使用Groovy按降序对Map值进行排序

Question

我有一个Map<String,Integer>其条目（键）需要按降序排序。 例如，如果地图如下所示：

"a" => 5
"b" => 3
"c" => 12
"d" => 9

排序后需要看起来像：

"c" => 12
"d" => 9
"a" => 5
"b" => 3

迄今为止我最好的尝试：

def test() {
    Map<String,Integer> toSort = new HashMap<String,Integer>()
    toSort.put("a", 5)
    toSort.put("b", 3)
    toSort.put("c", 12)
    toSort.put("d", 9)

    Map<String,Integer> sorted = sortMapDesc(toSort)
    sorted.each {
        println "${it.key} has a value of ${it.value}."
    }
}

def sortMapDesc(Map<String,Integer> toSort) {
    println "Sorting..."
    println toSort

    // The map of properly sorted entries.
    Map<String,Integer> sorted = new HashMap<String,Integer>()

    // Keep scanning the map for the key with the highest value. When we find
    // it, add it as the next entry to the 'sorted' map, and then zero it out
    // so it won't show up as the highest on subsequent scans/passes. Stop scanning
    // when the entire 'toSort' map contains keys with zeros.
    while(!mapIsAllZeros(toSort)) {
        int highest = -1
        String highestKey = ""
        toSort.each {
            if(it.value > highest) {
                highest = it.value
                highestKey = it.key
            }
        }

        toSort.put(highestKey, 0)
        sorted.put(highestKey, highest)
    }

    sorted
}

def mapIsAllZeros(Map<String,Integer> toCheck) {
    toCheck.values().every{!it}
}

当我运行test()我得到以下输出：

Sorting...
[d:9, b:3, c:12, a:5]
d has a value of 9.
b has a value of 3.
c has a value of 12.
a has a value of 5.

我在哪里错了？

Answer 1

做就是了：

​def m = [a:​5, b:12, c:3, d:9]
def sorted = m.sort { a, b -> b.value <=> a.value }

Answer 2

为了进行排序，Tim的实施是要走的路。 但是，如果您只是想知道为什么您的示例代码不能按预期工作，那么答案是变量'sorted'需要是LinkedHashMap类型，而不仅仅是HashMap。 您可以明确设置它：

Map<String,Integer> sorted = new LinkedHashMap<String,Integer>()

或者，只需这样做：

Map<String,Integer> sorted = [:]