How to return a list of keys corresponding to the smallest value in dictionary

Can do it as a two-pass:

>>> colour
{'blue': 5, 'purple': 6, 'green': 2, 'red': 2}
>>> min_val = min(colour.itervalues())
>>> [k for k, v in colour.iteritems() if v == min_val]
['green', 'red']

1. Find the min value of the dict's values
2. Then go back over and extract the key where it's that value...

An alternative (requires some imports, and means you could take the n many if wanted) - this code just takes the first though (which would be the min value):

from itertools import groupby
from operator import itemgetter

ordered = sorted(colour.iteritems(), key=itemgetter(1))
bykey = groupby(ordered, key=itemgetter(1))
print map(itemgetter(0), next(bykey)[1])
# ['green', 'red']

I would say that the best option is to make two passes:

min_value = min(dict.values())
result = [key for key, value in dict.iteritems() if value == min_value]

You can make a single pass by looping explicitly:

result = []
min_value = None
for key, value in dict.iteritems():
    if min_value is None or value < min_value:
        min_value = value
        result = []
    if value == min_value:
        result.append(key)

but this is going to be slower (except may be in PyPy)