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Accumulate JSON Values in Python Dictionary as Array

I have JSON File that has this format

{ 
"links": [
{"source":"0","target":"1","weight":1,"color":"white"},
{"source":"0","target":"2","weight":1,"color":"yellow"},
{"source":"0","target":"3","weight":1,"color":"white"},
]
}

I want to collect all target for a single source like this:

{"source": 0, "neighbors": ["1","2","3"]} where neighbors are all the collected target

Here's my Code

import json

with open("linksGr.json") as file:
    data = json.load(file)

collectDict = {}
for obj in data["links"]:
    if (collectDict["source"] == obj["source"]):
        collectDict["neighbour"] = obj["target"]

I just need a way to accumulate all targets for each source instead of there being multiple sources as I have done here

collectDict["source"] = obj["source"]
collectDict["neighbour"] = obj["target"]

Any help would be appreciated a lot. I am sure there is some basic concept and a simple way that I am missing here. Thanks for the help.

like image 305
obadul024 Avatar asked May 16 '26 18:05

obadul024


2 Answers

If I understand you correctly, you can use collections.defaultdict, to map from source to a list of targets, like this:

(I added some data to have multiple sources)

from collections import defaultdict

data = { 
"links": [
{"source":"0","target":"1","weight":1,"color":"white"},
{"source":"0","target":"2","weight":1,"color":"yellow"},
{"source":"0","target":"3","weight":1,"color":"white"},
{"source":"5","target":"7","weight":1,"color":"white"},
{"source":"5","target":"8","weight":1,"color":"yellow"},
{"source":"6","target":"9","weight":1,"color":"white"},
]
}

collectDict = defaultdict(list)
for obj in data["links"]:
    collectDict[obj["source"]].append(obj["target"])

print(dict(collectDict))

Output:

{'0': ['1', '2', '3'], '5': ['7', '8'], '6': ['9']}

EDIT: Here's another method using itertools.groupby, assuming the links are ordered by sources (otherwise, just sort it before)

from itertools import groupby

collectDict = {k: [t["target"] for t in g] for k,g in groupby(data["links"], lambda obj: obj["source"])}

print(collectDict)
like image 168
Adam.Er8 Avatar answered May 18 '26 07:05

Adam.Er8


data = { 
    "links": [
        {"source":"0","target":"1","weight":1,"color":"white"},
        {"source":"0","target":"2","weight":1,"color":"yellow"},
        {"source":"0","target":"3","weight":1,"color":"white"},
        {"source":"5","target":"7","weight":1,"color":"white"},
        {"source":"5","target":"8","weight":1,"color":"yellow"},
        {"source":"6","target":"9","weight":1,"color":"white"},
    ]
}

collected = []
for obj in data["links"]:
    source_matches = [item for item in collected if item["source"] == obj["source"]]
    if len(source_matches) == 0:
        source_match = {"source": obj["source"], "neighbour": [obj["target"]]}
        collected.append(source_match)
    elif len(source_matches) == 1:
        source_matches[0]["neighbour"].append(obj["target"])
    else:
        raise BaseException()

print(collected)  # [{'source': '0', 'neighbour': ['1', '2', '3']}, {'source': '5', 'neighbour': ['7', '8']}, {'source': '6', 'neighbour': ['9']}]

Not very elegant, but does the job.

If you don't really need the format {"source": 0, "neighbors": ["1","2","3"]} I recommend the above solution using defaultdict. If you need this format, you could also build it from the output of the defaultdict solution.

like image 26
Michael H. Avatar answered May 18 '26 07:05

Michael H.



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