Question

我正在尝试从地图中获取列表并基于id的列表，并希望根据年龄（年龄> 35）检索值（列表）

def people = [
"1": [[name:'Bob', age: 32, gender: 'M'],[name:'Johnny', age: 36, gender: 'M']],
"3": [[name:'Claire', age: 21, gender: 'F'],[name:'Amy', age: 54, gender:'F']],
"4": [[name:'John', age: 41, gender: 'F'],[name:'Sam', age: 54, gender:'F']]
]

def id = ["1","3"]


def age = people.subMap(id).values().findAll{it.age > 35}

但是，我收到以下错误：

Cannot compare java.util.ArrayList with value '[32, 36]' and java.lang.Integer with value '35'

如何获得年龄> 1的所有人的名单？ 35？

Answer 1

您需要UIColor.clear.cgColor .flatten()的结果，因为它会生成一系列地图列表：

people.subMap(id).values()

执行[[[name:Bob, age:32, gender:M], [name:Johnny, age:36, gender:M]], [[name:Claire, age:21, gender:F], [name:Amy, age:54, gender:F]]]时，您将获得地图列表：

people.subMap(id).values().flatten()

然后您可以申请[[name:Bob, age:32, gender:M], [name:Johnny, age:36, gender:M], [name:Claire, age:21, gender:F], [name:Amy, age:54, gender:F]]

.findAll { it.age > 35 }

输出：

def people = [
        "1": [[name: 'Bob', age: 32, gender: 'M'], [name: 'Johnny', age: 36, gender: 'M']],
        "3": [[name: 'Claire', age: 21, gender: 'F'], [name: 'Amy', age: 54, gender: 'F']],
        "4": [[name: 'John', age: 41, gender: 'F'], [name: 'Sam', age: 54, gender: 'F']]
]

def id = ["1", "3"]

def age = people.subMap(id).values().flatten().findAll { it.age > 35 }

println age

希望它有所帮助。

Answer 2

虽然不像其他答案那样 Groovy ，但是一个简单的方法是collect通过转换函数，并返回[]进行不匹配（可以是flattened出来。

就是这样：

def people = [
"1": [[name:'Bob', age: 32, gender: 'M'],[name:'Johnny', age: 36, gender: 'M']],
"3": [[name:'Claire', age: 21, gender: 'F'],[name:'Amy', age: 54, gender:'F']],
"4": [[name:'John', age: 41, gender: 'F'],[name:'Sam', age: 54, gender:'F']]
]

def targetIds = ["1","3"]
def targetAge = 35

然后：

def findAgeOverN = { def key, def list, def ids, def n ->
    (ids.contains(key)) ? list.findAll { it["age"] > n } : []
}

def result = people.collect { key, list -> findAgeOverN(key, list, targetIds, targetAge) }
                   .flatten()

assert [[name:'Johnny', age: 36, gender: 'M'], 
        [name:'Amy', age: 54, gender: 'F']]

从Map获取列表并使用groovy过滤

2 个答案: