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Asked: May 10, 20262026-05-10T17:24:05+00:00 2026-05-10T17:24:05+00:00

He’re an interesting problem that looks for the most Pythonic solution. Suppose I have

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He’re an interesting problem that looks for the most Pythonic solution. Suppose I have a list of mappings {'id': id, 'url': url}. Some ids in the list are duplicate, and I want to create a new list, with all the duplicates removed. I came up with the following function:

def unique_mapping(map):     d = {}     for res in map:         d[res['id']] = res['url']      return [{'id': id, 'url': d[id]} for id in d] 

I suppose it’s quite efficient. But is there a ‘more Pythonic’ way ? Or perhaps a more efficient way ?

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  1. 2026-05-10T17:24:05+00:00Added an answer on May 10, 2026 at 5:24 pm

    Your example can be rewritten slightly to construct the first dictionary using a generator expression and to remove necessity of construction of another mappings. Just reuse the old ones:

    def unique_mapping(mappings):     return dict((m['id'], m) for m in mappings).values() 

    Although this came out as a one-liner, I still think it’s quite readable.

    There are two things you have to keep in mind when using your original solution and mine:

    • the items will not always be returned in the same order they were originally
    • the later entry will overwrite previous entries with the same id

    If you don’t mind, then I suggest the solution above. In other case, this function preserves order and treats first-encountered ids with priority:

    def unique_mapping(mappings):     addedIds = set()     for m in mappings:         mId = m['id']         if mId not in addedIds:             addedIds.add(mId)             yield m 

    You might need to call it with list(unique_mappings(mappings)) if you need a list and not a generator.

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