Question

我有一个构建树的递归Python函数，我正在尝试将其转换为Groovy。

这是Python版本......

def get_tree(vertices):
    results = []

    if type(vertices) != list:
        vertices = [vertices]

    for vertex in vertices:
        results.append(vertex)
        children = get_children(vertex)
        if children:
            child_tree = get_tree(children)
            results.append(child_tree)

    return results

这是get_tree（1）...

的输出

  [1, [2, 3, 4, [5, 3]]]

这是我尝试将其转换为Groovy闭包......

_tree = { vertices ->

  results = []

  vertices.each() {
    results << it
    children = it."$direction"().toList()
    if (children) {
      child_tree = _tree(children)
      results << child_tree
    }
  }
  results
}

但这不起作用 - 这就是它返回的......

gremlin> g.v(1).outTree()    
==>[v[5], v[3], (this Collection), (this Collection)]

“这个收藏品”的内容是什么？

我只是粗略地理解了Groovy，我怀疑它与Groovy如何处理递归和闭包范围有关。

请赐教：）

Answer 1

解决方案是将def添加到results = []：

_tree = { vertices ->

  def results = []

  vertices.each() {
    results << it
    children = it."$direction"().toList()
    if (children) {
      child_tree = _tree(children)
      results << child_tree
    }
  }
  results
}

请参阅https://groups.google.com/d/msg/gremlin-users/iCPUifiU_wk/mrUhsjOM2h0J

Groovy如何处理闭包范围和递归？

1 个答案: