How to find list intersection

Question

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 achieve a boolean AND operation  list intersection  on two lists

User · Answer

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".

User · Answer

If order is not important and you don t need to worry about duplicates then you can use set intersection    gt  gt  gt  a    1 2 3 4 5   gt  gt  gt  b    1 3 5 6   gt  gt  gt  list set a   amp  set b    1  3  5

User · Answer

This way you get the intersection of two lists and also get the common duplicates   gt  gt  gt  from collections import Counter  gt  gt  gt  a   Counter  1 2 3 4 5    gt  gt  gt  b   Counter  1 3 5 6    gt  gt  gt  a  amp   b  gt  gt  gt  list a elements     1  3  5

User · Answer

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

User · Answer

You can also use a counter  It doesn t preserve the order  but it ll consider the duplicates    gt  gt  gt  from collections import Counter  gt  gt  gt  a    1 2 3 4 5   gt  gt  gt  b    1 3 5 6   gt  gt  gt  d1  d2   Counter a   Counter b   gt  gt  gt  c    n for n in d1 keys    amp  d2 keys   for   in range min d1 n   d2 n      gt  gt  gt  print c   1 3 5

User · Answer

It might be late but I just thought I should share for the case where you are required to do it manually  show working - haha  OR when you need all elements to appear as many times as possible or when you also need it to be unique   Kindly note that tests have also been written for it         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

User · Answer

If you convert the larger of the two lists into a set  you can get the intersection of that set with any iterable using intersection     a    1 2 3 4 5  b    1 3 5 6  set a  intersection b

User · Answer

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

User · Answer

This is an example when you need Each element in the result should appear as many times as it shows in both arrays   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   gt   len nums1  else  nums2  nums1   iterate will look into target      if len target     0              return         i   0     store          while i  lt  len iterate             element   iterate i            if element in target                 store append element                 target remove element            i    1       return store

User · Answer

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

User · Answer

a    1 2 3 4 5  b    1 3 5 6  c   list set a  intersection set b      Should work like a dream   And  if you can  use sets instead of lists to avoid all this type changing

User · Answer

A functional way can be achieved using filter and lambda operator   list1    1 2 3 4 5 6   list2    2 4 6 9 10    gt  gt  gt  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    gt  gt  gt  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

[python] How to find list intersection?

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