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This question asks how to determine if every element in a list is the same. How would I go about determining if 95% of the elements in a list are the same in a reasonably efficient way? For example:
>>> ninety_five_same([1,1,1,1,1,1,2,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1])
True
>>> ninety_five_same([1,1,1,1,1,1,2,1]) # only 80% the same
False
This would need to be somewhat efficient because the lists may be very large.
681
You can convert the list to a set. A set cannot have duplicates. So if all the elements in the original list are identical, the set will have just one element. if len(set(input_list)) == 1: # input_list has all identical elements.
There are 2 ways to understand check if the list contains elements of another list. First, use all() functions to check if a Python list contains all the elements of another list. And second, use any() function to check if the list contains any elements of another one.
>>> from collections import Counter
>>> lst = [1,1,1,1,1,1,2,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1]
>>> _, freq = Counter(lst).most_common(1)[0]
>>> len(lst)*.95 <= freq
True
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