Get the key corresponding to the minimum value within a dictionary

Question

If I have a Python dictionary, how do I get the key to the entry which contains the minimum value?

I was thinking about something to do with the min() function...

Given the input:

{320:1, 321:0, 322:3}

It would return 321.

Answer 1

This returns the key, value pair tuple after comparing values:

>>> d = {320:1, 321:0, 322:3}
>>> d.items()
dict_items([(320, 1), (321, 0), (322, 3)])  # Python 2.7 [(320, 1), (321, 0), (322, 3)]
>>> # find the minimum by comparing the second element of each tuple
>>> min(d.items(), key=lambda x: x[1]) 
(321, 0)

For Python 2.7, use d.iteritems() for larger dictionaries as it avoid copying. Python 3's dict.items() is already an itemview so no changes need.

Answer 2

my_dic = {320:1, 321:0, 322:3}
min_value = sorted(my_dic, key=lambda k: my_dic[k])[0]
print(min_value)

A solution with only the sorted method.

1. I sorted values from smallest to largest with sorted method
2. When we get the first index, it gives the smallest key.

Answer 3

min(zip(d.values(), d.keys()))[1]

Use the zip function to create an iterator of tuples containing values and keys. Then wrap it with a min function which takes the minimum based on the first key. This returns a tuple containing (value, key) pair. The index of [1] is used to get the corresponding key.

Answer 4

To create an orderable class you have to override six special functions, so that it would be called by the min() function.

These methods are__lt__ , __le__, __gt__, __ge__, __eq__ , __ne__ in order they are less than, less than or equal, greater than, greater than or equal, equal, not equal.

For example, you should implement __lt__ as follows:

def __lt__(self, other):
  return self.comparable_value < other.comparable_value

Then you can use the min function as follows:

minValue = min(yourList, key=(lambda k: yourList[k]))

This worked for me.

Answer 5

You can get the keys of the dict using the keys function, and you're right about using min to find the minimum of that list.

_{This is an answer to the OP's original question about the minimal key, not the minimal answer.}

Answer 6

Or __getitem__:

>>> d = {320: 1, 321: 0, 322: 3}
>>> min(d, key=d.__getitem__)
321

Answer 7

Best: min(d, key=d.get) -- no reason to interpose a useless lambda indirection layer or extract items or keys!

>>> d = {320: 1, 321: 0, 322: 3}
>>> min(d, key=d.get)
321

Answer 8

For multiple keys which have equal lowest value, you can use a list comprehension:

d = {320:1, 321:0, 322:3, 323:0}

minval = min(d.values())
res = [k for k, v in d.items() if v==minval]

[321, 323]

An equivalent functional version:

res = list(filter(lambda x: d[x]==minval, d))

Answer 9

I compared how the following three options perform:

    import random, datetime

myDict = {}
for i in range( 10000000 ):
    myDict[ i ] = random.randint( 0, 10000000 )



# OPTION 1

start = datetime.datetime.now()

sorted = []
for i in myDict:
    sorted.append( ( i, myDict[ i ] ) )
sorted.sort( key = lambda x: x[1] )
print( sorted[0][0] )

end = datetime.datetime.now()
print( end - start )



# OPTION 2

start = datetime.datetime.now()

myDict_values = list( myDict.values() )
myDict_keys = list( myDict.keys() )
min_value = min( myDict_values )
print( myDict_keys[ myDict_values.index( min_value ) ] )

end = datetime.datetime.now()
print( end - start )



# OPTION 3

start = datetime.datetime.now()

print( min( myDict, key=myDict.get ) )

end = datetime.datetime.now()
print( end - start )

Sample output:

#option 1
236230
0:00:14.136808

#option 2
236230
0:00:00.458026

#option 3
236230
0:00:00.824048

Answer 10

d={}
d[320]=1
d[321]=0
d[322]=3
value = min(d.values())
for k in d.keys(): 
    if d[k] == value:
        print k,d[k]

Answer 11

For the case where you have multiple minimal keys and want to keep it simple

def minimums(some_dict):
    positions = [] # output variable
    min_value = float("inf")
    for k, v in some_dict.items():
        if v == min_value:
            positions.append(k)
        if v < min_value:
            min_value = v
            positions = [] # output variable
            positions.append(k)

    return positions

minimums({'a':1, 'b':2, 'c':-1, 'd':0, 'e':-1})

['e', 'c']

Answer 12

Use min with an iterator (for python 3 use items instead of iteritems); instead of lambda use the itemgetter from operator, which is faster than lambda.

from operator import itemgetter
min_key, _ = min(d.iteritems(), key=itemgetter(1))

Answer 13

Another approach to addressing the issue of multiple keys with the same min value:

>>> dd = {320:1, 321:0, 322:3, 323:0}
>>>
>>> from itertools import groupby
>>> from operator import itemgetter
>>>
>>> print [v for k,v in groupby(sorted((v,k) for k,v in dd.iteritems()), key=itemgetter(0)).next()[1]]
[321, 323]

Answer 14

If you are not sure that you have not multiple minimum values, I would suggest:

d = {320:1, 321:0, 322:3, 323:0}
print ', '.join(str(key) for min_value in (min(d.values()),) for key in d if d[key]==min_value)

"""Output:
321, 323
"""

Answer 15

>>> d = {320:1, 321:0, 322:3}
>>> min(d, key=lambda k: d[k]) 
321

Answer 16

# python 
d={320:1, 321:0, 322:3}
reduce(lambda x,y: x if d[x]<=d[y] else y, d.iterkeys())
  321

Answer 17

min(d.items(), key=lambda x: x[1])[0]