Question

我有一个包含地图列表的JSON：

{
"aprLimits": [{
    "orgId": "Company 1",
    "rangeStart": "54936",
    "rangeEnd": "54936",
    "amountLimit": 100000
}, {
    "orgId": "Company 1",
    "rangeStart": "59434",
    "rangeEnd": "59434",
    "amountLimit": 0.01
}, {
    "orgId": "Company 1",
    "rangeStart": "54345",
    "rangeEnd": "54345",
    "amountLimit": 0
}, {
    "orgId": "Company 1",
    "rangeStart": "54345",
    "rangeEnd": "54345",
    "amountLimit": 250000
}, {
    "orgId": "Company 2",
    "rangeStart": "043",
    "rangeEnd": "043",
    "amountLimit": 0.01
}, {
    "orgId": "Company 2",
    "rangeStart": "042",
    "rangeEnd": "042",
    "amountLimit": 1000
}, {
    "orgId": "Company 2",
    "rangeStart": "032",
    "rangeEnd": "032",
    "amountLimit": 0
}]}

我的目标是通过将唯一的orgId转换为键并为其分配其余值来收集与唯一orgId相对应的数据，以便最终结果如下所示：

["Company 1": [[54936, 54936, 100000], [59434, 59434, 0.01], [54345, 54345, 0], [54345, 54345, 250000]], "Company 2": [[043, 043, 0.01], [042, 042, 1000], [032, 032, 0]]]

我已经获得了唯一的orgId并将数据收集到列表中，但是我很难弄清楚如何将唯一键与列表相关联

def slurper = new JsonSlurper()
def parsedJson = slurper.parseText(json)
def list = []
def orgIds = parsedJson.aprLimits.orgId.unique()

parsedJson.aprLimits.each {
if (orgIds.contains(it.orgId)) {
    data = new ArrayList(Arrays.asList(it.rangeStart, it.rangeEnd, it.amountLimit))
    list.add(data)
}}

Answer 1

这里最好的起点是groupBy，然后调整所需的数据。更简短的是 reduce （带有映射的默认值）。例如

parsedJson.aprLimits.inject(
    [:].withDefault{[]} // if a key is missing, add it and make the value an empty vector
){ r, v -> 
    r[v.orgId] << [v.rangeStart, v.rangeEnd, v.amountLimit]
    r
}

如何从重复的地图值产生唯一键

1 个答案: