python: union keys from multiple dictionary?

Question

I have 5 dictionaries and I want a union of their keys.

alldict =  [dict1, dict2, dict3, dict4, dict5]

I tried

allkey = reduce(lambda x, y: set(x.keys()).union(y.keys()), alldict)

but it gave me an error

AttributeError: 'set' object has no attribute 'keys'

Am I doing it wrong ? I using normal forloop but I wonder why the above code didn't work.

score 71 · Answer 1 · answered Mar 09 '11 at 08:42

71

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.

answered Mar 09 '11 at 08:42

Duncan

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chuck · Accepted Answer · 2011-03-09T07:05:30.380

10

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)))

edited Mar 09 '11 at 07:05

answered Mar 09 '11 at 06:49

chuck

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Almost... try set() as the initializer – itsadok Mar 09 '11 at 06:52
1

Or: `allkey = reduce(set.union, (set(d.keys()) for d in alldict))` – hughdbrown Jul 24 '12 at 03:22

joaquin · Answer 3 · 2014-01-23T10:54:33.597

5

A simple strategy for non-functional neurons (pun intended):

allkey = []

for dictio in alldict:
    for key in dictio:
        allkey.append(key)

allkey = set(allkey)

We can convert this code to a much sorter form using set comprehensions:

allkey = {key for dictio in alldict for key in dictio}

This one-liner is still very readable in comparison with the conventional for loop. The key to convert a nested loop to a list or set comprehension is to write the inner loop (the one that varies faster in the nested loop) as the last index (that is, for key in dictio).

edited Jan 23 '14 at 10:54

answered Mar 09 '11 at 07:36

joaquin

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@BjörnPollex You are right, the (set) comprehension successfully competes with functional alternatives both in readability and conciseness. – joaquin Jan 23 '14 at 11:01

score 2 · Answer 4 · answered Jul 16 '13 at 08:00

2

set().union(dict1.keys(),dict2.keys()...)

I tried the list and it didnt work so just putting it up here for anyone.

answered Jul 16 '13 at 08:00

Leon

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score 1 · Answer 5 · answered Mar 09 '11 at 07:55

1

Just one more way, 'cause what the hay:

a={}; [ a.update(b) for b 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?)

answered Mar 09 '11 at 07:55

Michael Lorton

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Try: `def f(a, b): return dict(a.items() + b.items())` – hughdbrown Jul 24 '12 at 03:24

score 1 · Answer 6 · answered Apr 20 '22 at 15:30

1

If you only want to union keys of 2 dicts you could use operator |.

Quote from docs:

Return a new set with elements from the set and all others.

Example:

all_keys = (dict1.keys() | dict2.keys())

answered Apr 20 '22 at 15:30

Alex Kosh

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python: union keys from multiple dictionary?

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