我试图找到每组的(最大+最小)/ 2。以下是我的架构
UrlXpathsCount: {url: chararray,leafpathstr: chararray,urlpath_count: long}
我试图通过url字段对其进行分组
byUrl = GROUP UrlXpathsCount by url;
我试图通过以下方式找到(最大+最小)/ 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FLATTEN(组),FLATTEN(urls_max),FLATTEN(urls_min)给了我很多我不想要的组合。
我希望每组得到最大+最小/ 2。
为此,我通过以下
投影max和min的urlpath_countcomputeMidRange = FOREACH midRangeByUrl generate urls_max::url as mid_url,((DOUBLE)urls_max::urlpath_count+(DOUBLE) urls_min::urlpath_count)/2 as midRange;
我将通过以下
加入这两个表格/* 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;
然后过滤应用过滤器
templates = FILTER midRangeOut by urlpath_count > midRange;
我想避开midRangeJoin。通过某种方式计算midRangeByUrl并在没有连接的情况下投射以下字段url,urlpath_count,leafpathstr,(min + max)/ 2。
请帮我解决这个问题。 感谢
答案 0 :(得分:2)
您可以使用内置的$('#myhidden').val(myarray.split("|")); //set "0|1".split("|") - creates array like [0,1]
myarray = $('#myhidden').val().join("|"); //get [0,1].join("|") - creates string like "0|1"
和MAX
UDF:
MIN
这将完全符合您的要求,没有嵌套的foreach或连接。我将计算分为UrlXpathsCount = load 'your_data' using PigStorage(',') as (url: chararray,leafpathstr: chararray,urlpath_count: long);
B = GROUP UrlXpathsCount by url;
C = foreach B generate group as url, MAX(UrlXpathsCount.urlpath_count) as max_count,
MIN(UrlXpathsCount.urlpath_count) as min_count;
D = foreach C generate url, ((double)max_count + (double)min_count)/2 as val;
和C
以避免极长的行,但您也可以在一行中完成。只需记住将值转换为D
,因为double
是urlpath_count
,所以如果你没有投出任何小数,你就不会得到任何小数。