Python:将元组列表转换为嵌套字典的字典

Question

所以我手头有点问题。我有一个元组列表(由级别编号和消息组成)最终将成为 HTML 列表。我的问题是，在发生这种情况之前，我想将元组值转换为嵌套字典的字典。所以这里是示例:

# I have this list of tuples in format of (level_number, message)
tuple_list = [(1, 'line 1'), (2, 'line 2'), (3, 'line 3'), (1, 'line 4')]

# And I want to turn it into this
a_dict = {
    'line 1': {
        'line 2': {
            'line 3': {}
        }
    }, 
    'line 4': {}
}

任何帮助将不胜感激，只要它是有效的 Python 3。谢谢!

Answer 1

正如我在评论中指出的那样，如果您完全可以控制它，您应该强烈考虑更改您的传入数据结构。一个连续的元组列表绝对不适合你在这里做的事情。然而，如果你把它当作一棵树来对待，这是可能的。让我们构建一个(理智的)数据结构来解析它

class Node(object):
    def __init__(self, name, level, parent=None):
        self.children = []
        self.name = name
        self.level = level
        self.parent = parent

    def make_child(self, othername, otherlevel):
        other = self.__class__(othername, otherlevel, self)
        self.children.append(other)
        return other

现在您应该能够以某种合理的方式迭代您的数据结构

def make_nodes(tuple_list):
    """Builds an ordered grouping of Nodes out of a list of tuples
    of the form (level, name). Returns the last Node.
    """

    curnode = Node("root", level=-float('inf'))
    # base Node who should always be first.

    for level, name in tuple_list:
        while curnode.level >= level:
            curnode = curnode.parent
            # if we've done anything but gone up levels, go
            # back up the tree to the first parent who can own this
        curnode = curnode.make_child(name, level)
        # then make the node and move the cursor to it
    return curnode

一旦您的结构完成，您就可以对其进行迭代。如果您采用深度优先或广度优先，这里并不重要，所以让我们做一个 DFS 只是为了便于实现。

def parse_tree(any_node):
    """Given any node in a singly-rooted tree, returns a dictionary
    of the form requested in the question
    """

    def _parse_subtree(basenode):
        """Actually does the parsing, starting with the node given
        as its root.
        """

        if not basenode.children:
            # base case, if there are no children then return an empty dict
            return {}
        subresult = {}
        for child in basenode.children:
            subresult.update({child.name: _parse_subtree(child)})
        return subresult

    cursor = any_node
    while cursor.parent:
        cursor = cursor.parent
        # finds the root node
    result = {}
    for child in cursor.children:
        result[child.name] = _parse_subtree(child)
    return result

然后输入元组列表瞧

tuple_list = [(1, 'line 1'), (2, 'line 2'), (3, 'line 3'), (1, 'line 4')]

last_node = make_nodes(tuple_list)
result = parse_tree(last_node)
# {'line 1': {'line 2': {'line 3': {}}}, 'line 4': {}}