Reputation: 3
I am given a list of dictionaries like this:
[
{'A': ['B', 'C', 'D']},
{'B': ['E', 'F']},
{'C': ['E']},
{'F': ['G', 'H']}
]
A key in the dictionary is a parent of a corresponding dictionary value, i.e., {parent: [child1, child2, child3]}
How can I construct a tree-like dictionary in the following format like the following:
{'A':{'B':{'E': None,
'F': {'G': None,
'H': None}
},
'C': {'E': None}}
If a node doesn't have a child, we will fill its value with None. I don't know how to write a recursive process to transform the list into a dictionary, any idea?
Thanks!
Upvotes: 0
Views: 664
Reputation: 71451
You can use recursion:
d = [{'A': ['B', 'C', 'D']}, {'B': ['E', 'F']}, {'C': ['E']}, {'F': ['G', 'H']}]
d1 = dict(list(i.items())[0] for i in d)
def flatten_structure(d, root = 'A'):
if root:
return {root:flatten_structure(d1[root], None)}
return {i:flatten_structure(d1[i], None) if i in d1 else None for i in d}
print(flatten_structure(d1['A']))
Output:
{'A': {'B': {'E': None, 'F': {'G': None, 'H': None}}, 'C': {'E': None}, 'D': None}}
Upvotes: 0
Reputation: 11073
You can do this with a recursive function:
def find_set(d, k, v):
for key, value in d.items():
if isinstance(value, dict):
find_set(d[key], k, v)
return
if key == k:
d[key] = {}
for i in v:
d[key][i] = None
return d
d[k] = {}
for i in v:
d[k][i] = None
return d
l = [{'A': ['B', 'C', 'D']}, {'B': ['E', 'F']}, {'C': ['E']}, {'F': ['G', 'H']}]
d = {}
for node in l:
for key, value in node.items():
find_set(d, key, value)
d
will be :
{'A': {'B': {'E': None, 'F': {'G': None, 'H': None}, 'C': {'E': None}}, 'C': None, 'D': None}}
Upvotes: 3