Given a string, remove any characters that are unique from the string.

Example: 

input: "abccdefee"

output: "cceee"

Difficulty: EASY
Starter code: def only_duplicates(string):
	

Solutions:
from collections import Counter

def only_duplicates(string):
	cs = Counter(string)
	return ''.join((c for c in string if cs[c] > 1))


def only_duplicates(string):
	return ''.join([x for x in string if string.count(x) > 1])


only_duplicates = lambda s: ''.join((c for c in s if s.count(c) > 1))


def only_duplicates(stg):
	return ''.join((c for c in stg if stg.count(c) > 1))


def only_duplicates(string):
	return ''.join((s for s in string if string.count(s) > 1))


def only_duplicates(string):
	sets = (unique, duplicate) = (set(), set())
	for c in string:
		sets[c in unique].add(c)
	return ''.join((c for c in string if c in duplicate))


def only_duplicates(a):
	b = []
	for j in range(0, len(a)):
		for i in range(j + 1, len(a)):
			if a[j] == a[i]:
				b.append(a[j])
	a = list(a)

	def seclect(i):
		for j in b:
			if j == i:
				return i
	return ''.join(filter(seclect, a))


def only_duplicates(string):
	string = list(string)
	foundStrings = []
	finalString = []
	for (index, item) in enumerate(string):
		if item not in string[index + 1:len(string)] and item not in string[0:index]:
			finalString.append('')
		else:
			finalString.append(item)
	return ''.join(finalString)


from collections import Counter
from re import sub

def only_duplicates(s):
	return sub('[%s]' % ''.join((k for (k, v) in Counter(s).items() if v == 1)), '', s)