Recursion 在Groovy或Java中从复杂数据结构中删除空值 我需要从复杂LinkedHashMap中删除空键值： 下面是一个简单的例子： 输入 期望输出：

我已经做了一半了，我只是停留在从列表中删除空值（人）。以下是我目前掌握的情况：

def removeNullValues(Object map) {
    map.collectEntries { k, v -> [k, v instanceof Map? removeNullValues(v) : v]}
            .findAll { k, v -> v != null}
}

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根据项目是列表、映射还是其他，可以使用多态性选择不同的方法：

def input = [
   f1:'abc',
   f2:'xyz',
   f3:'lmn',
   test:null,
   woo:[1, 2, null, 3],
   people:[
      [name:'James', City:'Atlanta'],
      [name:'Rachel', City:null]
   ],
   person:[name:'James', Phone:'4045555555', test:null]
]

def removeNulls(other) {
    other
}

def removeNulls(List list) {
    list.findResults { removeNulls(it) }
}

def removeNulls(Map map) {
    map.findAll { k, v -> v != null }.collectEntries { k, v ->
         [k, removeNulls(v)]
    }
}

println removeNulls(input)

将打印：

[f1:abc, f2:xyz, f3:lmn, woo:[1, 2, 3], people:[[name:James, City:Atlanta], [name:Rachel]], person:[name:James, Phone:4045555555]]

---

谢谢，这太棒了！

def input = [
   f1:'abc',
   f2:'xyz',
   f3:'lmn',
   test:null,
   woo:[1, 2, null, 3],
   people:[
      [name:'James', City:'Atlanta'],
      [name:'Rachel', City:null]
   ],
   person:[name:'James', Phone:'4045555555', test:null]
]

def removeNulls(other) {
    other
}

def removeNulls(List list) {
    list.findResults { removeNulls(it) }
}

def removeNulls(Map map) {
    map.findAll { k, v -> v != null }.collectEntries { k, v ->
         [k, removeNulls(v)]
    }
}

println removeNulls(input)

[f1:abc, f2:xyz, f3:lmn, woo:[1, 2, 3], people:[[name:James, City:Atlanta], [name:Rachel]], person:[name:James, Phone:4045555555]]