我有一个简短的groovy算法,可根据评分为食物分配排名。这可以在groovy控制台中运行。代码工作得很好,但我想知道是否有更多的Groovy或函数式编写代码。认为如果可能的话,最好摆脱previousItem
和rank
局部变量。
def food = [
[name:'Chocolate Brownie',rating:5.5, rank:null],
[name:'Fudge', rating:2.1, rank:null],
[name:'Pizza',rating:3.4, rank:null],
[name:'Icecream', rating:2.1, rank:null],
[name:'Cabbage', rating:1.4, rank:null]]
food.sort { -it.rating }
def previousItem = food[0]
def rank = 1
previousItem.rank = rank
food.each { item ->
if (item.rating == previousItem.rating) {
item.rank = previousItem.rank
} else {
item.rank = rank
}
previousItem = item
rank++
}
assert food[0].rank == 1
assert food[1].rank == 2
assert food[2].rank == 3
assert food[3].rank == 3 // Note same rating = same rank
assert food[4].rank == 5 // Note, 4 skipped as we have two at rank 3
建议?
答案 0 :(得分:2)
这是我的解决方案:
def rank = 1
def groupedByRating = food.groupBy { -it.rating }
groupedByRating.sort().each { rating, items ->
items.each { it.rank = rank }
rank += items.size()
}
答案 1 :(得分:0)
我没试过,但也许这可行。
food.eachWithIndex { item, i ->
if( i>0 && food[i-1].rating == item.rating )
item.rank = food[i-1].rank
else
item.rank = i + 1
}
答案 2 :(得分:0)
这是另一种不使用groovy注入方法的“local defs”的替代方法:
food.sort { -it.rating }.inject([index: 0]) { map, current ->
current.rank = (current.rating == map.lastRating ? map.lastRank : map.index + 1)
[ lastRating: current.rating, lastRank: current.rank, index: map.index + 1 ]
}