如何通过使用唯一值安全地合并 Python 中的嵌套列表？

Question

I have a nested list (list of list) and I want to merge if id from nested list is duplicates: Like :

[
  {'id': 2404, 'interfaces': [{'port': 78, 'module': 1 }]}
  {'id': 2404, 'interfaces': [{'port': 79, 'module': 1 }]} 
  {'id': 1234, 'interfaces': [{'port': 79, 'module': 1 }]} 
]

So at the final solution should be

[
  {'id': 2404, 'interfaces': [{'port': 78, 'module': 1 },{'port': 79, 'module': 1 } ]}
  {'id': 1234, 'interfaces': [{'port': 79, 'module': 1 }]} 
]

Answer 1

I defined your original nested list as the unmerged variable.
The merged variable will contain the output

unmerged = 
[
  {'id': 2404, 'interfaces': [{'port': 78, 'module': 1 }]},
  {'id': 2404, 'interfaces': [{'port': 79, 'module': 1 }]},
  {'id': 1234, 'interfaces': [{'port': 79, 'module': 1 }]},
]
merged = []

for unmerged_item in unmerged:
    match = next((item for item in merged if item['id'] == unmerged_item['id']), None)
    
    if match:
        match['interfaces'].extend(unmerged_item['interfaces'])
    else:
        merged.append(unmerged_item)

The output of the code will be as follows ( merged ):

[
    {'id': 2404, 'interfaces': [{'port': 78, 'module': 1}, {'port': 79, 'module': 1}]}, 
    {'id': 1234, 'interfaces': [{'port': 79, 'module': 1}]}
]

Answer 2

Use a dictionary to accumulate the interfaces with the same id:

data = [
  {'id': 2404, 'interfaces': [{'port': 78, 'module': 1 }]},
  {'id': 2404, 'interfaces': [{'port': 79, 'module': 1 }]},
  {'id': 1234, 'interfaces': [{'port': 79, 'module': 1 }]}
]

lookup = {}
for d in data:
    iid = d["id"]
    if iid not in lookup:
        lookup[iid] = []
    lookup[iid].extend(d["interfaces"])

res = [{ "id" : iid, "interfaces" : interfaces } for iid, interfaces in lookup.items()]
print(res)

Output

[{'id': 2404, 'interfaces': [{'port': 78, 'module': 1}, {'port': 79, 'module': 1}]}, {'id': 1234, 'interfaces': [{'port': 79, 'module': 1}]}]

Alternative solution using collections.defaultdict :

from collections import defaultdict
lookup = defaultdict(list)
for d in data:
    iid = d["id"]
    lookup[iid].extend(d["interfaces"])

res = [{ "id" : iid, "interfaces" : interfaces } for iid, interfaces in lookup.items()]

Special Case (the groups are contiguous)

If, and only if, the dictionaries id are contiguous, you could use itertools.groupby , as below:

from itertools import groupby, chain
from operator import itemgetter

res = []
for iid, vs in groupby(data, key=itemgetter("id")):
    interfaces = chain.from_iterable(v["interfaces"] for v in vs)
    res.append({"id": iid, "interfaces" : list(interfaces) })

print(res)

Output

[{'id': 2404, 'interfaces': [{'port': 78, 'module': 1}, {'port': 79, 'module': 1}]}, {'id': 1234, 'interfaces': [{'port': 79, 'module': 1}]}]

If the groups are not contiguous you could sort your data, but that will make the approach less efficient (O(nlogn) due to the sorting) that the dictionary alternatives O(n).