根据特定的键/值对聚类字典列表

Question

Say i have a list of dictionaries that all have the same keys. and i want to regroup these into several lists such that the values for certain attributes of my choosing are equal. here is an example:

Suppose i have the following list of dictionaries:

[  {'a': 0.0, 'b': 0.2, 'c': 0.1},
   {'a': 0.1, 'b': 0.7, 'c': 0.2},
   {'a': 0.0, 'b': 0.2, 'c': 0.3},
   {'a': 0.1, 'b': 0.7, 'c': 0.4},
   {'a': 0.0, 'b': 0.7, 'c': 0.5},
   {'a': 0.0, 'b': 0.7, 'c': 0.6}]

and i want to cluster it according the the keys a and b. Then the output would be the following list of list of dictionaries:

 [[{'a': 0.0, 'b': 0.2, 'c': 0.1},
   {'a': 0.0, 'b': 0.2, 'c': 0.3}]

   [{'a': 0.1, 'b': 0.7, 'c': 0.2},
    {'a': 0.1, 'b': 0.7, 'c': 0.4}]

   [{'a': 0.0, 'b': 0.7, 'c': 0.5},
    {'a': 0.0, 'b': 0.7, 'c': 0.6}]]

What is the best way of achieving this?

Answer 1

Sort this firstly, then use itertools.groupby .You may could try this below:

from itertools import groupby

t = [{'a': 0.0, 'b': 0.2, 'c': 0.1},
     {'a': 0.1, 'b': 0.7, 'c': 0.2},
     {'a': 0.0, 'b': 0.2, 'c': 0.3},
     {'a': 0.1, 'b': 0.7, 'c': 0.4},
     {'a': 0.0, 'b': 0.7, 'c': 0.5},
     {'a': 0.0, 'b': 0.7, 'c': 0.6}]

print([[*j] for i, j in groupby(sorted(t, key=lambda x: (x['a'], x['b'])), key=lambda x: (x['a'], x['b']))])

Result:

[[{'a': 0.0, 'b': 0.2, 'c': 0.1}, {'a': 0.0, 'b': 0.2, 'c': 0.3}], [{'a': 0.0, 'b': 0.7, 'c': 0.5}, {'a': 0.0, 'b': 0.7, 'c': 0.6}], [{'a': 0.1, 'b': 0.7, 'c': 0.2}, {'a': 0.1, 'b': 0.7, 'c': 0.4}]]

If you want to create a function to receive muitlple keys, you could try:

from itertools import groupby

def group_by(*args):
    return [[*j] for i, j in groupby(sorted(t, key=itemgetter(*args)), key=itemgetter(*args))]


t = [{'a': 0.0, 'b': 0.2, 'c': 0.1},
     {'a': 0.1, 'b': 0.7, 'c': 0.2},
     {'a': 0.0, 'b': 0.2, 'c': 0.3},
     {'a': 0.1, 'b': 0.7, 'c': 0.4},
     {'a': 0.0, 'b': 0.7, 'c': 0.5},
     {'a': 0.0, 'b': 0.7, 'c': 0.6}]

print(group_by('a', 'b'))