Swap keys for unique values in a dictionary in Python

Question

a = {0: \'PtpMotion\', 1: \'PtpMotion\', 2: \'LinMotion\', 3: \'LinMotion\', 4: \'LinMotion\', 5: \'LinMotion\', 6: \'LinMotion\', 7: \'LinMotion\', 8: \'LinMotion\', 9:         


        

Answer 1

Dictionary keys must be unique. If you wanted to keep them all you'd have to make each value for b[val] a list and add the values to those lists.

Answer 2

Perhaps you want lists in the output dictionary:

from collections import defaultdict
a = {0: 'PtpMotion', 1: 'PtpMotion', 2: 'LinMotion', 3: 'LinMotion', 4: 'LinMotion', 5: 'LinMotion', 6: 'LinMotion', 7: 'LinMotion', 8: 'LinMotion', 9: 'PtpMotion', 10: 'LinMotion', 11: 'Wait'}
b = defaultdict(list)
for key, val in a.items():
    b[val].append(key)
print b

yields:

defaultdict(<type 'list'>, {'LinMotion': [2, 3, 4, 5, 6, 7, 8, 10], 'PtpMotion': [0, 1, 9], 'Wait': [11]})

Answer 3

When you say

b[val] = key

and val already exists,it overrides the setting, getting what you see. To get all values, you must map the original values to lists of keys, such as

from collections import defaultdict

b = defaultdict(list)
for key, val in a.items():
    b[val].append(key)
print b

When I do it (python 2.5.1), I get

defaultdict(<type 'list'>, {'LinMotion': [2, 3, 4, 5, 6, 7, 8, 10], 
                            'PtpMotion': [0, 1, 9], 
                            'Wait': [11]})

Answer 4

Each key can only occur once in a dictionary. You could store a list of indices for each key:

import collections
b = collections.defaultdict(list)
for key, val in a.iteritems():
    b[val].append(key)
print b
# {'LinMotion': [2, 3, 4, 5, 6, 7, 8, 10], 'PtpMotion': [0, 1, 9], 'Wait': [11]}

Edit: As pointed out by ecik in the comments, you could also use a defaultdict(set) (and use .add() instead of .append() in the loop).