获取字典中具有最大值的键？

I have a dictionary: keys are strings, values are integers.

Example:

stats = {'a':1000, 'b':3000, 'c': 100}

I’d like to get 'b' as an answer, since it’s the key with a higher value.

I did the following, using an intermediate list with reversed key-value tuples:

inverse = [(value, key) for key, value in stats.items()]
print max(inverse)[1]

Is that one the better (or even more elegant) approach?

You can use operator.itemgetter for that:

import operator
stats = {'a':1000, 'b':3000, 'c': 100}
max(stats.iteritems(), key=operator.itemgetter(1))[0]

And instead of building a new list in memory use stats.iteritems(). The key parameter to the max() function is a function that computes a key that is used to determine how to rank items.

Please note that if you were to have another key-value pair ‘d’: 3000 that this method will only return one of the two even though they both have the maximum value.

>>> import operator
>>> stats = {'a':1000, 'b':3000, 'c': 100, 'd':3000}
>>> max(stats.iteritems(), key=operator.itemgetter(1))[0]
'b' 

If using Python3:

>>> max(stats.items(), key=operator.itemgetter(1))[0]
'b'

max(stats, key=stats.get)

I have tested MANY variants, and this is the fastest way to return the key of dict with the max value:

def keywithmaxval(d):
     """ a) create a list of the dict's keys and values; 
         b) return the key with the max value"""  
     v=list(d.values())
     k=list(d.keys())
     return k[v.index(max(v))]

To give you an idea, here are some candidate methods:

def f1():  
     v=list(d1.values())
     k=list(d1.keys())
     return k[v.index(max(v))]

def f2():
    d3={v:k for k,v in d1.items()}
    return d3[max(d3)]

def f3():
    return list(filter(lambda t: t[1]==max(d1.values()), d1.items()))[0][0]    

def f3b():
    # same as f3 but remove the call to max from the lambda
    m=max(d1.values())
    return list(filter(lambda t: t[1]==m, d1.items()))[0][0]        

def f4():
    return [k for k,v in d1.items() if v==max(d1.values())][0]    

def f4b():
    # same as f4 but remove the max from the comprehension
    m=max(d1.values())
    return [k for k,v in d1.items() if v==m][0]        

def f5():
    return max(d1.items(), key=operator.itemgetter(1))[0]    

def f6():
    return max(d1,key=d1.get)     

def f7():
     """ a) create a list of the dict's keys and values; 
         b) return the key with the max value"""    
     v=list(d1.values())
     return list(d1.keys())[v.index(max(v))]    

def f8():
     return max(d1, key=lambda k: d1[k])     

tl=[f1,f2, f3b, f4b, f5, f6, f7, f8, f4,f3]     
cmpthese.cmpthese(tl,c=100) 

The test dictionary:

d1={1: 1, 2: 2, 3: 8, 4: 3, 5: 6, 6: 9, 7: 17, 8: 4, 9: 20, 10: 7, 11: 15, 
    12: 10, 13: 10, 14: 18, 15: 18, 16: 5, 17: 13, 18: 21, 19: 21, 20: 8, 
    21: 8, 22: 16, 23: 16, 24: 11, 25: 24, 26: 11, 27: 112, 28: 19, 29: 19, 
    30: 19, 3077: 36, 32: 6, 33: 27, 34: 14, 35: 14, 36: 22, 4102: 39, 38: 22, 
    39: 35, 40: 9, 41: 110, 42: 9, 43: 30, 44: 17, 45: 17, 46: 17, 47: 105, 48: 12, 
    49: 25, 50: 25, 51: 25, 52: 12, 53: 12, 54: 113, 1079: 50, 56: 20, 57: 33, 
    58: 20, 59: 33, 60: 20, 61: 20, 62: 108, 63: 108, 64: 7, 65: 28, 66: 28, 67: 28, 
    68: 15, 69: 15, 70: 15, 71: 103, 72: 23, 73: 116, 74: 23, 75: 15, 76: 23, 77: 23, 
    78: 36, 79: 36, 80: 10, 81: 23, 82: 111, 83: 111, 84: 10, 85: 10, 86: 31, 87: 31, 
    88: 18, 89: 31, 90: 18, 91: 93, 92: 18, 93: 18, 94: 106, 95: 106, 96: 13, 9232: 35, 
    98: 26, 99: 26, 100: 26, 101: 26, 103: 88, 104: 13, 106: 13, 107: 101, 1132: 63, 
    2158: 51, 112: 21, 113: 13, 116: 21, 118: 34, 119: 34, 7288: 45, 121: 96, 122: 21, 
    124: 109, 125: 109, 128: 8, 1154: 32, 131: 29, 134: 29, 136: 16, 137: 91, 140: 16, 
    142: 104, 143: 104, 146: 117, 148: 24, 149: 24, 152: 24, 154: 24, 155: 86, 160: 11, 
    161: 99, 1186: 76, 3238: 49, 167: 68, 170: 11, 172: 32, 175: 81, 178: 32, 179: 32, 
    182: 94, 184: 19, 31: 107, 188: 107, 190: 107, 196: 27, 197: 27, 202: 27, 206: 89, 
    208: 14, 214: 102, 215: 102, 220: 115, 37: 22, 224: 22, 226: 14, 232: 22, 233: 84, 
    238: 35, 242: 97, 244: 22, 250: 110, 251: 66, 1276: 58, 256: 9, 2308: 33, 262: 30, 
    263: 79, 268: 30, 269: 30, 274: 92, 1300: 27, 280: 17, 283: 61, 286: 105, 292: 118, 
    296: 25, 298: 25, 304: 25, 310: 87, 1336: 71, 319: 56, 322: 100, 323: 100, 325: 25, 
    55: 113, 334: 69, 340: 12, 1367: 40, 350: 82, 358: 33, 364: 95, 376: 108, 
    377: 64, 2429: 46, 394: 28, 395: 77, 404: 28, 412: 90, 1438: 53, 425: 59, 430: 103, 
    1456: 97, 433: 28, 445: 72, 448: 23, 466: 85, 479: 54, 484: 98, 485: 98, 488: 23, 
    6154: 37, 502: 67, 4616: 34, 526: 80, 538: 31, 566: 62, 3644: 44, 577: 31, 97: 119, 
    592: 26, 593: 75, 1619: 48, 638: 57, 646: 101, 650: 26, 110: 114, 668: 70, 2734: 41, 
    700: 83, 1732: 30, 719: 52, 728: 96, 754: 65, 1780: 74, 4858: 47, 130: 29, 790: 78, 
    1822: 43, 2051: 38, 808: 29, 850: 60, 866: 29, 890: 73, 911: 42, 958: 55, 970: 99, 
    976: 24, 166: 112}

And the test results under Python 3.2:

    rate/sec       f4      f3    f3b     f8     f5     f2    f4b     f6     f7     f1
f4       454       --   -2.5% -96.9% -97.5% -98.6% -98.6% -98.7% -98.7% -98.9% -99.0%
f3       466     2.6%      -- -96.8% -97.4% -98.6% -98.6% -98.6% -98.7% -98.9% -99.0%
f3b   14,715  3138.9% 3057.4%     -- -18.6% -55.5% -56.0% -56.4% -58.3% -63.8% -68.4%
f8    18,070  3877.3% 3777.3%  22.8%     -- -45.4% -45.9% -46.5% -48.8% -55.5% -61.2%
f5    33,091  7183.7% 7000.5% 124.9%  83.1%     --  -1.0%  -2.0%  -6.3% -18.6% -29.0%
f2    33,423  7256.8% 7071.8% 127.1%  85.0%   1.0%     --  -1.0%  -5.3% -17.7% -28.3%
f4b   33,762  7331.4% 7144.6% 129.4%  86.8%   2.0%   1.0%     --  -4.4% -16.9% -27.5%
f6    35,300  7669.8% 7474.4% 139.9%  95.4%   6.7%   5.6%   4.6%     -- -13.1% -24.2%
f7    40,631  8843.2% 8618.3% 176.1% 124.9%  22.8%  21.6%  20.3%  15.1%     -- -12.8%
f1    46,598 10156.7% 9898.8% 216.7% 157.9%  40.8%  39.4%  38.0%  32.0%  14.7%     --

And under Python 2.7:

    rate/sec       f3       f4     f8    f3b     f6     f5     f2    f4b     f7     f1
f3       384       --    -2.6% -97.1% -97.2% -97.9% -97.9% -98.0% -98.2% -98.5% -99.2%
f4       394     2.6%       -- -97.0% -97.2% -97.8% -97.9% -98.0% -98.1% -98.5% -99.1%
f8    13,079  3303.3%  3216.1%     --  -5.6% -28.6% -29.9% -32.8% -38.3% -49.7% -71.2%
f3b   13,852  3504.5%  3412.1%   5.9%     -- -24.4% -25.8% -28.9% -34.6% -46.7% -69.5%
f6    18,325  4668.4%  4546.2%  40.1%  32.3%     --  -1.8%  -5.9% -13.5% -29.5% -59.6%
f5    18,664  4756.5%  4632.0%  42.7%  34.7%   1.8%     --  -4.1% -11.9% -28.2% -58.8%
f2    19,470  4966.4%  4836.5%  48.9%  40.6%   6.2%   4.3%     --  -8.1% -25.1% -57.1%
f4b   21,187  5413.0%  5271.7%  62.0%  52.9%  15.6%  13.5%   8.8%     -- -18.5% -53.3%
f7    26,002  6665.8%  6492.4%  98.8%  87.7%  41.9%  39.3%  33.5%  22.7%     -- -42.7%
f1    45,354 11701.5% 11399.0% 246.8% 227.4% 147.5% 143.0% 132.9% 114.1%  74.4%     -- 

You can see that f1 is the fastest under Python 3.2 and 2.7 (or, more completely, keywithmaxval at the top of this post)

If you need to know only a key with the max value you can do it without iterkeys or iteritems because iteration through dictionary in Python is iteration through it’s keys.

max_key = max(stats, key=lambda k: stats[k])

EDIT:

From comments, @user1274878 :

I am new to python. Can you please explain your answer in steps?

Yep…

max

max(iterable[, key])

max(arg1, arg2, *args[, key])

Return the largest item in an iterable or the largest of two or more arguments.

The optional key argument describes how to compare elements to get maximum among them:

lambda <item>: return <a result of operation with item> 

Returned values will be compared.

Dict

Python dict is a hash table. A key of dict is a hash of an object declared as a key. Due to performance reasons iteration though a dict implemented as iteration through it’s keys.

Therefore we can use it to rid operation of obtaining a keys list.

Closure

A function defined inside another function is called a nested function. Nested functions can access variables of the enclosing scope.

The stats variable available through __closure__ attribute of the lambda function as a pointer to the value of the variable defined in the parent scope.

Example:

stats = {'a':1000, 'b':3000, 'c': 100}

if you wanna find the max value with its key, maybe follwing could be simple, without any relevant functions.

max(stats, key=stats.get)

the output is the key which has the max value.

Here is another one:

stats = {'a':1000, 'b':3000, 'c': 100}
max(stats.iterkeys(), key=lambda k: stats[k])

The function key simply returns the value that should be used for ranking and max() returns the demanded element right away.

key, value = max(stats.iteritems(), key=lambda x:x[1])

If you don’t care about value (I’d be surprised, but) you can do:

key, _ = max(stats.iteritems(), key=lambda x:x[1])

I like the tuple unpacking better than a [0] subscript at the end of the expression. I never like the readability of lambda expressions very much, but find this one better than the operator.itemgetter(1) IMHO.

Given that more than one entry my have the max value. I would make a list of the keys that have the max value as their value.

>>> stats = {'a':1000, 'b':3000, 'c': 100, 'd':3000}
>>> [key for m in [max(stats.values())] for key,val in stats.iteritems() if val == m]
['b', 'd']

This will give you ‘b’ and any other max key as well.

Note: For python 3 use stats.items() instead of stats.iteritems()

You can use:

max(d, key = d.get) 
# which is equivalent to 
max(d, key = lambda k : d.get(k))

To return the key, value pair use:

max(d.items(), key = lambda k : k[1])

To get the maximum key/value of the dictionary stats:

stats = {'a':1000, 'b':3000, 'c': 100}

• Based on keys

>>> max(stats.items(), key = lambda x: x[0]) ('c', 100)

• Based on values

>>> max(stats.items(), key = lambda x: x[1]) ('b', 3000)

Of course, if you want to get only the key or value from the result, you can use tuple indexing. For Example, to get the key corresponding to the maximum value:

>>> max(stats.items(), key = lambda x: x[1])[0] 'b'

Explanation

The dictionary method in Python 3 returns a view object of the dictionary. When this view object is iterated over, by the max function, it yields the dictionary items as tuples of the form (key, value).

>>> list(stats.items()) [('c', 100), ('b', 3000), ('a', 1000)]

When you use the expression lambda x: x[1], in each iteration, x is one of these tuples (key, value). So, by choosing the right index, you select whether you want to compare by keys or by values.

Python 2

For Python 2.2+ releases, the same code will work. However, it is better to use iteritems() dictionary method instead of items() for performance.

Notes

• This answer is based on the comments on Climbs_lika_Spyder’s answer.

• The used code was tested on Python 3.5.2 and Python 2.7.10 .

d = {'A': 4,'B':10}

min_v = min(zip(d.values(), d.keys()))
# min_v is (4,'A')

max_v = max(zip(d.values(), d.keys()))
# max_v is (10,'B')

Per the iterated solutions via comments in the selected answer…

In Python 3:

max(stats.keys(), key=(lambda k: stats[k]))

In Python 2:

max(stats.iterkeys(), key=(lambda k: stats[k]))

I got here looking for how to return mydict.keys() based on the value of mydict.values(). Instead of just the one key returned, I was looking to return the top x number of values.

This solution is simpler than using the max() function and you can easily change the number of values returned:

stats = {'a':1000, 'b':3000, 'c': 100}

x = sorted(stats, key=(lambda key:stats[key]), reverse=True)
['b', 'a', 'c']

If you want the single highest ranking key, just use the index:

x[0]
['b']

If you want the top two highest ranking keys, just use list slicing:

x[:2]
['b', 'a']

I was not satisfied with any of these answers. max always picks the first key with the max value. The dictionary could have multiple keys with that value.

def keys_with_top_values(my_dict):
    return [key  for (key, value) in my_dict.items() if value == max(my_dict.values())]

Posting this answer in case it helps someone out. See the below SO post

Which maximum does Python pick in the case of a tie?

With collections.Counter you could do

>>> import collections
>>> stats = {'a':1000, 'b':3000, 'c': 100}
>>> stats = collections.Counter(stats)
>>> stats.most_common(1)
[('b', 3000)]

If appropriate, you could simply start with an empty collections.Counter and add to it

>>> stats = collections.Counter()
>>> stats['a'] += 1
:
etc. 

A heap queue is a generalised solution which allows you to extract the top n keys ordered by value:

from heapq import nlargest

stats = {'a':1000, 'b':3000, 'c': 100}

res1 = nlargest(1, stats, key=stats.__getitem__)  # ['b']
res2 = nlargest(2, stats, key=stats.__getitem__)  # ['b', 'a']

res1_val = next(iter(res1))                       # 'b'

Note dict.__getitem__ is the method called by the syntactic sugar dict[]. As opposed to dict.get, it will return KeyError if a key is not found, which here cannot occur.

max((value, key) for key, value in stats.items())[1]

+1 to @Aric Coady‘s simplest solution.
And also one way to random select one of keys with max value in the dictionary:

stats = {'a':1000, 'b':3000, 'c': 100, 'd':3000}

import random
maxV = max(stats.values())
# Choice is one of the keys with max value
choice = random.choice([key for key, value in stats.items() if value == maxV])

Counter = 0
for word in stats.keys():
    if stats[word]> counter:
        Counter = stats [word]
print Counter

How about:

 max(zip(stats.keys(), stats.values()), key=lambda t : t[1])[0]

I tested the accepted answer AND @thewolf’s fastest solution against a very basic loop and the loop was faster than both:

import time
import operator


d = {"a"+str(i): i for i in range(1000000)}

def t1(dct):
    mx = float("-inf")
    key = None
    for k,v in dct.items():
        if v > mx:
            mx = v
            key = k
    return key

def t2(dct):
    v=list(dct.values())
    k=list(dct.keys())
    return k[v.index(max(v))]

def t3(dct):
    return max(dct.items(),key=operator.itemgetter(1))[0]

start = time.time()
for i in range(25):
    m = t1(d)
end = time.time()
print ("Iterating: "+str(end-start))

start = time.time()
for i in range(25):
    m = t2(d)
end = time.time()
print ("List creating: "+str(end-start))

start = time.time()
for i in range(25):
    m = t3(d)
end = time.time()
print ("Accepted answer: "+str(end-start))

results:

Iterating: 3.8201940059661865
List creating: 6.928712844848633
Accepted answer: 5.464320182800293

For scientific python users, here is a simple solution using Pandas:

import pandas as pd
stats = {'a': 1000, 'b': 3000, 'c': 100}
series = pd.Series(stats)
series.idxmax()

>>> b

In the case you have more than one key with the same value, for example:

stats = {'a':1000, 'b':3000, 'c': 100, 'd':3000, 'e':3000}

You could get a collection with all the keys with max value as follow:

from collections import defaultdict
from collections import OrderedDict

groupedByValue = defaultdict(list)
for key, value in sorted(stats.items()):
    groupedByValue[value].append(key)

# {1000: ['a'], 3000: ['b', 'd', 'e'], 100: ['c']}

groupedByValue[max(groupedByValue)]
# ['b', 'd', 'e']

Much simpler to understand approach:

dict = { 'a':302, 'e':53, 'g':302, 'h':100 }
max_value_keys = [key for key in dict.keys() if dict[key] == max(dict.values())]
print(max_value_keys) # prints a list of keys with max value

Output: [‘a’, ‘g’]

Now you can choose only one key:

maximum = dict[max_value_keys[0]]