Python:父子层次结构的组合 [英] Python: Combinations of parent-child hierarchy

问题描述

对于子父关系表 (csv)，我正在尝试使用表中的所有数据收集可能的父子关系组合链.我正在尝试解决一个问题，即如果存在多个子父项(参见第 3 行和第 4 行)，则迭代中不包含第二个子父项组合(第 4 行).

For a child-parent relationship table (csv), I am trying to gather possible parent to child relationship combination chains using all data in the table. I am trying against a problem where if multiple sub-parents exist (see rows 3 & 4), the second sub-parent combination (row 4) is not included in the iteration.

数据示例:

孩子，父母

A,B
A,C
B,D
B,C
C,D

预期的连锁结果:

D|B|A
D|C|B|A
D|C|A

实际链结结果:

D|B|A
D|C|A

代码

find= 'A' #The child for which the code should find all possible parent relationships
sequence = ''
with open('testing.csv','r') as f:     #testing.csv = child,parent table (above example)
    for row in f:
        if row.strip().startswith(find):
            parent = row.strip().split(',')[1]
            sequence = parent + '|' + find
            f1 = open('testing.csv','r')
            for row in f1:
                if row.strip().startswith(parent):
                    parent2 = row.strip().split(',')[1]
                    sequence = parent2 + '|' + sequence
                    parent = parent2
        else:
            continue
        print sequence

推荐答案

你看过 this 精彩的文章?真正理解python中的模式是必不可少的阅读.您的问题可以被认为是图问题 - 查找关系基本上是查找从子节点到父节点的所有路径.

Have you looked at this fantastic essay? It is essential reading to really understand patterns in python. Your problem can be thought of as a graph problem - finding the relationships is basically finding all paths from a child node to the parent node.

由于可能存在任意数量的嵌套(child->parent1->parent2...)，因此您需要一个递归解决方案来查找所有路径.在您的代码中，您有 2 个 for 循环 - 正如您发现的那样，最多只会产生 3 级路径.

Since there could be an arbitrary amount of nesting (child->parent1->parent2...), you need a recursive solution to find all paths. In your code, you have 2 for loops - which will only result in 3level paths at most as you found out.

下面的代码改编自上面的链接，以解决您的问题.find_all_paths 函数需要一个图形作为输入.

The code below was adapted from the link above to fix your issue. The function find_all_paths requires a graph as an input.

让我们根据您的文件创建图表:

Let's create the graph from your file:

graph = {} # Graph is a dictionary to hold our child-parent relationships.
with open('testing.csv','r') as f:
    for row in f:
        child, parent = row.split(',')
        graph.setdefault(parent, []).append(child)

print graph

使用您的示例，应打印:

with your sample, this should print:

{'C': ['A', 'B'], 'B': ['A'], 'D': ['B', 'C']}

以下代码直接来自论文:

The following code is straight from the essay:

def find_all_paths(graph, start, end, path=[]):
    path = path + [start]
    if start == end:
        return [path]

    if not graph.has_key(start):
        return []

    paths = []

    for node in graph[start]:
        if node not in path:
            newpaths = find_all_paths(graph, node, end, path)
            for newpath in newpaths:
                paths.append(newpath)
    return paths

for path in find_all_paths(graph, 'D', 'A'):
    print '|'.join(path)

输出:

D|B|A
D|C|A
D|C|B|A

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