Group values of a key if another key has got same values that are list type in list of dictionary

I have a list of dictionary,

  data = [{"service":"1","unit":["A","B"]},{"service":"2","unit":["A","B"]},{"service":"1","unit":["C"]}]

I want the output to be:

  output = [{"service":["1","2"],"unit":["A","B"]},{"service":["1"],"unit":["C"]}]

I tried different codes that groups the key values if the value of another key is similar, but as unit are in list, I am unsure how to deal with this.

Tried Code:

 for dd in data:
   d.setdefault(dd['service'], set()).update(dd['unit'])

 newlst = [{'service':k, 'unit':list(v)} for k,v in d.items()]
 print(newlst)

Kindly help.

It is not great when the same place in the data structure is either a scalar or a vector; it would be easier to both construct and use later if service was consistently a list if it has a possibility to be multivalued.

However, to answer your question as asked, you can do this:

lookup = {}
for item in data:
    key = tuple(item["unit"])
    olditem = lookup.get(key)
    if olditem:
        if isinstance(olditem["service"], str):
            olditem["service"] = [olditem["service"]]
        olditem["service"].append(item["service"])
    else:
        lookup[key] = item
output = list(lookup.values())
print(output)
# => [{'service': ['1', '2'], 'unit': ['A', 'B']}, {'service': '1', 'unit': ['C']}]

Things to note: since you are grouping by the same unit, you should use the unit as the lookup key, not service. Furthermore, lists cannot be used as keys in a dictionary, but tuples can.

You need a temporary dictionary to consolidate the common values with the associated 'service'

Something like this:

data = [
    {"service":"1","unit":["A","B"]},
    {"service":"2","unit":["A","B"]},
    {"service":"1","unit":["C"]}
    ]

temp = {}

for d in data:
    temp.setdefault(tuple(d['unit']), []).append(d['service'])

output = [{'service': v, 'unit': list(k)} for k, v in temp.items()]

print(output)

Output:

[{'service': ['1', '2'], 'unit': ['A', 'B']}, {'service': ['1'], 'unit': ['C']}]