How to check if one of the following items is in a list

Question

I m trying to find a short way to see if any of the following items is in a list  but my first attempt does not work  Besides writing a function to accomplish this  is the any short way to check if one of multiple items is in a list    gt  gt  gt  a    2 3 4   gt  gt  gt  print  1 or 2  in a False  gt  gt  gt  print  2 or 1  in a True

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Think about what the code actually says    gt  gt  gt   1 or 2  1  gt  gt  gt   2 or 1  2   That should probably explain it      Python apparently implements  lazy or   which should come as no surprise   It performs it something like this   def or x  y       if x  return x     if y  return y     return False   In the first example  x    1 and y    2   In the second example  it s vice versa   That s why it returns different values depending on the order of them

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When you think  check to see if a in b   think hashes  in this case  sets   The fastest way is to hash the list you want to check  and then check each item in there   This is why Joe Koberg s answer is fast  checking set intersection is very fast   When you don t have a lot of data though  making sets can be a waste of time  So  you can make a set of the list and just check each item   tocheck    1 2    items to check a    2 3 4    the list  a   set a    convert to set  O len a    print  i for i in tocheck if i in a    check items  O len tocheck      When the number of items you want to check is small  the difference can be negligible  But check lots of numbers against a large list     tests   from timeit import timeit  methods       tocheck    1 2    items to check a    2 3 4    the list a   set a    convert to set  O n    i for i in tocheck if i in a    check items  O m           L1    2 3 4  L2    1 2   i for i in L1 if i in L2          S1   set  2 3 4   S2   set  1 2   S1 intersection S2          a    1 2  b    2 3 4  any x in a for x in b       for method in methods      print timeit method  number 10000   print  methods       tocheck   range 200 300    items to check a   range 2  10000    the list a   set a    convert to set  O n    i for i in tocheck if i in a    check items  O m           L1   range 2  10000  L2   range 200 300   i for i in L1 if i in L2          S1   set range 2  10000   S2   set range 200 300   S1 intersection S2          a   range 200 300  b   range 2  10000  any x in a for x in b       for method in methods      print timeit method  number 1000    speeds   M1  0 0170331001282   make one set M2  0 0164539813995   list comprehension M3  0 0286040306091   set intersection M4  0 0305438041687   any  M1  0 49850320816   make one set M2  25 2735087872   list comprehension M3  0 466138124466   set intersection M4  0 668627977371   any   The method that is consistently fast is to make one set  of the list   but the intersection works on large data sets the best

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This will do it in one line    gt  gt  gt  a  2 3 4   gt  gt  gt  b  1 2   gt  gt  gt  bool sum map lambda x  x in b  a    True

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Simple    new list      for item in a      if item in b           new list append item      else          pass

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a    2 3 4  if  1 2   amp  a      pass   Code golf version  Consider using a set if it makes sense to do so  I find this more readable than a list comprehension

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Ah  Tobias you beat me to it   I was thinking of this slight variation on your solution   gt  gt  gt  a    1 2 3 4   gt  gt  gt  b    2 7   gt  gt  gt  any x in a for x in b  True

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In python 3 we can start make use of the unpack asterisk  Given two lists   bool len   a   amp    b      Edit  incorporate alkanen s suggestion

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1 line without list comprehensions    gt  gt  gt  any map lambda each  each in  2 3 4    1 2    True  gt  gt  gt  any map lambda each  each in  2 3 4    1 5    False  gt  gt  gt  any map lambda each  each in  2 3 4    2 4    True

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gt  gt  gt  L1    2 3 4   gt  gt  gt  L2    1 2   gt  gt  gt   i for i in L1 if i in L2   2     gt  gt  gt  S1   set L1   gt  gt  gt  S2   set L2   gt  gt  gt  S1 intersection S2  set  2     Both empty lists and empty sets are False  so you can use the value directly as a truth value

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I collected several of the solutions mentioned in other answers and in comments  then ran a speed test  not set a  isdisjoint b  turned out the be the fastest  it also did not slowdown much when the result was False   Each of the three runs tests a small sample of the possible configurations of a and b  The times are in microseconds   Any with generator and max         2 093 1 997 7 879 Any with generator         0 907 0 692 2 337 Any with list         1 294 1 452 2 137 True in list         1 219 1 348 2 148 Set with  amp          1 364 1 749 1 412 Set intersection explcit set b          1 424 1 787 1 517 Set intersection implicit set b          0 964 1 298 0 976 Set isdisjoint explicit set b          1 062 1 094 1 241 Set isdisjoint implicit set b          0 622 0 621 0 753     import timeit  def printtimes t       print     3f   format t 10 0    setup1    a   range 10   b   range 9 15   setup2    a   range 10   b   range 10   setup3    a   range 10   b   range 10 20    print  Any with generator and max n t   printtimes timeit Timer  any x in max a b key len  for x in min b a key len    setup setup1  timeit 10000000   printtimes timeit Timer  any x in max a b key len  for x in min b a key len    setup setup2  timeit 10000000   printtimes timeit Timer  any x in max a b key len  for x in min b a key len    setup setup3  timeit 10000000   print  print  Any with generator n t   printtimes timeit Timer  any i in a for i in b   setup setup1  timeit 10000000   printtimes timeit Timer  any i in a for i in b   setup setup2  timeit 10000000   printtimes timeit Timer  any i in a for i in b   setup setup3  timeit 10000000   print  print  Any with list n t   printtimes timeit Timer  any  i in a for i in b    setup setup1  timeit 10000000   printtimes timeit Timer  any  i in a for i in b    setup setup2  timeit 10000000   printtimes timeit Timer  any  i in a for i in b    setup setup3  timeit 10000000   print  print  True in list n t   printtimes timeit Timer  True in  i in a for i in b   setup setup1  timeit 10000000   printtimes timeit Timer  True in  i in a for i in b   setup setup2  timeit 10000000   printtimes timeit Timer  True in  i in a for i in b   setup setup3  timeit 10000000   print  print  Set with  amp  n t   printtimes timeit Timer  bool set a   amp  set b    setup setup1  timeit 10000000   printtimes timeit Timer  bool set a   amp  set b    setup setup2  timeit 10000000   printtimes timeit Timer  bool set a   amp  set b    setup setup3  timeit 10000000   print  print  Set intersection explcit set b  n t   printtimes timeit Timer  bool set a  intersection set b     setup setup1  timeit 10000000   printtimes timeit Timer  bool set a  intersection set b     setup setup2  timeit 10000000   printtimes timeit Timer  bool set a  intersection set b     setup setup3  timeit 10000000   print  print  Set intersection implicit set b  n t   printtimes timeit Timer  bool set a  intersection b    setup setup1  timeit 10000000   printtimes timeit Timer  bool set a  intersection b    setup setup2  timeit 10000000   printtimes timeit Timer  bool set a  intersection b    setup setup3  timeit 10000000   print  print  Set isdisjoint explicit set b  n t   printtimes timeit Timer  not set a  isdisjoint set b    setup setup1  timeit 10000000   printtimes timeit Timer  not set a  isdisjoint set b    setup setup2  timeit 10000000   printtimes timeit Timer  not set a  isdisjoint set b    setup setup3  timeit 10000000   print  print  Set isdisjoint implicit set b  n t   printtimes timeit Timer  not set a  isdisjoint b   setup setup1  timeit 10000000   printtimes timeit Timer  not set a  isdisjoint b   setup setup1  timeit 10000000   printtimes timeit Timer  not set a  isdisjoint b   setup setup3  timeit 10000000   print

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In some cases  e g  unique list elements   set operations can be used    gt  gt  gt  a  2 3 4   gt  gt  gt  set a  - set  2 3      set a  True  gt  gt  gt     Or  using set isdisjoint      gt  gt  gt  not set a  isdisjoint set  2 3    True  gt  gt  gt  not set a  isdisjoint set  5 6    False  gt  gt  gt

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I have to say that my situation might not be what you are looking for  but it may provide an alternative to your thinking   I have tried both the set   and any   method but still have problems with speed  So I remembered Raymond Hettinger said everything in python is a dictionary and use dict whenever you can  So that s what I tried    I used a defaultdict with int to indicate negative results and used the item in the first list as the key for the second list  converted to defaultdict   Because you have instant lookup with dict  you know immediately whether that item exist in the defaultdict  I know you don t always get to change data structure for your second list  but if you are able to from the start  then it s much faster  You may have to convert list2  larger list  to a defaultdict  where key is the potential value you want to check from small list  and value is either 1  hit  or 0  no hit  default     from collections import defaultdict already indexed   defaultdict int   def check exist small list  default list       for item in small list          if default list item     1              return True     return False  if check exist small list  already indexed       continue else      for x in small list          already indexed x    1

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Maybe a bit more lazy   a    1 2 3 4  b    2 7   print any  True for x in a if x in b

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Best I could come up with   any  True for e in  1  2  if e in a

[python] How to check if one of the following items is in a list?

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