collections - 常规列表 : Group By element's count and find highest frequency elements

Question

我有一个常规列表如下

def certs = ['0xc1','0xc1','0xc1','0xc1','0xc2','0xc2','0xc3','0xc4','0xc4','0xc5','0xc5','0xc5','0xc5']

我试图通过其计数找到每个元素和组的出现。
我试过了

certs.groupBy { it }.findAll { it.value.size() }

但我得到以下输出

[0xc1:[0xc1, 0xc1, 0xc1, 0xc1], 0xc2:[0xc2, 0xc2], 0xc3:[0xc3], 0xc4:[0xc4, 0xc4], 0xc5:[0xc5, 0xc5, 0xc5, 0xc5]]

相反，我期待下面

[0xc1:4, 0xc2:2, 0xc3:1, 0xc4:2, 0xc5:4]

有人可以帮我弄这个吗？另外我想在我的例子中找到列表中出现的最大元素 0xc1和 0xc5
更新:

def myMap = certs.inject([:]) { m, x -> if (!m[x]) m[x] = 0; m[x] += 1; m }
def maxValue = myMap.values().max{it} 
def myKeys = []
myMap.findAll{ it.value == maxValue }.each{myKeys << it?.key}
println myKeys  // result = [0xc1:4, 0xc5:4]
//println myMap.sort { a, b -> b.value <=> a.value }

Answer 1

Map counts = certs.countBy { it }
counts.findAll { it.value == counts.values().max() }

或单线

certs.countBy { it }.groupBy { it.value }.max { it.key }.value.keySet()