Python alternative to using List as Dictionary Key?

Question

I currently have a dictionary that maps normalized mutual information scores to aligned sites in a multiple sequence alignment, which are stored in a list. However, I've realized that two list may happen to have the same score. Thus only one list will be stored in the dictionary since the score is the key. Clusters should be unique though, so I'm not sure what to do since a my list doesn't have a hash function.

This is what my data looks like. Ideally I would simply like to reverse the keys and values.

{
0.475: [10, 11]
0.434: [12, 14]
0.404: [16, 18]
0.341: [21, 22]
}

convert to a tuple, you can't hash a list – SuperStew Jan 13 '20 at 19:12 — SuperStew, Jan 13 '20 at 19:12
and how do you build your dictionary? – marcos Jan 13 '20 at 19:15 — marcos, Jan 13 '20 at 19:15

mathfux · Answer 1 · 2020-01-13T20:01:22.990

You can use comprehension of dictionary here:

d = {0.475: [10, 11], 0.434: [12, 14], 0.404: [16, 18], 0.341: [21, 22]}
print({tuple(d[n]):n for n in d})

Output

{(10, 11): 0.475, (12, 14): 0.434, (16, 18): 0.404, (21, 22): 0.341}

You can't reverse keys and values straightforwardly because lists are unhashable and therefore can't be kept as keys.

Update

In case you have duplicated lists, you might like to create enumeration of repeated lists. You can do it like this:

from collections import defaultdict
d = {0.475: [10, 11], 0.434: [12, 14], 0.404: [16, 18], 0.341: [12, 14]}
new_d = dict()
d_counts = defaultdict(int)
for n in d.keys():
    value = tuple(d[n])
    new_d[(value, d_counts[value])] = n
    d_counts[value] += 1
print(new_d)

Output2

{((10, 11), 0): 0.475, ((12, 14), 0): 0.434, ((16, 18), 0): 0.404, ((12, 14), 1): 0.341}

but the duplicates have already been lost when they built the original dict — Chris_Rands, Jan 13 '20 at 19:24

score 1 · Accepted Answer · edited Jan 13 '20 at 19:38

1

mydict = {0.475: [10, 11],
          0.434: [12, 14],
          0.404: [16, 18],
          0.341: [21, 22]}

newdict = dict()

for k, v in mydict.items():
    newdict[tuple(v)] = k

print(newdict)
# {(10, 11): 0.475, (12, 14): 0.434, (16, 18): 0.404, (21, 22): 0.341}

edited Jan 13 '20 at 19:38

wjandrea

28,235
9
60
81

answered Jan 13 '20 at 19:27

tbrk

156
8

A dict comprehension is simpler: `newdict = {tuple(v): k for k, v in mydict.items()}` – wjandrea Jan 13 '20 at 19:39
Same problem as @mathfux, this does not fix duplicate keys on original dict. – Gocht Jan 13 '20 at 19:43

score 0 · Answer 3 · answered Jan 13 '20 at 19:33

From this answer, you could try to create a new class from built-in dict:

class Dictlist(dict):
    def __setitem__(self, key, value):
        try:
            self[key]
        except KeyError:
            super(Dictlist, self).__setitem__(key, [])
        self[key].append(value)

So...

>>> d = Dictlist()
>>> d['100'] = [1, 2]
>>> d['100'] = [3, 4]
>>> d
{'100': [[1, 2], [3, 4]]}

score 0 · Answer 4 · answered Jan 20 '20 at 01:44

0

Dictionary comprehension should do it with no problem.

my_dict = {0.475: [10, 11], 0.434: [12, 14], 0.404: [16, 18], 0.341: [21, 22]}

my_dict = {my_dict[k]:k for k in my_dict}

Note: if you want to have an if-statement inside of it, you are able to.

answered Jan 20 '20 at 01:44

metageni

11
2

Python alternative to using List as Dictionary Key?

4 Answers4

Output

Update

Output2