如何找到列表交集？

Question

a = [1,2,3,4,5]
b = [1,3,5,6]
c = a and b
print c

实际输出：[1,3,5,6] 预期产出：[1,3,5]

我们如何在两个列表上实现布尔AND操作（列表交集）？

Answer 1

如果订单不重要且您不需要担心重复，那么您可以使用set intersection：

>>> a = [1,2,3,4,5]
>>> b = [1,3,5,6]
>>> list(set(a) & set(b))
[1, 3, 5]

Answer 2

如果将两个列表中较大的一个转换为一个集合，则可以使用intersection()获取该集合与任何可迭代的交集：

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

Answer 3

使用列表推导对我来说非常明显。不确定性能，但至少保留了清单。

[x for x in a if x in b]

或者＆＃34; A中的所有x值，如果X值在B＆＃34;中。

Answer 4

制作一个较大的一个：

_auxset = set(a)

然后，

c = [x for x in b if x in _auxset]

会做你想做的事情（保留b的排序，而不是a - 不一定能保留两者）并快速 。 （使用if x in a作为列表推导中的条件也可以，并且避免构建_auxset的需要，但不幸的是，对于长度很长的列表，它会慢得多。

如果您希望对结果进行排序，而不是保留列表的排序，则可能采用更简洁的方式：

c = sorted(set(a).intersection(b))

Answer 5

这里有一些Python 2 / Python 3代码，它们为查找两个列表的交集的基于列表和基于集合的方法生成时序信息。

纯列表推导算法是O（n ^ 2），因为列表上的in是线性搜索。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）（并且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，但对于小 n ，创建集合的开销使得它们比纯列表comp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单核机器生成2GB内存，运行Python 2.6.6，在Debian风格的Linux上运行（后台运行Firefox）。

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两个源列表中元素的比例的不同影响。

Answer 6

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

应该像梦一样工作。而且，如果可以的话，使用集合而不是列表来避免所有这种类型的变化！

Answer 7

您也可以使用 numpy.intersect1d(ar1, ar2)。
它返回两个数组中的唯一值和排序值。

Answer 8

使用filter和lambda运算符可以实现功能性方法。

list1 = [1,2,3,4,5,6]

list2 = [2,4,6,9,10]

>>> filter(lambda x:x in list1, list2)

[2, 4, 6]

编辑：它过滤掉了list1和list中存在的x，设置差异也可以通过以下方式实现：

>>> filter(lambda x:x not in list1, list2)
[9,10]

Answer 9

如果，通过布尔AND，表示出现在两个列表中的项目，例如交集，然后你应该看看Python的set和frozenset类型。

Answer 10

这是一个您需要时的示例，结果中的每个元素应出现在两个数组中的次数相同。

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

Answer 11

可能已经晚了，但我只是想和我分享一下您需要手动执行此操作（显示工作状态-哈哈）的情况，或者当您需要所有元素尽可能多地出现或您也需要它时独一无二。

请注意，也为此编写了测试。

    from nose.tools import assert_equal

    '''
    Given two lists, print out the list of overlapping elements
    '''

    def overlap(l_a, l_b):
        '''
        compare the two lists l_a and l_b and return the overlapping
        elements (intersecting) between the two
        '''

        #edge case is when they are the same lists
        if l_a == l_b:
            return [] #no overlapping elements

        output = []

        if len(l_a) == len(l_b):
            for i in range(l_a): #same length so either one applies
                if l_a[i] in l_b:
                    output.append(l_a[i])

            #found all by now
            #return output #if repetition does not matter
            return list(set(output))

        else:
            #find the smallest and largest lists and go with that
            sm = l_a if len(l_a)  len(l_b) else l_b

            for i in range(len(sm)):
                if sm[i] in lg:
                    output.append(sm[i])

            #return output #if repetition does not matter
            return list(set(output))

    ## Test the Above Implementation

    a = [1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89]
    b = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]
    exp = [1, 2, 3, 5, 8, 13]

    c = [4, 4, 5, 6]
    d = [5, 7, 4, 8 ,6 ] #assuming it is not ordered
    exp2 = [4, 5, 6]

    class TestOverlap(object):

        def test(self, sol):
            t = sol(a, b)
            assert_equal(t, exp)
            print('Comparing the two lists produces')
            print(t)

            t = sol(c, d)
            assert_equal(t, exp2)
            print('Comparing the two lists produces')
            print(t)

            print('All Tests Passed!!')

    t = TestOverlap()
    t.test(overlap)

Answer 12

您也可以使用柜台！它不会保留顺序，但会考虑重复项：

>>> from collections import Counter
>>> a = [1,2,3,4,5]
>>> b = [1,3,5,6]
>>> d1, d2 = Counter(a), Counter(b)
>>> c = [n for n in d1.keys() & d2.keys() for _ in range(min(d1[n], d2[n]))]
>>> print(c)
[1,3,5]

Answer 13

通过这种方式，您可以获得两个列表的交集，并且还获得了常见的重复项。

>>> 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]

Answer 14

这里的大多数解决方案都没有考虑列表中元素的顺序，而是将列表视为集合。另一方面，如果您想找到两个列表中包含的最长子序列之一，您可以尝试以下代码。

const example = require('./example');

jest.mock('./got');

const got = require('./got');

// set default event
let event = {
  process: 1
};

// set default context
const context = {};

// run before each test
beforeEach(() => {
  // set default got.post response
  got.post.mockReturnValue(Promise.resolve({
    body: {
      active: true
    }
  }));
});

// test artifact api
describe('[example]', () => {
  test('error calling process api', async () => {
    let error = 'error calling process';

    // set got mock response for this test to error
    got.post.mockReturnValue(Promise.reject(error));

    // function we want to test w/ mock data
    await expect(example.example(event, context)).rejects.toThrow(error);  // SUCCESS
  });
});

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