计算python dicitonaryarray数据结构的非空末页-递归算法？

计算python dicitonary / array数据结构的非空末页-递归算法？

也许这可以指导您正确的方向。

byPath

收集嵌套的字典项。调用之后，您基本上可以将结果列表弄平，并检查是否满足您的条件（例如

elem !=''

或类似条件

not elem

）：

x = #your x as posteddef byPath (tree, path):    try: head, tail = path.split ('.', 1)    except: return tree [path]    if head == 'XX': return [byPath (node, tail) for node in tree]    else: return byPath (tree [head], tail)print (byPath (x, 'top.middle.XX.nested') )print (byPath (x, 'top.last.XX.nested.XX.first') )print (byPath (x, 'top.last.XX.nested.XX.second') )print (byPath (x, 'other') )

编辑 ：这部分实际计数那些不是空字符串的元素：

def count (result):    if isinstance (result, list):        total = 0        positive = 0        for e in result: r = count (e) total += r [1] positive += r [0]        return (positive, total)    else: return (0 if result == '' else 1, 1)a = byPath (x, 'top.middle.XX.nested')b = byPath (x, 'top.last.XX.nested.XX.first')c = byPath (x, 'top.last.XX.nested.XX.second')d = byPath (x, 'other')for x in [a, b, c, d]: print (count (x) )

将所有内容放在一起：

def f (tree, path):    return count (byPath (tree, path) )for path in ['top.middle.XX.nested', 'top.last.XX.nested.XX.first', 'top.last.XX.nested.XX.second', 'other']:    print (path, f (x, path) )