Python：父子层次的组合 [英] Python: Combinations of parent-child hierarchy
问题描述
对于父子关系表（csv），我试图使用表中的所有数据收集可能的父子关系组合链。如果存在多个子父项（请参阅第3和4行），则第二个子父组合（第4行）不包含在迭代中。 b
数据示例：
小孩，父母
$ b $ C
$ b $ C $ b $ C $ b $ C $ b $ C $ b $ C D
预期的连锁结果：
$ b
D | B | A
D | C | B | A
D | C | A
实际连锁结果： 代码 您是否看过这篇精彩的文章？真正理解Python中的模式是必不可少的阅读。你的问题可以被认为是一个图形问题 - 找到关系基本上是找到从一个子节点到父节点的所有路径。 由于可以有任意的数量嵌套（child-> parent1-> parent2 ...），您需要一个递归解决方案来查找所有路径。在你的代码中，对于循环，你有2
< pre class =lang-none prettyprint-override> D | B | A
D | C | A
$ b
find ='A'#The子代码应该找到所有可能的父关系
sequence =''
with open（'testing.csv'，'r'）as f：＃testing.csv = child，parent table（above例如）
在f：
中的行如果row.strip（）。startswith（find）：
parent = row.strip（）。split（'，'）[1]
sequence = parent +'|'+ find
f1 = open（'testing.csv'，'r'）
用于f1中的行：
如果row.strip（）。startswith（父）：
parent2 = row.strip（） .blit（'，'）[1]
sequence = parent2 +'|'+ sequence
parent = parent2
else：
continue
print sequence
- 这只会在你发现的时候导致3级路径。
<下面的代码是根据上面的链接修改你的问题。函数 find_all_paths 需要一个图形作为输入。
让我们从您的文件中创建图表：
graph = { }＃图是一个字典来保存我们的孩子 - 父母关系。
with open（'testing.csv'，'r'）as f：
for row in f：
child，parent = row.split（'，'）
graph .setdefault（parent，[]）。append（child）
print graph
<与你的样品，这应该打印：
pre $ {c'：['A'，'B']，'' B'：['A']，'D'：['B'，'C']}
<下面的代码是直接的：
$ p $ def $ find_all_paths（graph，start，end，path = []）：
path = path + [start]
if start == end：
return [path]
如果不是graph.has_key（start）：
返回[]
路径= []
对于图中的节点[开始]：
如果节点不在路径中：
newpaths = find_all_paths（graph ，node，end，path）
用于newpaths中的newpath：
paths.append（newpath）
返回路径
用于find_all_paths中的路径（graph ，'D'，'A'）：
print'|'.join（path）
< h3>输出：
D | B | A
D | C | A
D | C | B | A
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.
Data Example:
child,parent
A,B
A,C
B,D
B,C
C,D
Expected chain results:
D|B|A
D|C|B|A
D|C|A
Actual chain results:
D|B|A
D|C|A
Code
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
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.
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.
The code below was adapted from the link above to fix your issue. The function find_all_paths requires a graph as an input.
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
with your sample, this should print:
{'C': ['A', 'B'], 'B': ['A'], 'D': ['B', 'C']}
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)
Output:
D|B|A
D|C|A
D|C|B|A
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