Unpacking a dictionary's tuple keys into individual keys using dictionary comprehension in Python
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Assuming a dictionary where the keys are tuples:
D = {
('a','b') : 1,
('x','y','z') : 2
}
How can I split each tuple-key into separate keys with the same value:
N = {
'a' : 1, 'b' : 1,
'x' : 2, 'y' : 2, 'z' : 2
}
In a single dictionary comprehension. I've drafted up the following line, but I'm wondering if it's possible to shorten it and not create a proxy value list of the same length as the key tuple.
N = { k:v for s in ( zip(keys,[value]*len(keys)) for keys,value in D.items() ) for k,v in s }
python dictionary key
asked Nov 16 '18 at 19:45
dexgeckodexgecko
1,4291817
add a comment |
Assuming a dictionary where the keys are tuples:
D = {
('a','b') : 1,
('x','y','z') : 2
}
How can I split each tuple-key into separate keys with the same value:
N = {
'a' : 1, 'b' : 1,
'x' : 2, 'y' : 2, 'z' : 2
}
In a single dictionary comprehension. I've drafted up the following line, but I'm wondering if it's possible to shorten it and not create a proxy value list of the same length as the key tuple.
N = { k:v for s in ( zip(keys,[value]*len(keys)) for keys,value in D.items() ) for k,v in s }
python dictionary key
asked Nov 16 '18 at 19:45
dexgeckodexgecko
1,4291817
(Note that for this question to be well defined the tuple-keys must be pairwise disjoint.)
– timgeb
Nov 16 '18 at 19:55
add a comment |
Assuming a dictionary where the keys are tuples:
D = {
('a','b') : 1,
('x','y','z') : 2
}
How can I split each tuple-key into separate keys with the same value:
N = {
'a' : 1, 'b' : 1,
'x' : 2, 'y' : 2, 'z' : 2
}
In a single dictionary comprehension. I've drafted up the following line, but I'm wondering if it's possible to shorten it and not create a proxy value list of the same length as the key tuple.
N = { k:v for s in ( zip(keys,[value]*len(keys)) for keys,value in D.items() ) for k,v in s }
python dictionary key
asked Nov 16 '18 at 19:45
dexgeckodexgecko
1,4291817
Assuming a dictionary where the keys are tuples:
D = {
('a','b') : 1,
('x','y','z') : 2
}
How can I split each tuple-key into separate keys with the same value:
N = {
'a' : 1, 'b' : 1,
'x' : 2, 'y' : 2, 'z' : 2
}
In a single dictionary comprehension. I've drafted up the following line, but I'm wondering if it's possible to shorten it and not create a proxy value list of the same length as the key tuple.
N = { k:v for s in ( zip(keys,[value]*len(keys)) for keys,value in D.items() ) for k,v in s }
python dictionary key
python dictionary key
asked Nov 16 '18 at 19:45
dexgeckodexgecko
1,4291817
asked Nov 16 '18 at 19:45
dexgeckodexgecko
1,4291817
edited Nov 16 '18 at 19:52
dexgecko
asked Nov 16 '18 at 19:45
dexgeckodexgecko
1,4291817
asked Nov 16 '18 at 19:45
dexgeckodexgecko
1,4291817
asked Nov 16 '18 at 19:45
dexgeckodexgecko
1,4291817
1,4291817
(Note that for this question to be well defined the tuple-keys must be pairwise disjoint.)
– timgeb
Nov 16 '18 at 19:55
add a comment |
(Note that for this question to be well defined the tuple-keys must be pairwise disjoint.)
– timgeb
Nov 16 '18 at 19:55
(Note that for this question to be well defined the tuple-keys must be pairwise disjoint.)
– timgeb
Nov 16 '18 at 19:55
(Note that for this question to be well defined the tuple-keys must be pairwise disjoint.)
– timgeb
Nov 16 '18 at 19:55
add a comment |
2 Answers
2
active
oldest
votes
The nested comprehension you are looking for looks like this:
>>> D = {
...: ('a','b') : 1,
...: ('x','y','z') : 2
...:}
>>>
>>> {k_i:v for k, v in D.items() for k_i in k}
>>> {'a': 1, 'b': 1, 'x': 2, 'y': 2, 'z': 2}
which could be written with traditional for loops like this:
>>> result = {}
>>> for k, v in D.items():
...: for k_i in k:
...: result[k_i] = v
...:
>>> result
>>> {'a': 1, 'b': 1, 'x': 2, 'y': 2, 'z': 2}
Bonus: itertools abuse!
>>> from itertools import repeat, chain
>>> dict(chain.from_iterable(zip(k, repeat(v)) for k, v in D.items()))
>>> {'a': 1, 'b': 1, 'x': 2, 'y': 2, 'z': 2}
answered Nov 16 '18 at 19:46
timgebtimgeb
51.4k126795
add a comment |
You can use a dict comprehension like this:
{k: v for t, v in D.items() for k in t}
answered Nov 16 '18 at 19:47
blhsingblhsing
44.2k51745
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
The nested comprehension you are looking for looks like this:
>>> D = {
...: ('a','b') : 1,
...: ('x','y','z') : 2
...:}
>>>
>>> {k_i:v for k, v in D.items() for k_i in k}
>>> {'a': 1, 'b': 1, 'x': 2, 'y': 2, 'z': 2}
which could be written with traditional for loops like this:
>>> result = {}
>>> for k, v in D.items():
...: for k_i in k:
...: result[k_i] = v
...:
>>> result
>>> {'a': 1, 'b': 1, 'x': 2, 'y': 2, 'z': 2}
Bonus: itertools abuse!
>>> from itertools import repeat, chain
>>> dict(chain.from_iterable(zip(k, repeat(v)) for k, v in D.items()))
>>> {'a': 1, 'b': 1, 'x': 2, 'y': 2, 'z': 2}
answered Nov 16 '18 at 19:46
timgebtimgeb
51.4k126795
add a comment |
The nested comprehension you are looking for looks like this:
>>> D = {
...: ('a','b') : 1,
...: ('x','y','z') : 2
...:}
>>>
>>> {k_i:v for k, v in D.items() for k_i in k}
>>> {'a': 1, 'b': 1, 'x': 2, 'y': 2, 'z': 2}
which could be written with traditional for loops like this:
>>> result = {}
>>> for k, v in D.items():
...: for k_i in k:
...: result[k_i] = v
...:
>>> result
>>> {'a': 1, 'b': 1, 'x': 2, 'y': 2, 'z': 2}
Bonus: itertools abuse!
>>> from itertools import repeat, chain
>>> dict(chain.from_iterable(zip(k, repeat(v)) for k, v in D.items()))
>>> {'a': 1, 'b': 1, 'x': 2, 'y': 2, 'z': 2}
answered Nov 16 '18 at 19:46
timgebtimgeb
51.4k126795
add a comment |
The nested comprehension you are looking for looks like this:
>>> D = {
...: ('a','b') : 1,
...: ('x','y','z') : 2
...:}
>>>
>>> {k_i:v for k, v in D.items() for k_i in k}
>>> {'a': 1, 'b': 1, 'x': 2, 'y': 2, 'z': 2}
which could be written with traditional for loops like this:
>>> result = {}
>>> for k, v in D.items():
...: for k_i in k:
...: result[k_i] = v
...:
>>> result
>>> {'a': 1, 'b': 1, 'x': 2, 'y': 2, 'z': 2}
Bonus: itertools abuse!
>>> from itertools import repeat, chain
>>> dict(chain.from_iterable(zip(k, repeat(v)) for k, v in D.items()))
>>> {'a': 1, 'b': 1, 'x': 2, 'y': 2, 'z': 2}
answered Nov 16 '18 at 19:46
timgebtimgeb
51.4k126795
The nested comprehension you are looking for looks like this:
>>> D = {
...: ('a','b') : 1,
...: ('x','y','z') : 2
...:}
>>>
>>> {k_i:v for k, v in D.items() for k_i in k}
>>> {'a': 1, 'b': 1, 'x': 2, 'y': 2, 'z': 2}
which could be written with traditional for loops like this:
>>> result = {}
>>> for k, v in D.items():
...: for k_i in k:
...: result[k_i] = v
...:
>>> result
>>> {'a': 1, 'b': 1, 'x': 2, 'y': 2, 'z': 2}
Bonus: itertools abuse!
>>> from itertools import repeat, chain
>>> dict(chain.from_iterable(zip(k, repeat(v)) for k, v in D.items()))
>>> {'a': 1, 'b': 1, 'x': 2, 'y': 2, 'z': 2}
answered Nov 16 '18 at 19:46
timgebtimgeb
51.4k126795
edited Nov 16 '18 at 20:02
answered Nov 16 '18 at 19:46
timgebtimgeb
51.4k126795
answered Nov 16 '18 at 19:46
timgebtimgeb
51.4k126795
answered Nov 16 '18 at 19:46
timgebtimgeb
51.4k126795
51.4k126795
add a comment |
add a comment |
You can use a dict comprehension like this:
{k: v for t, v in D.items() for k in t}
answered Nov 16 '18 at 19:47
blhsingblhsing
44.2k51745
add a comment |
You can use a dict comprehension like this:
{k: v for t, v in D.items() for k in t}
answered Nov 16 '18 at 19:47
blhsingblhsing
44.2k51745
add a comment |
You can use a dict comprehension like this:
{k: v for t, v in D.items() for k in t}
answered Nov 16 '18 at 19:47
blhsingblhsing
44.2k51745
You can use a dict comprehension like this:
{k: v for t, v in D.items() for k in t}
answered Nov 16 '18 at 19:47
blhsingblhsing
44.2k51745
answered Nov 16 '18 at 19:47
blhsingblhsing
44.2k51745
answered Nov 16 '18 at 19:47
blhsingblhsing
44.2k51745
answered Nov 16 '18 at 19:47
blhsingblhsing
44.2k51745
44.2k51745
add a comment |
add a comment |
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(Note that for this question to be well defined the tuple-keys must be pairwise disjoint.)
– timgeb
Nov 16 '18 at 19:55