[英]Top level length of nested dictionaries of lists with one function
我有一个包含列表的嵌套字典,需要获取字典中每个顶级值中所有列表的长度总和。 示例列表如下:
data = {0: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10],
1: {0: [11, 12],
1: {0: [13, 14],
1: [15, 16],
2: [17, 18, 19, 20],
3: {0: [43, 34],
1: [4],
2: [2, 3, 4, 5]}}}}
我相信我有一个使用以下两个功能的工作解决方案:
def TreeSum(tree):
return [len(v) if isinstance(v, list) else SubTreeSum(v) for v in tree.values()]
def SubTreeSum(tree):
return sum([len(v) if isinstance(v, list) else SubTreeSum(v) for v in tree.values()])
TreeSum(data)
的 output 是[10, 17]
而TreeSum(data[1][1])
是[2, 2, 4, 7]
。
我曾尝试将这些功能组合成一个 function,主要是出于美观原因,但我还没有找到解决方案。 有人可以建议解决该问题的方法吗?
您只需要:
def TreeSum(tree):
return [len(v) if isinstance(v, list) else sum(TreeSum(v)) for v in tree.values()]
您可以区分运行顶级 function 和递归调用的情况:
def TreeSum(tree, agg=False):
l = [len(v) if isinstance(v, list) else TreeSum(v, True) for v in tree.values()]
return sum(l) if agg else l
output:
>>> TreeSum(data)
[10, 17]
一个不传递行为改变标志的简单解决方案是只创建SubTreeSum
并嵌套 function 的TreeSum
。 瞧,单身 function:
def TreeSum(tree):
def SubTreeSum(tree):
return sum([len(v) if isinstance(v, list) else SubTreeSum(v) for v in tree.values()])
return [len(v) if isinstance(v, list) else SubTreeSum(v) for v in tree.values()]
也许更漂亮的方法是将通用功能剥离到另一个嵌套的 function 中以避免代码重复:
def TreeSum(tree):
def SubTreeList(tree):
return [len(v) if isinstance(v, list) else SubTreeSum(v) for v in tree.values()]
def SubTreeSum(tree):
return sum(SubTreeList(tree))
return SubTreeList(tree)
您可以通过使用字典键或列表值作为迭代驱动程序并计算 1(对于列表项)或长度列表的递归总和(对于字典)来在单个递归 function 中执行此操作:
def topCounts(C):
return [1 if type(C) is list else sum(topCounts(C[k])) for k in C]
Output:
data = {0: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10],
1: {0: [11, 12],
1: {0: [13, 14],
1: [15, 16],
2: [17, 18, 19, 20],
3: {0: [43, 34], 1: [4], 2: [2, 3, 4, 5]}}}}
print(topCounts(data))
print(topCounts(data[1][1]))
[10, 17]
[2, 2, 4, 7]
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