2

我有一组对象，我想按月份、名称和总和进行分组：

def things = [
    [id:1, name:"fred", total:10, date: "2012-01-01"],
    [id:2, name:"fred", total:10, date: "2012-01-03"],
    [id:3, name:"jane", total:10, date: "2012-01-04"],
    [id:4, name:"fred", total:10, date: "2012-02-11"],
    [id:5, name:"jane", total:10, date: "2012-01-01"],
    [id:6, name:"ted", total:10, date: "2012-03-21"],
    [id:7, name:"ted", total:10, date: "2012-02-09"]
];

我希望输出为：

[
 "fred":[[total:20, month:"January"],[total:10, month:"February"]],
 "jane":[[total:20,month:"January"]],
 "ted" :[[total:10, month:"February"],[total:10, month:"March"]]
]

或类似的规定。使用 groovy/grails 完成此任务的最佳方法是什么？

4

3 回答 3

3

以下几行

things.inject([:].withDefault { [:].withDefault { 0 } } ) { 
    map, v -> map[v.name][Date.parse('yyyy-MM-dd', v.date).format('MMMM')] += v.total; map 
}

会给你这个结果：

[fred:[January:20, February:10], jane:[January:20], ted:[March:10, February:10]]

（适用于 Groovy >= 1.8.7 和 2.0）

于 2012-11-08T20:36:58.087 回答
1

我结束了

things.collect { 
  // get the map down to name, total and month
  it.subMap( ['name', 'total' ] ) << [ month: Date.parse( 'yyyy-MM-dd', it.date ).format( 'MMMM' ) ]
  // Then group by name first and month second
}.groupBy( { it.name }, { it.month } ).collectEntries { k, v ->
  // Then for the names, collect
  [ (k):v.collectEntries { k2, v2 ->
    // For each month, the sum of the totals
    [ (k2): v2.total.sum() ]
  } ]
}

得到与安德烈相同的结果，更短，更好的答案;-)

编辑

有点短，但仍然没有那么好......

things.groupBy( { it.name }, { Date.parse( 'yyyy-MM-dd', it.date ).format( 'MMMM' ) } ).collectEntries { k, v ->
  [ (k):v.collectEntries { k2, v2 ->
    [ (k2): v2.total.sum() ]
  } ]
}
于 2012-11-08T20:39:28.617 回答
0

这是一个与其他解决方案执行相同操作的解决方案，但同时使用 GPars。可能有一个更严格的解决方案，但这个解决方案确实适用于测试输入。

@Grab(group='org.codehaus.gpars', module='gpars', version='1.0.0')

import static groovyx.gpars.GParsPool.*

//def things = [...]

withPool {
    def mapInner = { entrylist ->
         withPool{
             entrylist.getParallel()
                 .map{[Date.parse('yyyy-MM-dd', it.date).format('MMMM'), it.total]}
                 .combine(0) {acc, v -> acc + v}
         }
    }

    //for dealing with bug when only 1 list item
    def collectSingle = { entrylist ->
        def first = entrylist[0]
        return [(Date.parse('yyyy-MM-dd', first.date).format('MMMM')) : first.total]
    }

    def result = things.parallel
        .groupBy{it.name}.getParallel()
        .map{ [(it.key) : (it.value?.size())>1?mapInner.call(it.value):collectSingle.call(it.value) ] }
        .reduce([:]) {a, b -> a + b}


    println "result = $result"
}
于 2013-05-07T23:31:54.493 回答