If I have a dictionary with their corresponding frequency values:
numbers = {a: 1, b: 4, c: 1, d: 3, e: 3}
To find the highest, what I know
The collections.Counter
object is useful for this as well. It gives you a .most_common()
method which will given you the keys and counts of all available values:
from collections import Counter
numbers = Counter({'a': 1, 'b': 0, 'c': 1, 'd': 3, 'e': 3})
values = list(numbers.values())
max_value = max(values)
count = values.count(max_value)
numbers.most_common(n=count)
numbers = {'a': 1, 'b': 0, 'c': 1, 'd': 3, 'e': 3}
max_value = max(numbers.values())
[k for k,v in numbers.iteritems() if v == max_value]
prints
['e', 'd']
what it does is, loop over all entries via .iteritems
and then check if the value is the maximum and if so add the key to a list.
numbers = {'a': 1, 'b': 4, 'c': 1, 'd':4 , 'e': 3}
mx_tuple = max(numbers.items(),key = lambda x:x[1]) #max function will return a (key,value) tuple of the maximum value from the dictionary
max_list =[i[0] for i in numbers.items() if i[1]==mx_tuple[1]] #my_tuple[1] indicates maximum dictionary items value
print(max_list)
This code will work in O(n). O(n) in finding maximum value and O(n) in the list comprehension. So overall it will remain O(n).
Note : O(2n) is equivalent to O(n).