Question

我基本上想对地图进行分组，我必须按汽车和颜色分组并对其价格求和，如果任何一组价格大于25，则按汽车，颜色和电机分组结果并乘以总和* 3，部分已经完成，但是我被第二组困住了：

def row1 = ["car":'A',"color":'B',"motor":'C', "price": 12]
def row2 = ["car":'A',"color":'B',"motor":'C', "price": 12]
def row3 = ["car":'A',"color":'B',"motor":'D', "price": 2]
def row4 = ["car":'B',"color":'B',"motor":'D', "price": 13]
def arrayRows = []
arrayRows.add(row1)
arrayRows.add(row2)
arrayRows.add(row3)
arrayRows.add(row4)
println(arrayRows) //[[car:A, color:B, motor:C, price:12], [car:A, color:B, motor:C, price:12], [car:A, color:B, motor:D, price:2], [car:B, color:B, motor:D, price:13]]
def groups = arrayRows.groupBy {row -> [Car:row.car, Color:row.color] }.collect{k , v ->
    [
        car:k.Car,
        color:k.Color,
        motor:v.motor,
        price:v.collect { it.price }.sum()
    ]
}
println(groups) //[[car:A, color:B, motor:[C, C, D], price:26], [car:B, color:B, motor:[D], price:13]]
for(group in groups){
    if (group.price > 25){ //26>25
        println group //[car:A, color:B, motor:[C, C, D], price:26]
        //def groupByCarColorMotor = arrayRows.groupBy {[Car: group.car, Color: group.color, Motor:?]} | Must group by car, color and motor, and multiply * 3 it's price but since motor is an array, I'm not sure how to do so, I've tried groupBy { row -> group.it.motor} etc
    }
}

如果有以下情况，我应该如何第二次分组： [汽车：A，颜色：B，电机：[C，C，D] 我应该按以下条件分组： [汽车：A，颜色：B，马达：C] 和 [汽车：A，颜色：B，电动机：D]

预期输出应为： [[[“ car”：'A'，“ color”：'B'，“ motor”：'C'，“ price”：72，[“” car“：'A'，” color“：'B'，”电机”：“ D”，“价格”：6]]

编辑；我几乎设法做到了，问题是我得到了一组地图阵列，你也可能会得到它的支持。

def arrayRows = [ 
        ["car":'A',"color":'B',"motor":'C', "price": 12],
        ["car":'A',"color":'B',"motor":'C', "price": 12],
        ["car":'A',"color":'B',"motor":'D', "price": 2],
        ["car":'B',"color":'B',"motor":'D', "price": 13]
        ]

println(arrayRows) //[[car:A, color:B, motor:C, price:12], [car:A, color:B, motor:C, price:12], [car:A, color:B, motor:D, price:2], [car:B, color:B, motor:D, price:13]]
def groups = arrayRows.groupBy {row -> [Car:row.car, Color:row.color] }.collect{k , v ->
    [
            car:k.Car,
            color:k.Color,
            price:v.collect { it.price }.sum()
    ]
}.findAll{it.price > 25}

def groupByCarColor = []
for (group in groups){
    groupByCarColor.add(arrayRows.findAll{ row -> row.car == group.car && row.color == group.color}.groupBy {row -> [Car:group.car, Color:group.color, Motor:row.motor] }.collect{k , v ->
        [
                car:k.Car,
                color:k.Color,
                motor:k.Motor,
                price:v.collect { it.price }.sum()*3
        ]
    })
}

输出：[[[汽车：A，颜色：B，马达：C，价格：72]，[汽车：A，颜色：B，马达：D，价格：6]]]

Answer 1

直接完成您的家庭作业：

constructSearch(events)

Answer 2

您的目标是通过三元组找到组-但您想过滤掉谓词在子组上失败的那些组。例如

def arrayRows = [ 
    [car:'A', motor:'C', color:'B',price: 12],
    [car:'A', motor:'C', color:'B',price: 12],
    [car:'A', motor:'D', color:'B',price:  2],
    [car:'B', motor:'D', color:'B',price: 13],
]


def output = arrayRows.inject([:].withDefault{ 0 }){ acc, row -> // group by the triplet and sum up the price
    acc[row.subMap(['car','motor','color'])]+=row.price; acc
}.groupBy{ k, _ -> // group by the filter criteria tuple
    k.subMap(['car','color'])
}.findAll{ _, v -> // eliminate the tuple-groups where price is to low
    v.values().sum() > 25
}.collectMany{ _, vs -> // reshape the data
    vs.collect{ k, v ->
        k + [price: v * 3]
    }
}

assert output==[[car:"A", motor:"C", color:"B", price:72], [car:"A", motor:"D", color:"B", price:6]]

Groovy列表：按元素列表分组

2 个答案: