span class="dlikes">0</span>
            </div>
            <div class="feed">
                <p>This is my feed can someone like it</p>
                <i class="fa fa-thumbs-up voteup" aria-hidden="true" ></i>
                <span class="likes">0</span>
                <i class="fa fa-thumbs-down votedown" aria-hidden="true" ></i>
                <span class="dlikes">0</span>
            </div>
            <div class="feed">
                <p>This is my feed can someone like it</p>
                <i class="fa fa-thumbs-up voteup" aria-hidden="true" ></i>
                <span class="likes">0</span>
                <i class="fa fa-thumbs-down votedown" aria-hidden="true" ></i>
                <span class="dlikes">0</span>
            </div>
            <div class="feed">
                <p>This is my feed can someone like it</p>
                <i class="fa fa-thumbs-up voteup" aria-hidden="true" ></i>
                <span class="likes">0</span>
                <i class="fa fa-thumbs-down votedown" aria-hidden="true" ></i>
                <span class="dlikes">0</span>
            </div>
        </div>
    </div>


</body>


</html>

 

EDIT: 
 

Basically, I am just incrementing the posted variable count you do not have to do that you just need to update likes in the database once you send the ajax call and then count with an SQL query and show the output in the same format I have used.And about the $.parseJSON() you will notice that the ajax call used here has the dataType set to JSON if you have set the dataType you do not need to parse the response otherwise you should use var myData=$.parseJSON(data); and then use like myData.likes myData.success

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