我一直在尝试对Python中的BFS实现进行性能优化，而我最初的实现是使用双端队列来存储要扩展的节点队列和一个字典来存储相同的节点，这样我就可以高效地查找它是否已经打开。

我试图通过移至OrderedDict来优化（简单性和效率）。但是，这会花费更多时间。完成400个样本搜索时，双引号/字典需要2秒，而仅OrderedDict则需要3.5秒。

我的问题是，如果OrderedDict的功能与两个原始数据结构相同，那么它的性能至少不应该相似吗？还是我在这里想念什么？下面的代码示例。

仅使用OrderedDict：

open_nodes = OrderedDict()
closed_nodes = {}
current = Node(start_position, None, 0)
open_nodes[current.position] = current

while open_nodes:
  current = open_nodes.popitem(False)[1]
  closed_nodes[current.position] = (current)

  if goal(current.position):
    return trace_path(current, open_nodes, closed_nodes)

  # Nodes bordering current
  for neighbor in self.environment.neighbors[current.position]:
    new_node = Node(neighbor, current, current.depth + 1)
    open_nodes[new_node.position] = new_node

同时使用双端队列和字典：

open_queue = deque()
open_nodes = {}
closed_nodes = {}
current = Node(start_position, None, 0)
open_queue.append(current)
open_nodes[current.position] = current

while open_queue:
  current = open_queue.popleft()
  del open_nodes[current.position]
  closed_nodes[current.position] = (current)

  if goal_function(current.position):
    return trace_path(current, open_nodes, closed_nodes)

  # Nodes bordering current
  for neighbor in self.environment.neighbors[current.position]:
    new_node = Node(neighbor, current, current.depth + 1)
    open_queue.append(new_node)
    open_nodes[new_node.position] = new_node