Mixing
Convert dict to list of dict for each combinations
0
I have a dict looks like this :
my_dict = {
"a":[1, 2, 3],
"b":[10],
"c":[4, 5],
"d":[11]
}
And I would like to obtain a list containig all combinations keeping keys and value like this:
result = [
{"a":1, "b":10, "c":4, "d":11},
{"a":1, "b":10, "c":5, "d":11},
{"a":2, "b":10, "c":4, "d":11},
{"a":2, "b":10, "c":5, "d":11},
{"a":3, "b":10, "c":4, "d":11},
{"a":3, "b":10, "c":5, "d":11}
]
Do someone have a solution for this ?
Is there any existing solution to do this, or how should I proceed to do it myself ?
Thank you.
python list dictionary
edited Nov 22 '18 at 11:23
lospejos
1,46221426
asked Nov 22 '18 at 9:05
FurlingsFurlings
63
add a comment |
0
I have a dict looks like this :
my_dict = {
"a":[1, 2, 3],
"b":[10],
"c":[4, 5],
"d":[11]
}
And I would like to obtain a list containig all combinations keeping keys and value like this:
result = [
{"a":1, "b":10, "c":4, "d":11},
{"a":1, "b":10, "c":5, "d":11},
{"a":2, "b":10, "c":4, "d":11},
{"a":2, "b":10, "c":5, "d":11},
{"a":3, "b":10, "c":4, "d":11},
{"a":3, "b":10, "c":5, "d":11}
]
Do someone have a solution for this ?
Is there any existing solution to do this, or how should I proceed to do it myself ?
Thank you.
python list dictionary
edited Nov 22 '18 at 11:23
lospejos
1,46221426
asked Nov 22 '18 at 9:05
FurlingsFurlings
63
I hope using itertools.combinations you can generate various combinations of a list but not sure on generating for a dictionary.
– Sekar Ramu
Nov 22 '18 at 9:17
add a comment |
0
0
0
I have a dict looks like this :
my_dict = {
"a":[1, 2, 3],
"b":[10],
"c":[4, 5],
"d":[11]
}
And I would like to obtain a list containig all combinations keeping keys and value like this:
result = [
{"a":1, "b":10, "c":4, "d":11},
{"a":1, "b":10, "c":5, "d":11},
{"a":2, "b":10, "c":4, "d":11},
{"a":2, "b":10, "c":5, "d":11},
{"a":3, "b":10, "c":4, "d":11},
{"a":3, "b":10, "c":5, "d":11}
]
Do someone have a solution for this ?
Is there any existing solution to do this, or how should I proceed to do it myself ?
Thank you.
python list dictionary
edited Nov 22 '18 at 11:23
lospejos
1,46221426
asked Nov 22 '18 at 9:05
FurlingsFurlings
63
I have a dict looks like this :
my_dict = {
"a":[1, 2, 3],
"b":[10],
"c":[4, 5],
"d":[11]
}
And I would like to obtain a list containig all combinations keeping keys and value like this:
result = [
{"a":1, "b":10, "c":4, "d":11},
{"a":1, "b":10, "c":5, "d":11},
{"a":2, "b":10, "c":4, "d":11},
{"a":2, "b":10, "c":5, "d":11},
{"a":3, "b":10, "c":4, "d":11},
{"a":3, "b":10, "c":5, "d":11}
]
Do someone have a solution for this ?
Is there any existing solution to do this, or how should I proceed to do it myself ?
Thank you.
python list dictionary
python list dictionary
edited Nov 22 '18 at 11:23
lospejos
1,46221426
asked Nov 22 '18 at 9:05
FurlingsFurlings
63
edited Nov 22 '18 at 11:23
lospejos
1,46221426
asked Nov 22 '18 at 9:05
FurlingsFurlings
63
edited Nov 22 '18 at 11:23
lospejos
1,46221426
edited Nov 22 '18 at 11:23
lospejos
1,46221426
edited Nov 22 '18 at 11:23
lospejos
1,46221426
1,46221426
asked Nov 22 '18 at 9:05
FurlingsFurlings
63
asked Nov 22 '18 at 9:05
FurlingsFurlings
63
asked Nov 22 '18 at 9:05
FurlingsFurlings
63
63
I hope using itertools.combinations you can generate various combinations of a list but not sure on generating for a dictionary.
– Sekar Ramu
Nov 22 '18 at 9:17
add a comment |
I hope using itertools.combinations you can generate various combinations of a list but not sure on generating for a dictionary.
– Sekar Ramu
Nov 22 '18 at 9:17
I hope using itertools.combinations you can generate various combinations of a list but not sure on generating for a dictionary.
– Sekar Ramu
Nov 22 '18 at 9:17
I hope using itertools.combinations you can generate various combinations of a list but not sure on generating for a dictionary.
– Sekar Ramu
Nov 22 '18 at 9:17
add a comment |
5 Answers
5
active
oldest
votes
-1
Try:
def permute(d):
k = d.keys()
perms = itertools.product(*d.values())
return [dict(zip(k, v)) for v in perms]
>>> pprint(permute(my_dict))
[{'a': 1, 'b': 10, 'c': 4, 'd': 11},
{'a': 1, 'b': 10, 'c': 5, 'd': 11},
{'a': 2, 'b': 10, 'c': 4, 'd': 11},
{'a': 2, 'b': 10, 'c': 5, 'd': 11},
{'a': 3, 'b': 10, 'c': 4, 'd': 11},
{'a': 3, 'b': 10, 'c': 5, 'd': 11}]
answered Nov 22 '18 at 9:13
Mateen UlhaqMateen Ulhaq
11.5k114793
1
Thank you Mateen, that works perfectly !
– Furlings
Nov 22 '18 at 9:30
add a comment |
2
A task for itertools.product:
>>> from itertools import product
>>> for dict_items in product(*[product([k],v) for k, v in my_dict.items()]):
... print(dict(dict_items))
{'a': 1, 'b': 10, 'c': 4, 'd': 11}
{'a': 1, 'b': 10, 'c': 5, 'd': 11}
{'a': 2, 'b': 10, 'c': 4, 'd': 11}
{'a': 2, 'b': 10, 'c': 5, 'd': 11}
{'a': 3, 'b': 10, 'c': 4, 'd': 11}
{'a': 3, 'b': 10, 'c': 5, 'd': 11}
Small explanation:
The inner product(...) will expand the dict to a list such as [[(k1, v11), (k1, v12), ...], [(k2, v21), (k2, v22), ...], ...].
The outer product(...) will reassemble the items lists by choosing one tuple from each list.
dict(...) will create a dictionary from a sequence of (k1, v#), (k2, v#), ... tuples.
answered Nov 22 '18 at 9:18
fferrifferri
11.7k32251
Thanks for explanations ;)
– Furlings
Nov 22 '18 at 9:42
add a comment |
0
Assuming that you are only interested in my_dict having 4 keys, it is simple enough to use nested for loops:
my_dict = {
"a": [1, 2, 3],
"b": [10],
"c": [4, 5],
"d": [11]
}
result =
for a_val in my_dict['a']:
for b_val in my_dict['b']:
for c_val in my_dict['c']:
for d_val in my_dict['d']:
result.append({'a': a_val, 'b': b_val, 'c': c_val, 'd': d_val})
print(result)
This gives the expected result.
answered Nov 22 '18 at 9:13
DatHydroGuyDatHydroGuy
6812413
Thank you for answer but I can't use it because "my_dict" can contain different number of key with different name.
– Furlings
Nov 22 '18 at 9:19
No problem. Then theitertoolsapproach is definitely the right one!
– DatHydroGuy
Nov 22 '18 at 9:20
add a comment |
0
You can use:
from itertools import product
allNames = sorted(my_dict)
values= list(product(*(my_dict[Name] for Name in allNames)))
d = list(dict(zip(['a','b','c','d'],i)) for i in values)
Output:
[{'a': 1, 'c': 4, 'b': 10, 'd': 11},
{'a': 1, 'c': 5, 'b': 10, 'd': 11},
{'a': 2, 'c': 4, 'b': 10, 'd': 11},
{'a': 2, 'c': 5, 'b': 10, 'd': 11},
{'a': 3, 'c': 4, 'b': 10, 'd': 11},
{'a': 3, 'c': 5, 'b': 10, 'd': 11}]
answered Nov 22 '18 at 9:24
JoeJoe
6,05421429
add a comment |
0
itertools.product produces the combinations of a list of iterators.
dict.values() gets the list needed.
For each combination, zip up the dict.keys() with the combination.
Use a list comprehension to collect them up:
from itertools import product
from pprint import pprint
my_dict = {
"a":[1, 2, 3],
"b":[10],
"c":[4, 5],
"d":[11]
}
result = [dict(zip(my_dict,i)) for i in product(*my_dict.values())]
pprint(result)
Output:
[{'a': 1, 'b': 10, 'c': 4, 'd': 11},
{'a': 1, 'b': 10, 'c': 5, 'd': 11},
{'a': 2, 'b': 10, 'c': 4, 'd': 11},
{'a': 2, 'b': 10, 'c': 5, 'd': 11},
{'a': 3, 'b': 10, 'c': 4, 'd': 11},
{'a': 3, 'b': 10, 'c': 5, 'd': 11}]
answered Nov 22 '18 at 9:30
Mark TolonenMark Tolonen
94.4k12114176
Thank you very much for explanation ! It's more clear now ;)
– Furlings
Nov 22 '18 at 9:34
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
-1
Try:
def permute(d):
k = d.keys()
perms = itertools.product(*d.values())
return [dict(zip(k, v)) for v in perms]
>>> pprint(permute(my_dict))
[{'a': 1, 'b': 10, 'c': 4, 'd': 11},
{'a': 1, 'b': 10, 'c': 5, 'd': 11},
{'a': 2, 'b': 10, 'c': 4, 'd': 11},
{'a': 2, 'b': 10, 'c': 5, 'd': 11},
{'a': 3, 'b': 10, 'c': 4, 'd': 11},
{'a': 3, 'b': 10, 'c': 5, 'd': 11}]
answered Nov 22 '18 at 9:13
Mateen UlhaqMateen Ulhaq
11.5k114793
1
Thank you Mateen, that works perfectly !
– Furlings
Nov 22 '18 at 9:30
add a comment |
-1
Try:
def permute(d):
k = d.keys()
perms = itertools.product(*d.values())
return [dict(zip(k, v)) for v in perms]
>>> pprint(permute(my_dict))
[{'a': 1, 'b': 10, 'c': 4, 'd': 11},
{'a': 1, 'b': 10, 'c': 5, 'd': 11},
{'a': 2, 'b': 10, 'c': 4, 'd': 11},
{'a': 2, 'b': 10, 'c': 5, 'd': 11},
{'a': 3, 'b': 10, 'c': 4, 'd': 11},
{'a': 3, 'b': 10, 'c': 5, 'd': 11}]
answered Nov 22 '18 at 9:13
Mateen UlhaqMateen Ulhaq
11.5k114793
1
Thank you Mateen, that works perfectly !
– Furlings
Nov 22 '18 at 9:30
add a comment |
-1
-1
-1
Try:
def permute(d):
k = d.keys()
perms = itertools.product(*d.values())
return [dict(zip(k, v)) for v in perms]
>>> pprint(permute(my_dict))
[{'a': 1, 'b': 10, 'c': 4, 'd': 11},
{'a': 1, 'b': 10, 'c': 5, 'd': 11},
{'a': 2, 'b': 10, 'c': 4, 'd': 11},
{'a': 2, 'b': 10, 'c': 5, 'd': 11},
{'a': 3, 'b': 10, 'c': 4, 'd': 11},
{'a': 3, 'b': 10, 'c': 5, 'd': 11}]
answered Nov 22 '18 at 9:13
Mateen UlhaqMateen Ulhaq
11.5k114793
Try:
def permute(d):
k = d.keys()
perms = itertools.product(*d.values())
return [dict(zip(k, v)) for v in perms]
>>> pprint(permute(my_dict))
[{'a': 1, 'b': 10, 'c': 4, 'd': 11},
{'a': 1, 'b': 10, 'c': 5, 'd': 11},
{'a': 2, 'b': 10, 'c': 4, 'd': 11},
{'a': 2, 'b': 10, 'c': 5, 'd': 11},
{'a': 3, 'b': 10, 'c': 4, 'd': 11},
{'a': 3, 'b': 10, 'c': 5, 'd': 11}]
answered Nov 22 '18 at 9:13
Mateen UlhaqMateen Ulhaq
11.5k114793
edited Nov 22 '18 at 9:19
answered Nov 22 '18 at 9:13
Mateen UlhaqMateen Ulhaq
11.5k114793
answered Nov 22 '18 at 9:13
Mateen UlhaqMateen Ulhaq
11.5k114793
answered Nov 22 '18 at 9:13
Mateen UlhaqMateen Ulhaq
11.5k114793
11.5k114793
1
Thank you Mateen, that works perfectly !
– Furlings
Nov 22 '18 at 9:30
add a comment |
1
Thank you Mateen, that works perfectly !
– Furlings
Nov 22 '18 at 9:30
1
1
Thank you Mateen, that works perfectly !
– Furlings
Nov 22 '18 at 9:30
Thank you Mateen, that works perfectly !
– Furlings
Nov 22 '18 at 9:30
add a comment |
2
A task for itertools.product:
>>> from itertools import product
>>> for dict_items in product(*[product([k],v) for k, v in my_dict.items()]):
... print(dict(dict_items))
{'a': 1, 'b': 10, 'c': 4, 'd': 11}
{'a': 1, 'b': 10, 'c': 5, 'd': 11}
{'a': 2, 'b': 10, 'c': 4, 'd': 11}
{'a': 2, 'b': 10, 'c': 5, 'd': 11}
{'a': 3, 'b': 10, 'c': 4, 'd': 11}
{'a': 3, 'b': 10, 'c': 5, 'd': 11}
Small explanation:
The inner product(...) will expand the dict to a list such as [[(k1, v11), (k1, v12), ...], [(k2, v21), (k2, v22), ...], ...].
The outer product(...) will reassemble the items lists by choosing one tuple from each list.
dict(...) will create a dictionary from a sequence of (k1, v#), (k2, v#), ... tuples.
answered Nov 22 '18 at 9:18
fferrifferri
11.7k32251
Thanks for explanations ;)
– Furlings
Nov 22 '18 at 9:42
add a comment |
2
A task for itertools.product:
>>> from itertools import product
>>> for dict_items in product(*[product([k],v) for k, v in my_dict.items()]):
... print(dict(dict_items))
{'a': 1, 'b': 10, 'c': 4, 'd': 11}
{'a': 1, 'b': 10, 'c': 5, 'd': 11}
{'a': 2, 'b': 10, 'c': 4, 'd': 11}
{'a': 2, 'b': 10, 'c': 5, 'd': 11}
{'a': 3, 'b': 10, 'c': 4, 'd': 11}
{'a': 3, 'b': 10, 'c': 5, 'd': 11}
Small explanation:
The inner product(...) will expand the dict to a list such as [[(k1, v11), (k1, v12), ...], [(k2, v21), (k2, v22), ...], ...].
The outer product(...) will reassemble the items lists by choosing one tuple from each list.
dict(...) will create a dictionary from a sequence of (k1, v#), (k2, v#), ... tuples.
answered Nov 22 '18 at 9:18
fferrifferri
11.7k32251
Thanks for explanations ;)
– Furlings
Nov 22 '18 at 9:42
add a comment |
2
2
2
A task for itertools.product:
>>> from itertools import product
>>> for dict_items in product(*[product([k],v) for k, v in my_dict.items()]):
... print(dict(dict_items))
{'a': 1, 'b': 10, 'c': 4, 'd': 11}
{'a': 1, 'b': 10, 'c': 5, 'd': 11}
{'a': 2, 'b': 10, 'c': 4, 'd': 11}
{'a': 2, 'b': 10, 'c': 5, 'd': 11}
{'a': 3, 'b': 10, 'c': 4, 'd': 11}
{'a': 3, 'b': 10, 'c': 5, 'd': 11}
Small explanation:
The inner product(...) will expand the dict to a list such as [[(k1, v11), (k1, v12), ...], [(k2, v21), (k2, v22), ...], ...].
The outer product(...) will reassemble the items lists by choosing one tuple from each list.
dict(...) will create a dictionary from a sequence of (k1, v#), (k2, v#), ... tuples.
answered Nov 22 '18 at 9:18
fferrifferri
11.7k32251
A task for itertools.product:
>>> from itertools import product
>>> for dict_items in product(*[product([k],v) for k, v in my_dict.items()]):
... print(dict(dict_items))
{'a': 1, 'b': 10, 'c': 4, 'd': 11}
{'a': 1, 'b': 10, 'c': 5, 'd': 11}
{'a': 2, 'b': 10, 'c': 4, 'd': 11}
{'a': 2, 'b': 10, 'c': 5, 'd': 11}
{'a': 3, 'b': 10, 'c': 4, 'd': 11}
{'a': 3, 'b': 10, 'c': 5, 'd': 11}
Small explanation:
The inner product(...) will expand the dict to a list such as [[(k1, v11), (k1, v12), ...], [(k2, v21), (k2, v22), ...], ...].
The outer product(...) will reassemble the items lists by choosing one tuple from each list.
dict(...) will create a dictionary from a sequence of (k1, v#), (k2, v#), ... tuples.
answered Nov 22 '18 at 9:18
fferrifferri
11.7k32251
edited Nov 22 '18 at 9:27
answered Nov 22 '18 at 9:18
fferrifferri
11.7k32251
answered Nov 22 '18 at 9:18
fferrifferri
11.7k32251
answered Nov 22 '18 at 9:18
fferrifferri
11.7k32251
11.7k32251
Thanks for explanations ;)
– Furlings
Nov 22 '18 at 9:42
add a comment |
Thanks for explanations ;)
– Furlings
Nov 22 '18 at 9:42
Thanks for explanations ;)
– Furlings
Nov 22 '18 at 9:42
Thanks for explanations ;)
– Furlings
Nov 22 '18 at 9:42
add a comment |
0
Assuming that you are only interested in my_dict having 4 keys, it is simple enough to use nested for loops:
my_dict = {
"a": [1, 2, 3],
"b": [10],
"c": [4, 5],
"d": [11]
}
result =
for a_val in my_dict['a']:
for b_val in my_dict['b']:
for c_val in my_dict['c']:
for d_val in my_dict['d']:
result.append({'a': a_val, 'b': b_val, 'c': c_val, 'd': d_val})
print(result)
This gives the expected result.
answered Nov 22 '18 at 9:13
DatHydroGuyDatHydroGuy
6812413
Thank you for answer but I can't use it because "my_dict" can contain different number of key with different name.
– Furlings
Nov 22 '18 at 9:19
No problem. Then theitertoolsapproach is definitely the right one!
– DatHydroGuy
Nov 22 '18 at 9:20
add a comment |
0
Assuming that you are only interested in my_dict having 4 keys, it is simple enough to use nested for loops:
my_dict = {
"a": [1, 2, 3],
"b": [10],
"c": [4, 5],
"d": [11]
}
result =
for a_val in my_dict['a']:
for b_val in my_dict['b']:
for c_val in my_dict['c']:
for d_val in my_dict['d']:
result.append({'a': a_val, 'b': b_val, 'c': c_val, 'd': d_val})
print(result)
This gives the expected result.
answered Nov 22 '18 at 9:13
DatHydroGuyDatHydroGuy
6812413
Thank you for answer but I can't use it because "my_dict" can contain different number of key with different name.
– Furlings
Nov 22 '18 at 9:19
No problem. Then theitertoolsapproach is definitely the right one!
– DatHydroGuy
Nov 22 '18 at 9:20
add a comment |
0
0
0
Assuming that you are only interested in my_dict having 4 keys, it is simple enough to use nested for loops:
my_dict = {
"a": [1, 2, 3],
"b": [10],
"c": [4, 5],
"d": [11]
}
result =
for a_val in my_dict['a']:
for b_val in my_dict['b']:
for c_val in my_dict['c']:
for d_val in my_dict['d']:
result.append({'a': a_val, 'b': b_val, 'c': c_val, 'd': d_val})
print(result)
This gives the expected result.
answered Nov 22 '18 at 9:13
DatHydroGuyDatHydroGuy
6812413
Assuming that you are only interested in my_dict having 4 keys, it is simple enough to use nested for loops:
my_dict = {
"a": [1, 2, 3],
"b": [10],
"c": [4, 5],
"d": [11]
}
result =
for a_val in my_dict['a']:
for b_val in my_dict['b']:
for c_val in my_dict['c']:
for d_val in my_dict['d']:
result.append({'a': a_val, 'b': b_val, 'c': c_val, 'd': d_val})
print(result)
This gives the expected result.
answered Nov 22 '18 at 9:13
DatHydroGuyDatHydroGuy
6812413
answered Nov 22 '18 at 9:13
DatHydroGuyDatHydroGuy
6812413
answered Nov 22 '18 at 9:13
DatHydroGuyDatHydroGuy
6812413
answered Nov 22 '18 at 9:13
DatHydroGuyDatHydroGuy
6812413
6812413
Thank you for answer but I can't use it because "my_dict" can contain different number of key with different name.
– Furlings
Nov 22 '18 at 9:19
No problem. Then theitertoolsapproach is definitely the right one!
– DatHydroGuy
Nov 22 '18 at 9:20
add a comment |
Thank you for answer but I can't use it because "my_dict" can contain different number of key with different name.
– Furlings
Nov 22 '18 at 9:19
No problem. Then theitertoolsapproach is definitely the right one!
– DatHydroGuy
Nov 22 '18 at 9:20
Thank you for answer but I can't use it because "my_dict" can contain different number of key with different name.
– Furlings
Nov 22 '18 at 9:19
Thank you for answer but I can't use it because "my_dict" can contain different number of key with different name.
– Furlings
Nov 22 '18 at 9:19
No problem. Then the
itertools approach is definitely the right one!– DatHydroGuy
Nov 22 '18 at 9:20
No problem. Then the
itertools approach is definitely the right one!– DatHydroGuy
Nov 22 '18 at 9:20
add a comment |
0
You can use:
from itertools import product
allNames = sorted(my_dict)
values= list(product(*(my_dict[Name] for Name in allNames)))
d = list(dict(zip(['a','b','c','d'],i)) for i in values)
Output:
[{'a': 1, 'c': 4, 'b': 10, 'd': 11},
{'a': 1, 'c': 5, 'b': 10, 'd': 11},
{'a': 2, 'c': 4, 'b': 10, 'd': 11},
{'a': 2, 'c': 5, 'b': 10, 'd': 11},
{'a': 3, 'c': 4, 'b': 10, 'd': 11},
{'a': 3, 'c': 5, 'b': 10, 'd': 11}]
answered Nov 22 '18 at 9:24
JoeJoe
6,05421429
add a comment |
0
You can use:
from itertools import product
allNames = sorted(my_dict)
values= list(product(*(my_dict[Name] for Name in allNames)))
d = list(dict(zip(['a','b','c','d'],i)) for i in values)
Output:
[{'a': 1, 'c': 4, 'b': 10, 'd': 11},
{'a': 1, 'c': 5, 'b': 10, 'd': 11},
{'a': 2, 'c': 4, 'b': 10, 'd': 11},
{'a': 2, 'c': 5, 'b': 10, 'd': 11},
{'a': 3, 'c': 4, 'b': 10, 'd': 11},
{'a': 3, 'c': 5, 'b': 10, 'd': 11}]
answered Nov 22 '18 at 9:24
JoeJoe
6,05421429
add a comment |
0
0
0
You can use:
from itertools import product
allNames = sorted(my_dict)
values= list(product(*(my_dict[Name] for Name in allNames)))
d = list(dict(zip(['a','b','c','d'],i)) for i in values)
Output:
[{'a': 1, 'c': 4, 'b': 10, 'd': 11},
{'a': 1, 'c': 5, 'b': 10, 'd': 11},
{'a': 2, 'c': 4, 'b': 10, 'd': 11},
{'a': 2, 'c': 5, 'b': 10, 'd': 11},
{'a': 3, 'c': 4, 'b': 10, 'd': 11},
{'a': 3, 'c': 5, 'b': 10, 'd': 11}]
answered Nov 22 '18 at 9:24
JoeJoe
6,05421429
You can use:
from itertools import product
allNames = sorted(my_dict)
values= list(product(*(my_dict[Name] for Name in allNames)))
d = list(dict(zip(['a','b','c','d'],i)) for i in values)
Output:
[{'a': 1, 'c': 4, 'b': 10, 'd': 11},
{'a': 1, 'c': 5, 'b': 10, 'd': 11},
{'a': 2, 'c': 4, 'b': 10, 'd': 11},
{'a': 2, 'c': 5, 'b': 10, 'd': 11},
{'a': 3, 'c': 4, 'b': 10, 'd': 11},
{'a': 3, 'c': 5, 'b': 10, 'd': 11}]
answered Nov 22 '18 at 9:24
JoeJoe
6,05421429
answered Nov 22 '18 at 9:24
JoeJoe
6,05421429
answered Nov 22 '18 at 9:24
JoeJoe
6,05421429
answered Nov 22 '18 at 9:24
JoeJoe
6,05421429
6,05421429
add a comment |
add a comment |
0
itertools.product produces the combinations of a list of iterators.
dict.values() gets the list needed.
For each combination, zip up the dict.keys() with the combination.
Use a list comprehension to collect them up:
from itertools import product
from pprint import pprint
my_dict = {
"a":[1, 2, 3],
"b":[10],
"c":[4, 5],
"d":[11]
}
result = [dict(zip(my_dict,i)) for i in product(*my_dict.values())]
pprint(result)
Output:
[{'a': 1, 'b': 10, 'c': 4, 'd': 11},
{'a': 1, 'b': 10, 'c': 5, 'd': 11},
{'a': 2, 'b': 10, 'c': 4, 'd': 11},
{'a': 2, 'b': 10, 'c': 5, 'd': 11},
{'a': 3, 'b': 10, 'c': 4, 'd': 11},
{'a': 3, 'b': 10, 'c': 5, 'd': 11}]
answered Nov 22 '18 at 9:30
Mark TolonenMark Tolonen
94.4k12114176
Thank you very much for explanation ! It's more clear now ;)
– Furlings
Nov 22 '18 at 9:34
add a comment |
0
itertools.product produces the combinations of a list of iterators.
dict.values() gets the list needed.
For each combination, zip up the dict.keys() with the combination.
Use a list comprehension to collect them up:
from itertools import product
from pprint import pprint
my_dict = {
"a":[1, 2, 3],
"b":[10],
"c":[4, 5],
"d":[11]
}
result = [dict(zip(my_dict,i)) for i in product(*my_dict.values())]
pprint(result)
Output:
[{'a': 1, 'b': 10, 'c': 4, 'd': 11},
{'a': 1, 'b': 10, 'c': 5, 'd': 11},
{'a': 2, 'b': 10, 'c': 4, 'd': 11},
{'a': 2, 'b': 10, 'c': 5, 'd': 11},
{'a': 3, 'b': 10, 'c': 4, 'd': 11},
{'a': 3, 'b': 10, 'c': 5, 'd': 11}]
answered Nov 22 '18 at 9:30
Mark TolonenMark Tolonen
94.4k12114176
Thank you very much for explanation ! It's more clear now ;)
– Furlings
Nov 22 '18 at 9:34
add a comment |
0
0
0
itertools.product produces the combinations of a list of iterators.
dict.values() gets the list needed.
For each combination, zip up the dict.keys() with the combination.
Use a list comprehension to collect them up:
from itertools import product
from pprint import pprint
my_dict = {
"a":[1, 2, 3],
"b":[10],
"c":[4, 5],
"d":[11]
}
result = [dict(zip(my_dict,i)) for i in product(*my_dict.values())]
pprint(result)
Output:
[{'a': 1, 'b': 10, 'c': 4, 'd': 11},
{'a': 1, 'b': 10, 'c': 5, 'd': 11},
{'a': 2, 'b': 10, 'c': 4, 'd': 11},
{'a': 2, 'b': 10, 'c': 5, 'd': 11},
{'a': 3, 'b': 10, 'c': 4, 'd': 11},
{'a': 3, 'b': 10, 'c': 5, 'd': 11}]
answered Nov 22 '18 at 9:30
Mark TolonenMark Tolonen
94.4k12114176
itertools.product produces the combinations of a list of iterators.
dict.values() gets the list needed.
For each combination, zip up the dict.keys() with the combination.
Use a list comprehension to collect them up:
from itertools import product
from pprint import pprint
my_dict = {
"a":[1, 2, 3],
"b":[10],
"c":[4, 5],
"d":[11]
}
result = [dict(zip(my_dict,i)) for i in product(*my_dict.values())]
pprint(result)
Output:
[{'a': 1, 'b': 10, 'c': 4, 'd': 11},
{'a': 1, 'b': 10, 'c': 5, 'd': 11},
{'a': 2, 'b': 10, 'c': 4, 'd': 11},
{'a': 2, 'b': 10, 'c': 5, 'd': 11},
{'a': 3, 'b': 10, 'c': 4, 'd': 11},
{'a': 3, 'b': 10, 'c': 5, 'd': 11}]
answered Nov 22 '18 at 9:30
Mark TolonenMark Tolonen
94.4k12114176
answered Nov 22 '18 at 9:30
Mark TolonenMark Tolonen
94.4k12114176
answered Nov 22 '18 at 9:30
Mark TolonenMark Tolonen
94.4k12114176
answered Nov 22 '18 at 9:30
Mark TolonenMark Tolonen
94.4k12114176
94.4k12114176
Thank you very much for explanation ! It's more clear now ;)
– Furlings
Nov 22 '18 at 9:34
add a comment |
Thank you very much for explanation ! It's more clear now ;)
– Furlings
Nov 22 '18 at 9:34
Thank you very much for explanation ! It's more clear now ;)
– Furlings
Nov 22 '18 at 9:34
Thank you very much for explanation ! It's more clear now ;)
– Furlings
Nov 22 '18 at 9:34
add a comment |
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I hope using itertools.combinations you can generate various combinations of a list but not sure on generating for a dictionary.
– Sekar Ramu
Nov 22 '18 at 9:17