Question

原始列表：

def list = [[category: "DLF", name: "some name one"],
        [category: "EF", name: "some name two"],
        [category: "RW", name: "some name five"],
        [category: "EF", name: "some name four"],
        [category: "DLF", name: "some name two"],
        [category: "RW", name: "some name three"]]

有没有办法按类别对其进行分组并按类别进行排序，但应该按照＆＃34; EF＆＃34;，＆＃34; DLF＆＃34;，＆＃34; RW＆＃34;进行排序。我还想在类别的每个最后一个索引中插入新元素，就像这个：

def list = [[category: "EF", name: "some name two"],
            [category: "EF", name: "some name four"],
            [category: "EF", name: "others"],
            [category: "DLF", name: "some name one"],
            [category: "DLF", name: "some name two"],
            [category: "DLF", name: "others"],
            [category: "RW", name: "some name three"],
            [category: "RW", name: "some name five"],
            [category: "RW", name: "others"]]

Answer 1

因此，您可以定义结果的顺序：

def requiredOrder = ['EF', 'DLF', 'RW']

然后，您可以对其他元素进行分组，排序和添加：

def result = list.groupBy { it.category }
    .sort { requiredOrder.indexOf(it.key) }
    .collectMany { k, v -> v + [category: k, name: 'others'] }

所需的结果将在变量result

中

Answer 2

与@tim_yates几乎相同，但稍微更直接：

def requiredOrder = ['EF', 'DLF', 'RW']

def result = ( list + requiredOrder.collect{ [ category:it, name: "others"] } )
             .sort{ requiredOrder.indexOf it.category }

Groovy - 在组的每个最后一个条目中插入字符串

2 个答案: