获取分支数据/图数据

Question

假设我有以下数据，我想在“免疫缺陷”的保护伞下找到所有的东西。

startNode - relationship - endNode
immunodeficiencies - includes - B cell disorders
B cell disorders - includes - Bruton Agammaglobulinemia
B cell disorders - includes - Hyper-IgM syndrome
immunodeficiencies - includes - T cell disorders
T cell disorders - includes - DiGeorge Syndrome
immunodeficiencies - includes - combined B and T cell disorders
combined B and T cell disorders - includes - Ataxia-Telangiectasia

我该怎么做？数据是JSON格式的，所以它就像这个{"startNode": "immunodeficiencies", "relationship": "includes", "endNode": "B cell disorders"}

对于一个非嵌套的类别来说，这很容易。我做了以下工作：

for x in range(len(peds_diseaseR))
if peds_diseaseR[x]["startNode"] == "immunodeficiencies" && peds_diseaseR[x]["relationship"] == "includes":
    List.append(x)

对于一种嵌套的疾病，我真的迷路了。我觉得我可能需要一个递归函数，但我不知道。我本来打算把我写的代码打印出来，但我觉得它比任何东西都更让人困惑，而且不起作用。

编辑:刚添加了这个，但是它不起作用。

var endNodes = []

func findEndNodes(start):
    for x in range(len(peds_diseaseR)):
        if peds_diseaseR[x]["startNode"] == start && peds_diseaseR[x]["relationship"] == "includes":
            endNodes.append(peds_diseaseR[x]["endNode"])
        if endNodes != []:
            for disease in endNodes:
                findEndNodes(disease)
    print(endNodes)
    return endNodes

Answer 1

你们很亲密。只要稍微重新安排一下，你就会找到解决办法：

peds_diseaseR = [] 
def findEndNodes(start):
    
    # declared inside the function, because you need a fresh one for every recursion
    childnodes = [] 
    for relationship in peds_diseaseR:
        if relationship["startNode"] == start && relationship["relationship"] == "includes":
            disease = relationship["endNode"]
            child_nodes.append(disease) 
            # If you process it now, you do not need to check if you have values in your list. 
            # Thus, less code, but more importantly, less states your code can be in.
            # It is very easy to get lost trying to make a mental map when doing recursion. 
            # Keep it as simple as possible! 
            recursion_nodes = findEndNodes(disease)
            child_nodes.extend(recursion_nodes)
    print(child_nodes)
    return child_nodes

如您所见，我只将func替换为def，并删除了var关键字。在python中是不存在的。您只需分配一个值。附加和扩展之间有细微但微妙的区别。extend将append列表中的所有值。