CODEWITHSUNDEEP
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Lopsdie
Lopsdie

Reputation: 11

Python merge multiple dicts with common keys, generating new keys for the respective source dict

Is there an elegant way in python to go from

d1 = {"key1": ["a", "b"], "key2": ["c", "d"]}  
d2 = {"key1": ["e", "f"], ...}

to:

dd =
[
   { "old_key": "key1",
     "d1_key": ["a", "b"],
     "d2_key": ["e", "f"]
   },
   { "old_key": "key2",
     "d1_key": ["c", "d"]
   }
]

I'm not fussed if there is a d2_key: "" if there exists no old_key in d2

EDITED: from pseudo code

Upvotes: 0

Views: 37

Answers (2)

Lopsdie
Lopsdie

Reputation: 11

Courtesy of a very helpful colleague I think have a solution I think solves it. The union operation is something I have never used/considered before.

all_keys = (d1.keys() | d2.keys())
dd = []
for key in all_keys:
    dd.append({"old_key": key,
               "d1_key": d1.get(key, ""),
               "d2_key": d2.get(key, "")})

Upvotes: 1

Stragas
Stragas

Reputation: 36

You can do something like this:

dd = list()
for k, v in d1.items():
    temp_dict = dict()
    temp_dict["old_key"] = k
    temp_dict["d1_"+k] = v
    if k in d2.keys():
        temp_dict["d2_"+k] = d2[k]
    dd.append(temp_dict)

Beware that this will ignore all the keys that are in d2 but not in d1 for if they are neccesary you can also add this for loop afterwards:

for k, v in d2.items():
    temp_dict = dict()
    if k not in d1.keys():
        temp_dict["d2_" + k] = d2[k]
    dd.append(temp_dict)

Upvotes: 0

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