Python: union keys from multiple dictionary?

Your solution works for the first two elements in the list, but then dict1 and dict2 got reduced into a set and that set is put into your lambda as the x So now x does not have the method keys() anymore The solution is to make x be a set from the very beginning by initializing the reduction with an empty set (which happens to be the neutral element of the union) Try it with an initializer: allkey = reduce(lambda x, y: x. Union(y.keys()), alldict, set()) An alternative without any lambdas would be: allkey = reduce(set. Union, map(set, map(dict.

Keys, alldict))).

Your solution works for the first two elements in the list, but then dict1 and dict2 got reduced into a set and that set is put into your lambda as the x. So now x does not have the method keys() anymore. The solution is to make x be a set from the very beginning by initializing the reduction with an empty set (which happens to be the neutral element of the union).

Try it with an initializer: allkey = reduce(lambda x, y: x. Union(y.keys()), alldict, set()) An alternative without any lambdas would be: allkey = reduce(set. Union, map(set, map(dict.

Keys, alldict))).

Almost... try set() as the initializer – itsadok Mar 9 at 6:52.

I think @chuck already answered the question why it doesn't work, but a simpler way to do this would be to remember that the union method can take multiple arguments: allkey = set(). Union(*alldict) does what you want without any loops or lambdas.

Good call. Have an upvote, Sir! – chuck Mar 9 at 10:06.

A simple strategy for non-functional neurons (pun intended): allkey = for dictio in alldict: for key in dictio: allkey. Append(key) allkey = set(allkey).

If this were a list comprehension, i'd upvote it. – Björn Pollex Mar 9 at 8:01.

Just one more way, 'cause what the hay: a={}; a. Update(b) for be in alldict and a.keys() or the slightly-more-mysterious reduce(lambda a, b: a. Update(b) or a, alldict, {}).keys() (I'm bummed that there's no built-in function equivalent to def f(a,b): r = {} r.

Update(a) r. Update(b) return r is there? ).

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