如何确定列表中是否有任何值超过两次？

Question

I have a list and would like to determine if any value occurs more than twice. I've tried using collections and counter but I cannot get those to evaluate to a single True or False value.

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

I would like it to return: True if any value occurs more than twice .

Any help is appreciated and it would really help if the solution was fast. I'm checking hundreds of thousands of lists. I'm new to programming and this is my first post.

EDIT: my attempts, also I'm new to stackoverflow UI

import collections

arr= [1,2,3,5,6]

Counter(arr)

Returns: Counter({1: 1, 2: 1, 3: 1, 5: 1, 6: 1})

Answer 1

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

Answer 2

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.

Answer 3

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

Answer 4

Try it using set()

def OccursMoreThanTwice(myArray):
    for e in myArray:
        if myArray.count(e) > 2:
           return True
    return False

print OccursMoreThanTwice([1,1,1,1,1,1,1,1,2])