Question

我有一个用户列表

def items = [[name:"tony", age:4], [name:"tony", age: 5], [name:"alan", age:16]]

我想按名称对它们进行分组，但只想要列表中的年龄，以便我想要

["tony": [4, 5], "alan": [16]]

当我这样做时

def groups = items.groupBy {it.name}

我得到： [托尼：[[姓名：托尼，年龄：4]，[姓名：托尼，年龄：5]]，阿兰：[[姓名：阿兰，年龄：16]]]

为了得到我想要的东西，还有很多工作要做。有什么提示吗？

Answer 1

你也可以在1循环而不是2循环中完成：

def items = [[name:"tony", age:4], [name:"tony", age: 5], [name:"alan", age:16]]
def groupped = items.inject( [:].withDefault{ [] } ){ res, curr ->
  res[ curr.name ] << curr.age
  res
}

Answer 2

尝试：

def items = [[name:"tony", age:4], [name:"tony", age: 5], [name:"alan", age:16]]
def t = items.groupBy { it.name }.collectEntries { [(it.key):(it.value*.age)] }
assert t == ['tony':[4,5],'alan':[16]]

在Groovy中使用groupBy

2 个答案: