How to find list intersection?

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别那么骄傲
别那么骄傲 2020-11-22 05:21
a = [1,2,3,4,5]
b = [1,3,5,6]
c = a and b
print c

actual output: [1,3,5,6] expected output: [1,3,5]

How can we ac

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  • 2020-11-22 05:30

    Using list comprehensions is a pretty obvious one for me. Not sure about performance, but at least things stay lists.

    [x for x in a if x in b]

    Or "all the x values that are in A, if the X value is in B".

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  • 2020-11-22 05:31

    A functional way can be achieved using filter and lambda operator.

    list1 = [1,2,3,4,5,6]
    
    list2 = [2,4,6,9,10]
    
    >>> list(filter(lambda x:x in list1, list2))
    
    [2, 4, 6]
    

    Edit: It filters out x that exists in both list1 and list, set difference can also be achieved using:

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

    Edit2: python3 filter returns a filter object, encapsulating it with list returns the output list.

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  • 2020-11-22 05:33

    Here's some Python 2 / Python 3 code that generates timing information for both list-based and set-based methods of finding the intersection of two lists.

    The pure list comprehension algorithms are O(n^2), since in on a list is a linear search. The set-based algorithms are O(n), since set search is O(1), and set creation is O(n) (and converting a set to a list is also O(n)). So for sufficiently large n the set-based algorithms are faster, but for small n the overheads of creating the set(s) make them slower than the pure list comp algorithms.

    #!/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)
    

    output

    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
    

    Generated using a 2GHz single core machine with 2GB of RAM running Python 2.6.6 on a Debian flavour of Linux (with Firefox running in the background).

    These figures are only a rough guide, since the actual speeds of the various algorithms are affected differently by the proportion of elements that are in both source lists.

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  • 2020-11-22 05:39

    If, by Boolean AND, you mean items that appear in both lists, e.g. intersection, then you should look at Python's set and frozenset types.

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  • 2020-11-22 05:41

    If order is not important and you don't need to worry about duplicates then you can use set intersection:

    >>> a = [1,2,3,4,5]
    >>> b = [1,3,5,6]
    >>> list(set(a) & set(b))
    [1, 3, 5]
    
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  • 2020-11-22 05:41

    Make a set out of the larger one:

    _auxset = set(a)
    

    Then,

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

    will do what you want (preserving b's ordering, not a's -- can't necessarily preserve both) and do it fast. (Using if x in a as the condition in the list comprehension would also work, and avoid the need to build _auxset, but unfortunately for lists of substantial length it would be a lot slower).

    If you want the result to be sorted, rather than preserve either list's ordering, an even neater way might be:

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