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'
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(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.
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.
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.
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()
To get the maximum key/value of the dictionary stats:
stats = {'a':1000, 'b':3000, 'c': 100}
>>> max(stats.items(), key = lambda x: x[0])
('c', 100)
>>> 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 items() 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 lambda 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 .
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
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']
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.
Much simpler to understand approach:
mydict = { 'a':302, 'e':53, 'g':302, 'h':100 }
max_value_keys = [key for key in mydict.keys() if mydict[key] == max(mydict.values())]
print(max_value_keys) # prints a list of keys with max value
Output: ['a', 'g']
Now you can choose only one key:
maximum = mydict[max_value_keys[0]]
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.
+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])
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
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']
max(stats)? – John Redmax(stats)will use the labels as keys (it will return'c', given that's the maximum label),max(stats, key=lambda key: stats[key])is what OP was after (which will return'b', label of maximal indexed value). Is it any clearer? – Atcold