Group list of dictionaries by value

Question

I have a list of dictionaries. How can i group that list by valaues.

list = [{a:1},{b:2},{c:1},{d:3},{e:2}]

Now my result should be like below

1:a,c
2:b,e
3:d

I tried using groupby from itertools. But i couldn't get the required result. I am using python 2.7.

Could you help me achieve this?

Don't name things like builtin types (`list`), that can have bad side effects. Now, you can either go the "how can I use `itertools.groupby`" way or you could roll your own. In either case, you will have to be much more specific in your question and also show some effort. Remember, nobody here is going to do your homework for you. — Ulrich Eckhardt, Mar 09 '18 at 09:38
Goupby only works if the data is sorted by the same key function, so sort first. — tobias_k, Mar 09 '18 at 09:47
Also, why do you have a list of dicts, each having a single, unique key? Can't you just have a single dict? — tobias_k, Mar 09 '18 at 09:49

score 7 · Accepted Answer · answered Mar 09 '18 at 09:55

If you want to use groupby, the list has to be sorted by the same key you want to group by.

>>> lst = [{'a':1}, {'b':2}, {'c':1}, {'d':3}, {'e':2}]

>>> keyfunc = lambda d: next(iter(d.values()))

>>> sorted(lst, key=keyfunc)
[{'a': 1}, {'c': 1}, {'b': 2}, {'e': 2}, {'d': 3}]

>>> {k: [x for d in g for x in d] 
...  for k, g in itertools.groupby(sorted(lst, key=keyfunc), key=keyfunc)}
{1: ['a', 'c'], 2: ['b', 'e'], 3: ['d']}

score 4 · Answer 2 · answered Mar 09 '18 at 10:13

Here's a possible solution without using any library.

def get_dict(list):
    res = {}

    for elem in list:
        k, v = elem.keys(), elem.values()

        if v[0] in res:
            res[v[0]].append(k[0])
        else:
            res[v[0]] = [k[0]]

    return res

With a list like yours, this would output a dictionary with the following format:

{ 1:[a,c], 2:[b, e], 3:[c] }

This is considering you're always going to have the same format as input. If not, you could just adjust what is read and saved.

Rakesh · Answer 3 · 2018-03-09T12:13:48.577

2

This might help.

list = [{"a":1},{"b":2},{"c":1},{"d":3},{"e":2}]
d = {}
for i in list:
    key, value = i.items()[0]
    if value not in d:
        d[value] = [key]
    else:
        d[value].append(key)
print(d)

Output:

{1: ['a', 'c'], 2: ['b', 'e'], 3: ['d']}

Tested in python2.7

edited Mar 09 '18 at 12:13

answered Mar 09 '18 at 09:53

Rakesh

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Fails in Py 3.. – Chris_Rands Mar 09 '18 at 09:54
Do you get an error? – Rakesh Mar 09 '18 at 09:54
Well I haven't tried it but you can't index `dict_items`, of course `next(iter())` syntax is possible – Chris_Rands Mar 09 '18 at 09:55

score 0 · Answer 4 · answered Mar 09 '18 at 09:55

Here is a way to do what you are looking for:

list_ = [{"a":1},{"b":2},{"c":1},{"d":3},{"e":2}]
values = set(value for dic in list_ for value in dic.values())

for value in values:
  keys = [list(dic.keys())[0] for dic in list_ if value in dic.values()]
  print("{}: {}".format(value, keys))

Output:

1: ['a', 'c']
2: ['b', 'e']
3: ['d']

score 0 · Answer 5 · answered Mar 09 '18 at 09:59

Here's a solution that uses defaultdict.

from __future__ import print_function
from collections import defaultdict

lst = [{'a': 1}, {'b': 2}, {'c': 1}, {'d': 3}, {'e': 2}]
d = defaultdict(list)

for l in lst:
    val, key = l.items()[0]
    d[key].append(val)

print(d)

Output: defaultdict(<type 'list'>, {1: ['a', 'c'], 2: ['b', 'e'], 3: ['d']})

Group list of dictionaries by value

5 Answers5