CODEWITHSUNDEEP
pythondictionarydata-structures
itsMe
itsMe

Reputation: 785

python data structure: list of dict to one dict

I have a data structure. It looks as follows:

data = [[{'id': '1', 'name': 'parent', 'steps': 0},
 {'id': '2A', 'name': 'child', 'steps': 1},
 {'id': '3A', 'name': 'grandChild-A', 'steps': 2}],

[{'id': '1', 'name': 'parent', 'steps': 0},
 {'id': '2A', 'name': 'child', 'steps': 1},
 {'id': '3B', 'name': 'grandChild-B', 'steps': 2}],

[{'id': '1', 'name': 'parent', 'steps': 0},
 {'id': '2B', 'name': 'child', 'steps': 1},
 {'id': '3A', 'name': 'grandChild-C', 'steps': 2}],

[{'id': '1', 'name': 'parent', 'steps': 0},
 {'id': '2B', 'name': 'child', 'steps': 1},
 {'id': '3B', 'name': 'grandChild-D', 'steps': 2}],

[{'id': '1', 'name': 'parent', 'steps': 0},
 {'id': '2B', 'name': 'child', 'steps': 1},
 {'id': '3C', 'name': 'grandChild-E', 'steps': 2},
 {'id': '4A', 'name': 'final', 'steps': 3}
 ],

[{'id': '1', 'name': 'parent', 'steps': 0},
 {'id': '2', 'name': 'child', 'steps': 1},
 ]
]

How my expected output is

expected output

output = {
    "1" : {
        "2A": {
            "3A": "grandChild-A",
            "3B": "grandChild-B"

        },
        "2B": {
            "3A": "grandChild-C",
            "3B": "grandChild-D",
            "3C": {
                "4A": "final"
            }

        },
        "2":"child"

    }
}

How can I do that? I wanted to use the enumerator, But I always everything inside 1. Thanks in advance

Update:

I have tried the following code:

parent = data[0][0]["id"]
dict_new = {}
dict_new[parent] = {}

for e in data:
    for idx, item in enumerate(e):
        display(item)
        if idx>0:
            dict_new[parent][e[idx]["id"]] = e[idx]["name"]

Upvotes: 1

Views: 83

Answers (2)

jferard
jferard

Reputation: 8180

You can try:

d = {}
root = d
for L in data:
    d = root
    for M in L[:-1]:
        d = d.setdefault(M["id"], {})
    d[L[-1]["id"]] = L[-1]['name']

The idea is to follow each list to build a tree (thus d.setdefault(M["id"], {}). The leaf is handled differently, because it has to be the value of 'name'.

from pprint import pprint
pprint(root)

Output:

{'1': {'2': 'child',
       '2A': {'3A': 'grandChild-A', '3B': 'grandChild-B'},
       '2B': {'3A': 'grandChild-C',
              '3B': 'grandChild-D',
              '3C': {'4A': 'final'}}}}

The solution above won't work for the following input:

data = [[{'id': '1', 'name': 'parent', 'steps': 0},
 {'id': '2B', 'name': 'child', 'steps': 1}]],

[{'id': '1', 'name': 'parent', 'steps': 0},
 {'id': '2B', 'name': 'child', 'steps': 1},
 {'id': '3A', 'name': 'grandChild-C', 'steps': 2}]]

Iterating over the second list will try to add a new element 3A -> grandChild-C to the d['1']['2B'] dict. But d['1']['2B'] is not a dict but the 'child' string here, because of the first list.

When we iterate over the elements, we check if the key is already mapped and otherwise create a new dict (that's the setdefault job). We can also check if the key was mapped to a str, and if that's the case, replace the string by a fresh new dict:

...
for M in L[:-1]:
    if M["id"] not in d or isinstance(d[M["id"]], str):
        d[M["id"]] = {}
    d = d[M["id"]]
...

Output:

{'1': {'2B': {'3A': 'grandChild-C'}}}

Upvotes: 2

quamrana
quamrana

Reputation: 39404

I fixed your data: (missing comma)

data = [[{'id': '1', 'name': 'parent', 'steps': 0},
 {'id': '2A', 'name': 'child', 'steps': 1},
 {'id': '3A', 'name': 'grandChild-A', 'steps': 2}],

[{'id': '1', 'name': 'parent', 'steps': 0},
 {'id': '2A', 'name': 'child', 'steps': 1},
 {'id': '3B', 'name': 'grandChild-B', 'steps': 2}],

[{'id': '1', 'name': 'parent', 'steps': 0},
 {'id': '2B', 'name': 'child', 'steps': 1},
 {'id': '3A', 'name': 'grandChild-C', 'steps': 2}],

[{'id': '1', 'name': 'parent', 'steps': 0},
 {'id': '2B', 'name': 'child', 'steps': 1},
 {'id': '3B', 'name': 'grandChild-D', 'steps': 2}],

[{'id': '1', 'name': 'parent', 'steps': 0},
 {'id': '2B', 'name': 'child', 'steps': 1},
 {'id': '3C', 'name': 'grandChild-E', 'steps': 2},
 {'id': '4A', 'name': 'final', 'steps': 3}
 ],

[{'id': '1', 'name': 'parent', 'steps': 0},
 {'id': '2', 'name': 'child', 'steps': 1},
 ]
]

And I came up with this code:

output = {}
#print(data)

for lis in data:
    o = output
    ln = len(lis) - 1
    for idx,d in enumerate(lis):
        id = d['id']
        if idx == ln:
            o[id] = d['name']
        else:
            if id not in o:
                o[id] = {}
            o = o[id]

print('Result:')
print(output)

Upvotes: 1

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