这样可以得到两个列表的交集，也可以得到公共重复项。

>>> from collections import Counter
>>> a = Counter([1,2,3,4,5])
>>> b = Counter([1,3,5,6])
>>> a &= b
>>> list(a.elements())
[1, 3, 5]

这是一个示例，当您需要在结果中的每个元素出现的次数应该与它在两个数组中显示的次数相同。

def intersection(nums1, nums2):
    #example:
    #nums1 = [1,2,2,1]
    #nums2 = [2,2]
    #output = [2,2]
    #find first 2 and remove from target, continue iterating

    target, iterate = [nums1, nums2] if len(nums2) >= len(nums1) else [nums2, nums1] #iterate will look into target

    if len(target) == 0:
            return []

    i = 0
    store = []
    while i < len(iterate):

         element = iterate[i]

         if element in target:
               store.append(element)
               target.remove(element)

         i += 1


    return store

下面是一些Python 2 / Python 3代码，用于为基于列表和基于集的方法生成时间信息，以查找两个列表的交集。

纯粹的列表理解算法是O(n²)，因为在列表上进行搜索是一个线性搜索。基于集合的算法是O(n)，因为集合搜索是O(1)，集合创建是O(n)(将集合转换为列表也是O(n))。因此，对于足够大的n，基于集合的算法更快，但对于较小的n，创建集合的开销使它们比纯列表比较算法慢。

#!/usr/bin/env python

''' Time list- vs set-based list intersection
    See http://stackoverflow.com/q/3697432/4014959
    Written by PM 2Ring 2015.10.16
'''

from __future__ import print_function, division
from timeit import Timer

setup = 'from __main__ import a, b'
cmd_lista = '[u for u in a if u in b]'
cmd_listb = '[u for u in b if u in a]'

cmd_lcsa = 'sa=set(a);[u for u in b if u in sa]'

cmd_seta = 'list(set(a).intersection(b))'
cmd_setb = 'list(set(b).intersection(a))'

reps = 3
loops = 50000

def do_timing(heading, cmd, setup):
    t = Timer(cmd, setup)
    r = t.repeat(reps, loops)
    r.sort()
    print(heading, r)
    return r[0]

m = 10
nums = list(range(6 * m))

for n in range(1, m + 1):
    a = nums[:6*n:2]
    b = nums[:6*n:3]
    print('\nn =', n, len(a), len(b))
    #print('\nn = %d\n%s %d\n%s %d' % (n, a, len(a), b, len(b)))
    la = do_timing('lista', cmd_lista, setup) 
    lb = do_timing('listb', cmd_listb, setup) 
    lc = do_timing('lcsa ', cmd_lcsa, setup)
    sa = do_timing('seta ', cmd_seta, setup)
    sb = do_timing('setb ', cmd_setb, setup)
    print(la/sa, lb/sa, lc/sa, la/sb, lb/sb, lc/sb)

输出

n = 1 3 2
lista [0.082171916961669922, 0.082588911056518555, 0.0898590087890625]
listb [0.069530963897705078, 0.070394992828369141, 0.075379848480224609]
lcsa  [0.11858987808227539, 0.1188349723815918, 0.12825107574462891]
seta  [0.26900982856750488, 0.26902294158935547, 0.27298116683959961]
setb  [0.27218389511108398, 0.27459001541137695, 0.34307217597961426]
0.305460649521 0.258469975867 0.440838458259 0.301898526833 0.255455833892 0.435697630214

n = 2 6 4
lista [0.15915989875793457, 0.16000485420227051, 0.16551494598388672]
listb [0.13000702857971191, 0.13060092926025391, 0.13543915748596191]
lcsa  [0.18650484085083008, 0.18742108345031738, 0.19513416290283203]
seta  [0.33592700958251953, 0.34001994132995605, 0.34146714210510254]
setb  [0.29436492919921875, 0.2953648567199707, 0.30039691925048828]
0.473793098554 0.387009751735 0.555194537893 0.540689066428 0.441652573672 0.633583767462

n = 3 9 6
lista [0.27657914161682129, 0.28098297119140625, 0.28311991691589355]
listb [0.21585917472839355, 0.21679902076721191, 0.22272896766662598]
lcsa  [0.22559309005737305, 0.2271728515625, 0.2323150634765625]
seta  [0.36382699012756348, 0.36453008651733398, 0.36750602722167969]
setb  [0.34979605674743652, 0.35533690452575684, 0.36164689064025879]
0.760194128313 0.59330170819 0.62005595016 0.790686848184 0.61710008036 0.644927481902

n = 4 12 8
lista [0.39616990089416504, 0.39746403694152832, 0.41129183769226074]
listb [0.33485794067382812, 0.33914685249328613, 0.37850618362426758]
lcsa  [0.27405810356140137, 0.2745978832244873, 0.28249192237854004]
seta  [0.39211201667785645, 0.39234519004821777, 0.39317893981933594]
setb  [0.36988520622253418, 0.37011313438415527, 0.37571001052856445]
1.01034878821 0.85398540833 0.698928091731 1.07106176249 0.905302334456 0.740927452493

n = 5 15 10
lista [0.56792402267456055, 0.57422614097595215, 0.57740211486816406]
listb [0.47309303283691406, 0.47619009017944336, 0.47628307342529297]
lcsa  [0.32805585861206055, 0.32813096046447754, 0.3349759578704834]
seta  [0.40036201477050781, 0.40322518348693848, 0.40548801422119141]
setb  [0.39103078842163086, 0.39722800254821777, 0.43811702728271484]
1.41852623806 1.18166313332 0.819398061028 1.45237674242 1.20986133789 0.838951479847

n = 6 18 12
lista [0.77897095680236816, 0.78187918663024902, 0.78467702865600586]
listb [0.629547119140625, 0.63210701942443848, 0.63321495056152344]
lcsa  [0.36563992500305176, 0.36638498306274414, 0.38175487518310547]
seta  [0.46695613861083984, 0.46992206573486328, 0.47583580017089844]
setb  [0.47616910934448242, 0.47661614418029785, 0.4850609302520752]
1.66818870637 1.34819326075 0.783028414812 1.63591241329 1.32210827369 0.767878297495

n = 7 21 14
lista [0.9703209400177002, 0.9734041690826416, 1.0182771682739258]
listb [0.82394003868103027, 0.82625699043273926, 0.82796716690063477]
lcsa  [0.40975093841552734, 0.41210508346557617, 0.42286920547485352]
seta  [0.5086359977722168, 0.50968098640441895, 0.51014018058776855]
setb  [0.48688101768493652, 0.4879908561706543, 0.49204087257385254]
1.90769222837 1.61990115188 0.805587768483 1.99293236904 1.69228211566 0.841583309951

n = 8 24 16
lista [1.204819917678833, 1.2206029891967773, 1.258256196975708]
listb [1.014998197555542, 1.0206191539764404, 1.0343101024627686]
lcsa  [0.50966787338256836, 0.51018595695495605, 0.51319599151611328]
seta  [0.50310111045837402, 0.50556015968322754, 0.51335406303405762]
setb  [0.51472997665405273, 0.51948785781860352, 0.52113485336303711]
2.39478683834 2.01748351664 1.01305257092 2.34068341135 1.97190418975 0.990165516871

n = 9 27 18
lista [1.511646032333374, 1.5133969783782959, 1.5639569759368896]
listb [1.2461750507354736, 1.254518985748291, 1.2613379955291748]
lcsa  [0.5565330982208252, 0.56119203567504883, 0.56451296806335449]
seta  [0.5966339111328125, 0.60275578498840332, 0.64791703224182129]
setb  [0.54694414138793945, 0.5508568286895752, 0.55375313758850098]
2.53362406013 2.08867620074 0.932788243907 2.76380331728 2.27843203069 1.01753187594

n = 10 30 20
lista [1.7777848243713379, 2.1453688144683838, 2.4085969924926758]
listb [1.5070111751556396, 1.5202279090881348, 1.5779800415039062]
lcsa  [0.5954139232635498, 0.59703707695007324, 0.60746097564697266]
seta  [0.61563014984130859, 0.62125110626220703, 0.62354087829589844]
setb  [0.56723213195800781, 0.57257509231567383, 0.57460403442382812]
2.88774814689 2.44791645689 0.967161734066 3.13413984189 2.6567803378 1.04968299523

使用2GHz的单核机器和2GB的RAM在Debian风格的Linux上运行Python 2.6.6 (Firefox在后台运行)生成。

这些数字只是一个粗略的指南，因为各种算法的实际速度受到两个源列表中元素的比例的不同影响。

a = [1,2,3,4,5]
b = [1,3,5,6]
c = list(set(a).intersection(set(b)))

应该像做梦一样工作。并且，如果可以的话，使用集合而不是列表来避免所有这些类型更改!

这里的大多数解决方案都不考虑列表中元素的顺序，而是将列表视为集合。另一方面，如果希望找到两个列表中包含的一个最长子序列，则可以尝试以下代码。

def intersect(a, b):
    if a == [] or b == []: 
        return []
    inter_1 = intersect(a[1:], b)
    if a[0] in b:
        idx = b.index(a[0])
        inter_2 = [a[0]] + intersect(a[1:], b[idx+1:])        
        if len(inter_1) <= len(inter_2):
            return inter_2
    return inter_1

对于a=[1,2,3]和b=[3,1,4,2]，返回[1,2]而不是[1,2,3]。注意，这样的子序列不是唯一的，因为[1]，[2]，[3]都是a=[1,2,3]和b=[3,2,1]的解。

当我们使用tuple时，我们想要交叉

a=([1,2,3,4,5,20], [8,3,9,5,1,4,20])

for i in range(len(a)):

    b=set(a[i-1]).intersection(a[i])
print(b)
{1, 3, 4, 5, 20}