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我有以下斯卡拉列表 -基於使用scala對元素進行分組的元素總和?
List((192.168.1.1,8590298237), (192.168.1.1,8590122837), (192.168.1.1,4016236988),
(192.168.1.1,1018539117), (192.168.1.1,2733649135), (192.168.1.2,16755417009),
(192.168.1.1,3315423529), (192.168.1.2,1523080027), (192.168.1.1,1982762904),
(192.168.1.2,6148851261), (192.168.1.1,1070935897), (192.168.1.2,276531515092),
(192.168.1.1,17180030107), (192.168.1.1,8352532280), (192.168.1.3,8590120563),
(192.168.1.3,24651063), (192.168.1.3,4431959144), (192.168.1.3,8232349877),
(192.168.1.2,17493253102), (192.168.1.2,4073818556), (192.168.1.2,42951186251))
我想下面的輸出 -
List((192.168.1.1, sum of all values of 192.168.1.1),
(192.168.1.2, sum of all values of 192.168.1.2),
(192.168.1.3, sum of all values of 192.168.1.3))
如何使用Scala的第一個元素分組得到名單第二元素的總和?