另一种解决方案如下所示，不使用任何集合库。

a = [1,2,3,5,4,6,4,21,4,6,3,32,5,2,23,5]
duplicates = []

for i in a:
    if a.count(i) > 1 and i not in duplicates:
        duplicates.append(i)

print(duplicates)

输出是[2,3,5,4,6]

还有其他测试。当然要做……

set([x for x in l if l.count(x) > 1])

.．.代价太大了。使用下一个final方法大约快500倍(数组越长结果越好):

def dups_count_dict(l):
    d = {}

    for item in l:
        if item not in d:
            d[item] = 0

        d[item] += 1

    result_d = {key: val for key, val in d.iteritems() if val > 1}

    return result_d.keys()

只有2个循环，没有非常昂贵的l.count()操作。

下面是一个比较方法的代码。代码如下，输出如下:

dups_count: 13.368s # this is a function which uses l.count()
dups_count_dict: 0.014s # this is a final best function (of the 3 functions)
dups_count_counter: 0.024s # collections.Counter

测试代码:

import numpy as np
from time import time
from collections import Counter

class TimerCounter(object):
    def __init__(self):
        self._time_sum = 0

    def start(self):
        self.time = time()

    def stop(self):
        self._time_sum += time() - self.time

    def get_time_sum(self):
        return self._time_sum


def dups_count(l):
    return set([x for x in l if l.count(x) > 1])


def dups_count_dict(l):
    d = {}

    for item in l:
        if item not in d:
            d[item] = 0

        d[item] += 1

    result_d = {key: val for key, val in d.iteritems() if val > 1}

    return result_d.keys()


def dups_counter(l):
    counter = Counter(l)    

    result_d = {key: val for key, val in counter.iteritems() if val > 1}

    return result_d.keys()



def gen_array():
    np.random.seed(17)
    return list(np.random.randint(0, 5000, 10000))


def assert_equal_results(*results):
    primary_result = results[0]
    other_results = results[1:]

    for other_result in other_results:
        assert set(primary_result) == set(other_result) and len(primary_result) == len(other_result)


if __name__ == '__main__':
    dups_count_time = TimerCounter()
    dups_count_dict_time = TimerCounter()
    dups_count_counter = TimerCounter()

    l = gen_array()

    for i in range(3):
        dups_count_time.start()
        result1 = dups_count(l)
        dups_count_time.stop()

        dups_count_dict_time.start()
        result2 = dups_count_dict(l)
        dups_count_dict_time.stop()

        dups_count_counter.start()
        result3 = dups_counter(l)
        dups_count_counter.stop()

        assert_equal_results(result1, result2, result3)

    print 'dups_count: %.3f' % dups_count_time.get_time_sum()
    print 'dups_count_dict: %.3f' % dups_count_dict_time.get_time_sum()
    print 'dups_count_counter: %.3f' % dups_count_counter.get_time_sum()

假设我们有这个元素列表:

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

我们可以只使用集合来找到唯一的元素:

unique = set()
for num in a:
    if num not in unique:
        unique.add(num)
    else:
        unique = unique - set([num])

最后:

>>> unique
{3, 6}

如果你想要得到副本，你可以简单地做:

>>> duplicates = set(a) - unique
>>> duplicates
{1, 2, 5}

注:

集合中的元素查找是O(1) 从集合中移除的元素是O(1)

试试这个检查副本

>>> def checkDuplicate(List):
    duplicate={}
    for i in List:
            ## checking whether the item is already present in dictionary or not
            ## increasing count if present
            ## initializing count to 1 if not present

        duplicate[i]=duplicate.get(i,0)+1

    return [k for k,v in duplicate.items() if v>1]

>>> checkDuplicate([1,2,3,"s",1,2,3])
[1, 2, 3]

使用sort()函数。重复项可以通过遍历它并检查l1[i] == l1[i+1]来识别。

使用toolz时:

from toolz import frequencies, valfilter

a = [1,2,2,3,4,5,4]
>>> list(valfilter(lambda count: count > 1, frequencies(a)).keys())
[2,4]