Groovy：解析列表元素

Question

出于学习目的，我喜欢将以下查找表解析如下： 我想检查分支名称并获取相应的名称值，但我想获取每个分支的名字，例如： [branch: "test"] => name "a" than [branch: "test-1" ] => 名称 "d" ;等等，

LUT = [
    [branch: "test", name: 'a', image_name: 'abc'],
    [branch: "test", name: 'b', image_name: 'abc'],
    [branch: "test", name: 'c', image_name: 'abc'],
    [branch: "test-1", name: 'd', image_name: 'abc'],
    [branch: "test-1", name: 'e', image_name: 'abc'],
    [branch: "test-2", name: 'f', image_name: 'abc'],
    [branch: "test-2", name: 'g', image_name: 'abc'],
    [branch: "test-2", name: 'h', image_name: 'abc'],
    [branch: "test-3", name: 'i', image_name: 'abc'],
    [branch: "test-3", name: 'j', image_name: 'abc'],
    [branch: "test-4", name: 'k', image_name: 'abc'],
    [branch: "test-5", name: 'l', image_name: 'abc'],
]

例如，结果可能如下所示：

result = {
 "test": [{ "name":"a" }],
 "test-1": [{ "name":"d" }],
 "test-2": [{ "name":"f" }],
 "test-3": [{ "name":"i" }], 
"test-4": [{ "name":"k" }], 
"test-5": [{ "name":"l" }], 
} 

此外，稍后我可能想为键添加其他值，例如在 test-1 中： "test-1": [{ "name":"a",“新名称”：“新”}]

Answer 1

我不确定你的大括号，但这会创建一个地图：

def original = [
    [branch: "test", name: 'a', image_name: 'abc'],
    [branch: "test", name: 'b', image_name: 'abc'],
    [branch: "test", name: 'c', image_name: 'abc'],
    [branch: "test-1", name: 'd', image_name: 'abc'],
    [branch: "test-1", name: 'e', image_name: 'abc'],
    [branch: "test-2", name: 'f', image_name: 'abc'],
    [branch: "test-2", name: 'g', image_name: 'abc'],
    [branch: "test-2", name: 'h', image_name: 'abc'],
    [branch: "test-3", name: 'i', image_name: 'abc'],
    [branch: "test-3", name: 'j', image_name: 'abc'],
    [branch: "test-4", name: 'k', image_name: 'abc'],
    [branch: "test-5", name: 'l', image_name: 'abc'],
]

println original
    .groupBy { branch -> branch.branch }
    .collectEntries { key,value -> [(key) : [name : value.get(0).name]] }

Answer 2

不确定如何按照您的意愿格式化结果，但是找到每个分支的第一个实例相当容易。

result = [:]
for (map in LUT)
{
    if (!result.containsKey(map['branch']))
    {
        println map['name'] // prints the unique name
        result.put(map['branch'], map['name'])
    }
}