How to determine if any value occurs more than twice in a list?

You can use collections.Counter for this:

from collections import Counter
print any(count > 2 for count in Counter(myArray).itervalues())   # True

Or, if you are using Python 3:

from collections import Counter
print(any(count > 2 for count in Counter(myArray).values()))   # True

Here's one way using collections.defaultdict. Like @HoriaComan's method, this solution doesn't require iterating your entire list.

myArray = [1,1,1,1,1,1,1,1,2]

from collections import defaultdict

def count_limit(L, k=2):
    d = defaultdict(int)
    for item in L:
        if d[item] == k:
            return True
        else:
            d[item] += 1
    return False

res = count_limit(myArray)  # False

Performance benchmarking

To demonstrate the impact, we can compare with Counter on a larger iterable:

myArray = [1,1,1,1,1,1,1,1,2]*10000

from collections import defaultdict, Counter

def count_limit(L, k=2):
    d = defaultdict(int)
    for item in L:
        if d[item] == k:
            return True
        else:
            d[item] += 1
    return False

def count_limit_counter(myArray):
    return any(count > 2 for count in Counter(myArray).values())

%timeit count_limit(myArray)          # 1.52 µs per loop
%timeit count_limit_counter(myArray)  # 6.64 ms per loop

You could always construct a histogram of the values and see if any entry is greater than two. It could look something like this:

def is_more_than_twice(l):
   hist = {}
   for el in l:
       if el in hist:
           hist[el] += 1
       else:
           hist[el] = 1
       if hist[el] > 2:
           return True
   return False

You don't need to iterate until the end of the list, just until you've met the condition of seeing an element el appearing more than twice.