我试图为每组找到（最大+最小）/2。下面是我的模式

UrlXpathsCount: {url: chararray,leafpathstr: chararray,urlpath_count: long}

我正在尝试按url字段对其进行分组

byUrl = GROUP UrlXpathsCount by url;

我试着用下面的方法找到（max+min）/2。

midRangeByUrl = FOREACH byUrl{
    urls_desc = order UrlXpathsCount by urlpath_count desc;
    urls_max = limit urls_desc 1;
    urls_asc = order UrlXpathsCount by urlpath_count asc;
    urls_min = limit urls_asc 1;

    GENERATE FLATTEN(urls_max),FLATTEN(urls_min);
};

以下是midrangebyurl的架构

midRangeByUrl: {urls_max::url: chararray,urls_max::leafpathstr: chararray,urls_max::urlpath_count: long,urls_min::url: chararray,urls_min::leafpathstr: chararray,urls_min::urlpath_count: long}

我现在面临的问题是，添加一个flatten（group），flatten（url\u max），flatten（url\u min）给了我很多我不想要的组合。
我想得到每个组的最大值+最小值/2。
为此，我将最大值和最小值的urlpath\u计数投影如下

computeMidRange = FOREACH midRangeByUrl generate urls_max::url as mid_url,((DOUBLE)urls_max::urlpath_count+(DOUBLE) urls_min::urlpath_count)/2 as midRange;

我把两张table连接起来

/* Join computeMidRange  and UrlXpathsCount */
midRangeJoin = join UrlXpathsCount by url , computeMidRange by mid_url using 'replicated';
midRangeOut = FOREACH midRangeJoin GENERATE UrlXpathsCount::url as url,UrlXpathsCount::leafpathstr as leafpathstr,
    UrlXpathsCount::urlpath_count as urlpath_count,computeMidRange::midRange as midRange;

然后用过滤器过滤

templates = FILTER midRangeOut by urlpath_count > midRange;

我想避开中档。通过某种方式计算midrangebyurl并投影以下字段url，urlpath\u count，leafpathstr，（min+max）/2而不使用join。
请帮我弄清楚这件事。谢谢