在python中查找树中所有可能的路径

Question

我正在尝试创建一个包含树中所有可能路径的列表。我给出了以下结构（DB的子集）：

text = """
1,Product1,INVOICE_FEE,
3,Product3,INVOICE_FEE,
7,Product7,DEFAULT,
2,Product2,DEFAULT,7
4,Product4,DEFAULT,7
5,Product5,DEFAULT,2
"""

其中列为：ID、产品名称、发票类型、引用到父级ID。 我想创建包含所有可能路径的列表，如示例中所示：

[[Product1],[Product3],[Product7,Product2,Product5],[Product7,Product4]]

我谨此陈辞：

lines = [ l.strip() for l in text.strip().splitlines() ]
hierarchy = [ tuple(l.split(',')) for l in lines ]

parents = defaultdict(list)
for p in hierarchy:
    parents[p[3]].append(p)

为了创建树，我想找到所有路径：

def pathsMet(parents, node=''):
    childNodes = parents.get(node)
    if not childNodes:
        return []
    paths = []
    for ID, productName, invoiceType, parentID in childNodes:
        paths.append([productName] + pathsMet(parents, ID))
    return paths

print(pathsMet(parents))

我得到的结果如下：

[['FeeCashFlow1'], ['FeeCashFlow3'], ['PrincipalCashFlow7', ['AmortisationCashFlow3', ['AmortisationCashFlow2']], ['AmortisationCashFlow4']]]

如何更正代码以获得以下输出：

[['FeeCashFlow1'], ['FeeCashFlow3'], ['PrincipalCashFlow7', 'AmortisationCashFlow3', 'AmortisationCashFlow2'], ['PrincipalCashFlow7','AmortisationCashFlow4']]

Answer 1

为此，您可以先构建数据节点树，然后遍历所有分支以构建路径列表：

text = """
1,Product1,INVOICE_FEE,
3,Product3,INVOICE_FEE,
7,Product7,DEFAULT,
2,Product2,DEFAULT,7
4,Product4,DEFAULT,7
5,Product5,DEFAULT,2
"""

data  = [ line.split(",") for line in text.split("\n") if line.strip() ]
keys  = { k:name for k,name,*_ in data }  # to get names from keys
tree  = { k:{} for k in keys }            # initial tree structure with all keys
root  = tree[""] = dict()                 # tree root
for k,_,_,parent in data: 
    tree[parent].update({k:tree[k]})      # connect children to their parent

nodes = [[k] for k in root]  # cumulative paths of keys
paths = []                   # list of paths by name
while nodes:
    kPath = nodes.pop(0)
    subs  = tree[kPath[-1]]                         # get children
    if subs: nodes.extend(kPath+[k] for k in subs)  # accumulate nodes
    else   : paths.append([keys[k] for k in kPath]) # return path if leaf node

输出：

print(paths)
[['Product1'], ['Product3'], ['Product7', 'Product4'], ['Product7', 'Product2', 'Product5']]

Answer 2

除了将整个列表附加到paths变量而不是列表元素之外，您的代码似乎是正确的

尝试此修改：

def pathsMet(parents, node=''):
    childNodes = parents.get(node)
    if not childNodes:
        return [[]]
    paths = []
    for ID, productName, invoiceType, parentID in childNodes:
        for p in pathsMet(parents, ID):
            paths.append([productName] + p)
    return paths