How to find common occurrences of dictionary in a list of list of dictionaries
I have a list of list of dictionaries and I want to find the common dictionaries between the two list.
Eg:
dict_list = [[{'1' : 1,'2' : 2, '3' :3}, {'6' : 6,'5' : 5, '4' : 4}],
[{'1' : 1,'2' : 2, '3' :3}, {'7' : 7,'8' : 8, '9' : 9}]]
The result should be [{'1' : 1,'2' : 2, '3' :3}]
I tried using set intersections but dictionaries are unhashable in python.
How to solve this?
python dictionary
edited Nov 14 '18 at 10:57
Zlytherin
1,7071728
asked Nov 14 '18 at 10:54
Perseus14Perseus14
302316
add a comment |
I have a list of list of dictionaries and I want to find the common dictionaries between the two list.
Eg:
dict_list = [[{'1' : 1,'2' : 2, '3' :3}, {'6' : 6,'5' : 5, '4' : 4}],
[{'1' : 1,'2' : 2, '3' :3}, {'7' : 7,'8' : 8, '9' : 9}]]
The result should be [{'1' : 1,'2' : 2, '3' :3}]
I tried using set intersections but dictionaries are unhashable in python.
How to solve this?
python dictionary
edited Nov 14 '18 at 10:57
Zlytherin
1,7071728
asked Nov 14 '18 at 10:54
Perseus14Perseus14
302316
5
[dict(k) for k,v in Counter(tuple(x.items()) for x in chain.from_iterable(dict_list)).items() if v > 1]
– Chris_Rands
Nov 14 '18 at 11:00
1
@Chris_Rands I didn't see your solution at all when I posted mine. Do you want me to remove it?
– RoadRunner
Nov 14 '18 at 11:13
2
@RoadRunner no don't remove it, people do have the same idea sometimes, thanks for asking though :)
– Chris_Rands
Nov 14 '18 at 11:18
after converting a list of dict to list of tuple, he can use set.intersection too, I guess
– khelili miliana
Nov 14 '18 at 11:21
add a comment |
I have a list of list of dictionaries and I want to find the common dictionaries between the two list.
Eg:
dict_list = [[{'1' : 1,'2' : 2, '3' :3}, {'6' : 6,'5' : 5, '4' : 4}],
[{'1' : 1,'2' : 2, '3' :3}, {'7' : 7,'8' : 8, '9' : 9}]]
The result should be [{'1' : 1,'2' : 2, '3' :3}]
I tried using set intersections but dictionaries are unhashable in python.
How to solve this?
python dictionary
edited Nov 14 '18 at 10:57
Zlytherin
1,7071728
asked Nov 14 '18 at 10:54
Perseus14Perseus14
302316
I have a list of list of dictionaries and I want to find the common dictionaries between the two list.
Eg:
dict_list = [[{'1' : 1,'2' : 2, '3' :3}, {'6' : 6,'5' : 5, '4' : 4}],
[{'1' : 1,'2' : 2, '3' :3}, {'7' : 7,'8' : 8, '9' : 9}]]
The result should be [{'1' : 1,'2' : 2, '3' :3}]
I tried using set intersections but dictionaries are unhashable in python.
How to solve this?
python dictionary
python dictionary
edited Nov 14 '18 at 10:57
Zlytherin
1,7071728
asked Nov 14 '18 at 10:54
Perseus14Perseus14
302316
edited Nov 14 '18 at 10:57
Zlytherin
1,7071728
asked Nov 14 '18 at 10:54
Perseus14Perseus14
302316
edited Nov 14 '18 at 10:57
Zlytherin
1,7071728
edited Nov 14 '18 at 10:57
Zlytherin
1,7071728
edited Nov 14 '18 at 10:57
Zlytherin
1,7071728
1,7071728
asked Nov 14 '18 at 10:54
Perseus14Perseus14
302316
asked Nov 14 '18 at 10:54
Perseus14Perseus14
302316
asked Nov 14 '18 at 10:54
Perseus14Perseus14
302316
302316
5
[dict(k) for k,v in Counter(tuple(x.items()) for x in chain.from_iterable(dict_list)).items() if v > 1]
– Chris_Rands
Nov 14 '18 at 11:00
1
@Chris_Rands I didn't see your solution at all when I posted mine. Do you want me to remove it?
– RoadRunner
Nov 14 '18 at 11:13
2
@RoadRunner no don't remove it, people do have the same idea sometimes, thanks for asking though :)
– Chris_Rands
Nov 14 '18 at 11:18
after converting a list of dict to list of tuple, he can use set.intersection too, I guess
– khelili miliana
Nov 14 '18 at 11:21
add a comment |
5
[dict(k) for k,v in Counter(tuple(x.items()) for x in chain.from_iterable(dict_list)).items() if v > 1]
– Chris_Rands
Nov 14 '18 at 11:00
1
@Chris_Rands I didn't see your solution at all when I posted mine. Do you want me to remove it?
– RoadRunner
Nov 14 '18 at 11:13
2
@RoadRunner no don't remove it, people do have the same idea sometimes, thanks for asking though :)
– Chris_Rands
Nov 14 '18 at 11:18
after converting a list of dict to list of tuple, he can use set.intersection too, I guess
– khelili miliana
Nov 14 '18 at 11:21
5
5
[dict(k) for k,v in Counter(tuple(x.items()) for x in chain.from_iterable(dict_list)).items() if v > 1] – Chris_Rands
Nov 14 '18 at 11:00
[dict(k) for k,v in Counter(tuple(x.items()) for x in chain.from_iterable(dict_list)).items() if v > 1] – Chris_Rands
Nov 14 '18 at 11:00
1
1
@Chris_Rands I didn't see your solution at all when I posted mine. Do you want me to remove it?
– RoadRunner
Nov 14 '18 at 11:13
@Chris_Rands I didn't see your solution at all when I posted mine. Do you want me to remove it?
– RoadRunner
Nov 14 '18 at 11:13
2
2
@RoadRunner no don't remove it, people do have the same idea sometimes, thanks for asking though :)
– Chris_Rands
Nov 14 '18 at 11:18
@RoadRunner no don't remove it, people do have the same idea sometimes, thanks for asking though :)
– Chris_Rands
Nov 14 '18 at 11:18
after converting a list of dict to list of tuple, he can use set.intersection too, I guess
– khelili miliana
Nov 14 '18 at 11:21
after converting a list of dict to list of tuple, he can use set.intersection too, I guess
– khelili miliana
Nov 14 '18 at 11:21
add a comment |
1 Answer
1
active
oldest
votes
A list comprehension could work here:
>>> [x for x in dict_list[0] if x in dict_list[1]]
[{'1': 1, '2': 2, '3': 3}]
But this is not a very general solution, since it assumes only two nested lists are exististent.
A more general solution would be to count the occurences with collections.Counter(), and storing the dictionary items() with hashable/immutable types such as frozenset() or tuple(). Then all you need to do is filter the occurences that count more than 1.
Example:
>>> from itertools import chain
>>> from collections import Counter
>>> [dict(k) for k, v in Counter(frozenset(x.items()) for x in chain.from_iterable(dict_list)).items() if v > 1]
[{'1': 1, '2': 2, '3': 3}]
Which is very similar to the approach @Chris_Rands posted in the comments.
answered Nov 14 '18 at 10:59
RoadRunnerRoadRunner
11.2k31340
1
The main point here is that even though dictionaries are not hashable they are comparable. More about comparing dictionaries in python.
– sophros
Nov 14 '18 at 11:01
add a comment |
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1 Answer
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1 Answer
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active
oldest
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votes
A list comprehension could work here:
>>> [x for x in dict_list[0] if x in dict_list[1]]
[{'1': 1, '2': 2, '3': 3}]
But this is not a very general solution, since it assumes only two nested lists are exististent.
A more general solution would be to count the occurences with collections.Counter(), and storing the dictionary items() with hashable/immutable types such as frozenset() or tuple(). Then all you need to do is filter the occurences that count more than 1.
Example:
>>> from itertools import chain
>>> from collections import Counter
>>> [dict(k) for k, v in Counter(frozenset(x.items()) for x in chain.from_iterable(dict_list)).items() if v > 1]
[{'1': 1, '2': 2, '3': 3}]
Which is very similar to the approach @Chris_Rands posted in the comments.
answered Nov 14 '18 at 10:59
RoadRunnerRoadRunner
11.2k31340
1
The main point here is that even though dictionaries are not hashable they are comparable. More about comparing dictionaries in python.
– sophros
Nov 14 '18 at 11:01
add a comment |
A list comprehension could work here:
>>> [x for x in dict_list[0] if x in dict_list[1]]
[{'1': 1, '2': 2, '3': 3}]
But this is not a very general solution, since it assumes only two nested lists are exististent.
A more general solution would be to count the occurences with collections.Counter(), and storing the dictionary items() with hashable/immutable types such as frozenset() or tuple(). Then all you need to do is filter the occurences that count more than 1.
Example:
>>> from itertools import chain
>>> from collections import Counter
>>> [dict(k) for k, v in Counter(frozenset(x.items()) for x in chain.from_iterable(dict_list)).items() if v > 1]
[{'1': 1, '2': 2, '3': 3}]
Which is very similar to the approach @Chris_Rands posted in the comments.
answered Nov 14 '18 at 10:59
RoadRunnerRoadRunner
11.2k31340
1
The main point here is that even though dictionaries are not hashable they are comparable. More about comparing dictionaries in python.
– sophros
Nov 14 '18 at 11:01
add a comment |
A list comprehension could work here:
>>> [x for x in dict_list[0] if x in dict_list[1]]
[{'1': 1, '2': 2, '3': 3}]
But this is not a very general solution, since it assumes only two nested lists are exististent.
A more general solution would be to count the occurences with collections.Counter(), and storing the dictionary items() with hashable/immutable types such as frozenset() or tuple(). Then all you need to do is filter the occurences that count more than 1.
Example:
>>> from itertools import chain
>>> from collections import Counter
>>> [dict(k) for k, v in Counter(frozenset(x.items()) for x in chain.from_iterable(dict_list)).items() if v > 1]
[{'1': 1, '2': 2, '3': 3}]
Which is very similar to the approach @Chris_Rands posted in the comments.
answered Nov 14 '18 at 10:59
RoadRunnerRoadRunner
11.2k31340
A list comprehension could work here:
>>> [x for x in dict_list[0] if x in dict_list[1]]
[{'1': 1, '2': 2, '3': 3}]
But this is not a very general solution, since it assumes only two nested lists are exististent.
A more general solution would be to count the occurences with collections.Counter(), and storing the dictionary items() with hashable/immutable types such as frozenset() or tuple(). Then all you need to do is filter the occurences that count more than 1.
Example:
>>> from itertools import chain
>>> from collections import Counter
>>> [dict(k) for k, v in Counter(frozenset(x.items()) for x in chain.from_iterable(dict_list)).items() if v > 1]
[{'1': 1, '2': 2, '3': 3}]
Which is very similar to the approach @Chris_Rands posted in the comments.
answered Nov 14 '18 at 10:59
RoadRunnerRoadRunner
11.2k31340
edited Nov 19 '18 at 15:23
answered Nov 14 '18 at 10:59
RoadRunnerRoadRunner
11.2k31340
answered Nov 14 '18 at 10:59
RoadRunnerRoadRunner
11.2k31340
answered Nov 14 '18 at 10:59
RoadRunnerRoadRunner
11.2k31340
11.2k31340
1
The main point here is that even though dictionaries are not hashable they are comparable. More about comparing dictionaries in python.
– sophros
Nov 14 '18 at 11:01
add a comment |
1
The main point here is that even though dictionaries are not hashable they are comparable. More about comparing dictionaries in python.
– sophros
Nov 14 '18 at 11:01
1
1
The main point here is that even though dictionaries are not hashable they are comparable. More about comparing dictionaries in python.
– sophros
Nov 14 '18 at 11:01
The main point here is that even though dictionaries are not hashable they are comparable. More about comparing dictionaries in python.
– sophros
Nov 14 '18 at 11:01
add a comment |
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5
[dict(k) for k,v in Counter(tuple(x.items()) for x in chain.from_iterable(dict_list)).items() if v > 1]– Chris_Rands
Nov 14 '18 at 11:00
1
@Chris_Rands I didn't see your solution at all when I posted mine. Do you want me to remove it?
– RoadRunner
Nov 14 '18 at 11:13
2
@RoadRunner no don't remove it, people do have the same idea sometimes, thanks for asking though :)
– Chris_Rands
Nov 14 '18 at 11:18
after converting a list of dict to list of tuple, he can use set.intersection too, I guess
– khelili miliana
Nov 14 '18 at 11:21