Python遍历嵌套字典 [英] Python iterating through nested dictionaries

问题描述

首先，这里是问题和编写的代码:

First of all, here is the question and the code written:

def family_lineage(familytree, lineage):
    '''(dict, list of strs) -> boolean
    Return True if lineage specifies a list of names who are directly related
    in a chain of parent-child relationships, and NOT child-parent, parent-grandchild..etc,
    beginning from the first generation listed.

    >>> trace_lineage({'Gina': {'Sam': {'Tina': {}},
                           'Li': {}},
                   'Guy': {}},
                      ['Gina'])
    True

    >>> trace_lineage({'Gina': {'Sam': {'Tina': {}},
                               'Li': {}},
                       'Guy': {}},
                      ['Gina', 'Sam', 'Tina'])
    True
    >>> trace_lineage({'Gina': {'Sam': {'Tina': {}},
                               'Li': {}},
                       'Guy': {}},
                      ['Gina', 'Tina'])
    False
    '''

因此，在上面的示例中，它显示"Guy"没有孩子，而"Gina"有两个孩子"Sam"和"Li". 山姆"有一个孩子，蒂娜".

So, in the above example, it shows the 'Guy' has no children, and 'Gina' has two children, 'Sam', and 'Li'. 'Sam' has one child, 'Tina'.

for k, v in familytree.items():
    for n, m in v.items():
        if lineage[0] == any(k) and len(lineage) == 1:
            return True
        elif lineage[0] == k and lineage[1] == n and len(lineage) ==2:
            return True
        elif lineage[0] == k and lineage[1] == n and lineage[2] == m and \
        len(lineage) == 3:
            return True
        else:
            return False

所以，我的问题是，如果家族树超过三代，我该怎么写?有没有更简洁的代码编写方式?

So, my question is, how would I write this if the familytree extended beyond three generations? Is there a more concise way of writing this code?

推荐答案

以下是一种迭代方法，即使lineage不在家族树的顶部也可以使用:

Here is an iterative approach that will work even if lineage does not start at the top of the family tree:

def family_lineage(familytree, lineage):
    trees = [familytree]
    while trees:
        tree = trees.pop()
        trees.extend(t for t in tree.values() if t)
        for name in lineage:
            if name not in tree:
                break
            tree = tree[name]
        else:
            return True
    return False

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