Question

我想检查谁能想出最好的Groovy-sh方式来实现这个目标 -

def m1 = [["id":"1","o":"11"],["id":"1","o":"12"],["id":"2","o":"21"]]
def m2 = [["o":"11","t":"t1"],["o":"11","t":"t2"],["o":"21","t":"t1"]]

我想要结果

[["id":"1","t":"t1"],["id":"1","t":"t2"],["id":"2","t":"t1"]]

我目前正在迭代地图并执行此操作。我正在寻找使用Gpath和findAll

的解决方案

谢谢， Sreehari。

Answer 1

您可以transpose列出每个列表并从每个列表中获取条目（id或t）：

def fn = { m1, m2 ->
    return [m1,m2]
        .transpose()
        .collect { [ id: it.first().id, t: it.last().t ] }
}


def m1 = [["id":"1","o":"11"],["id":"1","o":"12"],["id":"2","o":"21"]]
def m2 = [["o":"11","t":"t1"],["o":"11","t":"t2"],["o":"21","t":"t1"]]


assert fn(m1, m2) == 
        [["id":"1","t":"t1"],["id":"1","t":"t2"],["id":"2","t":"t1"]]

Answer 2

您可以使用转置将地图压缩成对，然后将这些对组合并按地图键过滤：

[m1, m2]
    .transpose()
    .collect { (it[0] + it[1]).subMap(['id', 't']) }

评估为

[[id:1, t:t1], [id:1, t:t2], [id:2, t:t1]]

这适用于使用groovy-2.4.4的groovysh，使用jdk7或jdk8。

Groovy - 加入过滤器映射

2 个答案: