使用熊猫:

>>> import pandas as pd
>>> a = [1, 2, 1, 3, 3, 3, 0]
>>> pd.Series(a)[pd.Series(a).duplicated()].values
array([1, 3, 3])

我们可以使用itertools。Groupby，以便找到所有有dup的项:

from itertools import groupby

myList  = [2, 4, 6, 8, 4, 6, 12]
# when the list is sorted, groupby groups by consecutive elements which are similar
for x, y in groupby(sorted(myList)):
    #  list(y) returns all the occurences of item x
    if len(list(y)) > 1:
        print x  

输出将是:

4
6

你可以使用iteration_utilities.duplicate:

>>> from iteration_utilities import duplicates

>>> list(duplicates([1,1,2,1,2,3,4,2]))
[1, 1, 2, 2]

或者如果你只想要一个副本，可以结合iteration_utilities.unique_everseen:

>>> from iteration_utilities import unique_everseen

>>> list(unique_everseen(duplicates([1,1,2,1,2,3,4,2])))
[1, 2]

它也可以处理不可哈希的元素(但是以性能为代价):

>>> list(duplicates([[1], [2], [1], [3], [1]]))
[[1], [1]]

>>> list(unique_everseen(duplicates([[1], [2], [1], [3], [1]])))
[[1]]

这是只有少数其他方法可以处理的问题。

基准

我做了一个快速的基准测试，其中包含了这里提到的大部分(但不是全部)方法。

第一个基准测试只包含了很小范围的列表长度，因为一些方法具有O(n**2)行为。

在图表中，y轴代表时间，所以值越低越好。它还绘制了log-log，以便更好地可视化广泛的值范围:

除去O(n**2)方法，我在一个列表中做了另一个多达50万个元素的基准测试:

正如您所看到的iteration_utilities。duplicate方法比任何其他方法都快，甚至连链接unique_everseen(duplicate(…))也比其他方法更快或同样快。

这里需要注意的另一件有趣的事情是，熊猫方法对于小列表非常慢，但可以轻松地竞争较长的列表。

然而，由于这些基准测试显示大多数方法的性能大致相同，因此使用哪一种并不重要(除了有O(n**2)运行时的3种方法)。

from iteration_utilities import duplicates, unique_everseen
from collections import Counter
import pandas as pd
import itertools

def georg_counter(it):
    return [item for item, count in Counter(it).items() if count > 1]

def georg_set(it):
    seen = set()
    uniq = []
    for x in it:
        if x not in seen:
            uniq.append(x)
            seen.add(x)

def georg_set2(it):
    seen = set()
    return [x for x in it if x not in seen and not seen.add(x)]   

def georg_set3(it):
    seen = {}
    dupes = []

    for x in it:
        if x not in seen:
            seen[x] = 1
        else:
            if seen[x] == 1:
                dupes.append(x)
            seen[x] += 1

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

def moooeeeep(seq):
    seen = set()
    seen_add = seen.add
    # adds all elements it doesn't know yet to seen and all other to seen_twice
    seen_twice = set( x for x in seq if x in seen or seen_add(x) )
    # turn the set into a list (as requested)
    return list( seen_twice )

def F1Rumors_implementation(c):
    a, b = itertools.tee(sorted(c))
    next(b, None)
    r = None
    for k, g in zip(a, b):
        if k != g: continue
        if k != r:
            yield k
            r = k

def F1Rumors(c):
    return list(F1Rumors_implementation(c))

def Edward(a):
    d = {}
    for elem in a:
        if elem in d:
            d[elem] += 1
        else:
            d[elem] = 1
    return [x for x, y in d.items() if y > 1]

def wordsmith(a):
    return pd.Series(a)[pd.Series(a).duplicated()].values

def NikhilPrabhu(li):
    li = li.copy()
    for x in set(li):
        li.remove(x)

    return list(set(li))

def firelynx(a):
    vc = pd.Series(a).value_counts()
    return vc[vc > 1].index.tolist()

def HenryDev(myList):
    newList = set()

    for i in myList:
        if myList.count(i) >= 2:
            newList.add(i)

    return list(newList)

def yota(number_lst):
    seen_set = set()
    duplicate_set = set(x for x in number_lst if x in seen_set or seen_set.add(x))
    return seen_set - duplicate_set

def IgorVishnevskiy(l):
    s=set(l)
    d=[]
    for x in l:
        if x in s:
            s.remove(x)
        else:
            d.append(x)
    return d

def it_duplicates(l):
    return list(duplicates(l))

def it_unique_duplicates(l):
    return list(unique_everseen(duplicates(l)))

基准1

from simple_benchmark import benchmark
import random

funcs = [
    georg_counter, georg_set, georg_set2, georg_set3, RiteshKumar_count, moooeeeep, 
    F1Rumors, Edward, wordsmith, NikhilPrabhu, firelynx,
    HenryDev, yota, IgorVishnevskiy, it_duplicates, it_unique_duplicates
]

args = {2**i: [random.randint(0, 2**(i-1)) for _ in range(2**i)] for i in range(2, 12)}

b = benchmark(funcs, args, 'list size')

b.plot()

基准2

funcs = [
    georg_counter, georg_set, georg_set2, georg_set3, moooeeeep, 
    F1Rumors, Edward, wordsmith, firelynx,
    yota, IgorVishnevskiy, it_duplicates, it_unique_duplicates
]

args = {2**i: [random.randint(0, 2**(i-1)) for _ in range(2**i)] for i in range(2, 20)}

b = benchmark(funcs, args, 'list size')
b.plot()

免责声明

1这是我写的一个第三方库:iteration_utilities。

不需要转换为列表，可能最简单的方法是如下所示。 在面试中，当他们要求不要使用集合时，这可能会很有用

a=[1,2,3,3,3]
dup=[]
for each in a:
  if each not in dup:
    dup.append(each)
print(dup)

======= else获取唯一值和重复值的2个单独列表

a=[1,2,3,3,3]
uniques=[]
dups=[]

for each in a:
  if each not in uniques:
    uniques.append(each)
  else:
    dups.append(each)
print("Unique values are below:")
print(uniques)
print("Duplicate values are below:")
print(dups)

我会用熊猫做这个，因为我经常用熊猫

import pandas as pd
a = [1,2,3,3,3,4,5,6,6,7]
vc = pd.Series(a).value_counts()
vc[vc > 1].index.tolist()

给了

[3,6]

可能不是很有效，但它肯定比许多其他答案的代码更少，所以我想我可以贡献一下

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]