当使用foldleft发现重复的scala时，从List of case类结构中删除元素

Question

如果在scala 中发现重复，则删除case类的两个对象

示例列表如下：

val pets = List(
Pet("cat", "sassy", 2), Pet("cat", "bella", 3),
Pet("dog", "poppy", 3), Pet("dog", "bodie", 4), Pet("dog", "poppy", 2),
Pet("bird", "coco", 2), Pet("bird", "kiwi", 1),
Pet("hen", "kiwi", 2), Pet("horse", "kiwi", 1),
Pet("cow", "mow", 2)
)

我想检查列表中是否有重复项，并将它们从列表中删除。在这种情况下，我的列表将只有一个值，在本例中，它将是值Pet("cow", "mow", 2)，其他值将消失。这里不考虑Pet("hen", "kiwi", 2)、Pet("horse", "kiwi", 1)；猕猴桃"；重复。

case class Pet(species: String, name: String, age: Int)
val pets = List(
Pet("cat", "sassy", 2), Pet("cat", "bella", 3),
Pet("dog", "poppy", 3), Pet("dog", "bodie", 4), Pet("dog", "poppy", 2),
Pet("bird", "coco", 2), Pet("bird", "kiwi", 1),
Pet("hen", "kiwi", 2), Pet("horse", "kiwi", 1),
Pet("cow", "mow", 2)
)
def customEqualsPet(x : Pet, y: Pet) = (x.species == y.species || x.name== y.name)
val petsOut = pets.foldLeft(Nil : List[Pet]) {(list, item) =>
val exists = list.find(x => customEqualsPet(item, x))

if (exists.isEmpty) {
item :: list
} else {
println("once......"+list)
println("once......"+item)
list.filter(u=>{(u.name!=item.name) || (u.species!=item.species)})
}
}.reverse
println(cc)

似乎干扰了输出。感谢您的帮助预期输出

List( Pet("cow", "mow", 2))

Answer 1

您的另一个选择是，首先查找要删除的所有名称和物种：

val speciesToRemove = pets.groupBy(_.species).filter(_._2.length > 1).keySet
val namesToRemove = pets.groupBy(_.name).filter(_._2.length > 1).keySet

然后通过它们进行过滤：

pets.filterNot(pet => speciesToRemove.contains(pet.species) || namesToRemove.contains(pet.name))

代码在Scastie中运行。

Answer 2

效率不高，但简洁易懂。(不需要foldLeft()。(

val species = pets.groupBy(_.species)
val names   = pets.groupBy(_.name)
pets.filter(p => species(p.species).length + 
names(p.name).length == 2)
//res0: List[Pet] = List(Pet(cow,mow,2))

如果species指定不能出现在name字符串中，而name字符串不能作为species出现，则可以将其浓缩。

val groups = pets.groupBy(_.species) ++ pets.groupBy(_.name)
pets.filter(p => groups(p.species).length + 
groups(p.name).length == 2)

---

如果您想要唯一元素和所有非唯一元素的列表。。。

val (unique, dups) =
pets.partition(p => groups(p.species).length +
groups(p.name).length == 2)