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]

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

将min与迭代器一起使用(对于python 3使用items而不是iteritems);使用itemgetter from操作符代替lambda，这比lambda快。

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

或__getitem__:

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

我比较了以下三个选项的表现:

    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 )

样例输出:

#option 1
236230
0:00:14.136808

#option 2
236230
0:00:00.458026

#option 3
236230
0:00:00.824048

对于具有相同最小值的多个键，您可以使用列表推导式:

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]

等价的功能版本:

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