Python find elements in one list that are not in the other

Question

I need to compare two lists in order to create a new list of specific elements found in one list but not in the other  For example   main list    list 1   a    b    c    d    e   list 2   a    f    c    m      I want to loop through list 1 and append to main list all the elements from list 2 that are not found in list 1   The result should be   main list   f    m     How can I do it with python

User · Accepted Answer

TL DR   SOLUTION  1    import numpy as np main list   np setdiff1d list 2 list 1    yields the elements in  list 2  that are NOT in  list 1    SOLUTION  2  You want a sorted list  def setdiff sorted array1 array2 assume unique False       ans   np setdiff1d array1 array2 assume unique  tolist       if assume unique          return sorted ans      return ans main list   setdiff sorted list 2 list 1        EXPLANATIONS    1  You can use NumPy s setdiff1d  array1 array2 assume unique False     assume unique asks the user IF the arrays ARE ALREADY UNIQUE   If False  then the unique elements are determined first   If True  the function will assume that the elements are already unique AND function will skip determining the unique elements   This yields the unique values in array1 that are not in array2  assume unique is False by default   If you are concerned with the unique elements  based on the response of Chinny84   then simply use  where assume unique False    the default value    import numpy as np list 1     a    b    c    d    e   list 2     a    f    c    m    main list   np setdiff1d list 2 list 1    yields the elements in  list 2  that are NOT in  list 1      2  For those who want answers to be sorted  I ve made a custom function   import numpy as np def setdiff sorted array1 array2 assume unique False       ans   np setdiff1d array1 array2 assume unique  tolist       if assume unique          return sorted ans      return ans   To get the answer  run   main list   setdiff sorted list 2 list 1      SIDE NOTES    a  Solution 2  custom function setdiff sorted  returns a list  compared to an array in solution 1     b  If you aren t sure if the elements are unique  just use the default setting of NumPy s setdiff1d in both solutions A and B  What can be an example of a complication  See note  c     c  Things will be different if either of the two lists is not unique    Say list 2 is not unique  list2     a    f    c    m    m    Keep list1 as is  list 1     a    b    c    d    e    Setting the default value of assume unique yields   f    m    in both solutions   HOWEVER  if you set assume unique True  both solutions give   f    m    m    Why  This is because the user ASSUMED that the elements are unique   Hence  IT IS BETTER TO KEEP assume unique to its default value  Note that both answers are sorted   pythonnumpy

User · Answer

Not sure why the above explanations are so complicated when you have native methods available   main list   list set list 2 -set list 1

User · Answer

If you want a one-liner solution  ignoring imports  that only requires O max n  m   work for inputs of length n and m  not O n   m  work  you can do so with the itertools module  from itertools import filterfalse  main list   list filterfalse set list 1    contains    list 2    This takes advantage of the functional functions taking a callback function on construction  allowing it to create the callback once and reuse it for every element without needing to store it somewhere  because filterfalse stores it internally   list comprehensions and generator expressions can do this  but it s ugly     That gets the same results in a single line as  main list    x for x in list 2 if x not in list 1   with the speed of  set 1   set list 1  main list    x for x in list 2 if x not in set 1   Of course  if the comparisons are intended to be positional  so  list 1    1  2  3  list 2    2  3  4   should produce  main list    2  3  4    because no value in list 2 has a match at the same index in list 1   you should definitely go with Patrick s answer  which involves no temporary lists or sets  even with sets being roughly O 1   they have a higher  quot constant quot  factor per check than simple equality checks  and involves O min n  m   work  less than any other answer  and if your problem is position sensitive  is the only correct solution when matching elements appear at mismatched offsets       The way to do the same thing with a list comprehension as a one-liner would be to abuse nested looping to create and cache value s  in the  quot outermost quot  loop  e g   main list    x for set 1 in  set list 1    for x in list 2 if x not in set 1   which also gives a minor performance benefit on Python 3  because now set 1 is locally scoped in the comprehension code  rather than looked up from nested scope for each check  on Python 2 that doesn t matter  because Python 2 doesn t use closures for list comprehensions  they operate in the same scope they re used in

User · Answer

You can use sets   main list   list set list 2  - set list 1     Output    gt  gt  gt  list 1   a    b    c    d    e    gt  gt  gt  list 2   a    f    c    m    gt  gt  gt  set list 2  - set list 1  set   m    f     gt  gt  gt  list set list 2  - set list 1     m    f     Per  JonClements  comment  here is a tidier version    gt  gt  gt  list 1   a    b    c    d    e    gt  gt  gt  list 2   a    f    c    m    gt  gt  gt  list set list 2  difference list 1     m    f

User · Answer

I used two methods and I found one method useful over other  Here is my answer    My input data    crkmod mpp     M13   M18   M19   M24   testmod mpp     M13   M14   M15   M16   M17   M18   M19   M20   M21   M22   M23   M24     Method1  np setdiff1d I like this approach over other because it preserves the position  test  list np setdiff1d testmod mpp crkmod mpp   print test    M15    M16    M22    M23    M20    M14    M17    M21     Method2  Though it gives same answer as in Method1 but disturbs the order   test   list set testmod mpp  difference set crkmod mpp    print test    POA23    POA15    POA17    POA16    POA22    POA18    POA24    POA21     Method1 np setdiff1d meets my requirements perfectly   This answer for information

User · Answer

main list    list 1   a    b    c    d    e   list 2   a    f    c    m    for i in list 2      if i not in list 1          main list append i   print main list    output     f    m

User · Answer

If the number of occurences should be taken into account you probably need to use something like collections Counter   list 1   a    b    c    d    e   list 2   a    f    c    m    from collections import Counter cnt1   Counter list 1  cnt2   Counter list 2  final    key for key  counts in cnt2 items   if cnt1 get key  0     counts    gt  gt  gt  final   f    m     As promised this can also handle differing number of occurences as  difference    list 1   a    b    c    d    e    a   cnt1   Counter list 1  cnt2   Counter list 2  final    key for key  counts in cnt2 items   if cnt1 get key  0     counts    gt  gt  gt  final   a    f    m

User · Answer

I would zip the lists together to compare them element by element   main list    b for a  b in zip list1  list2  if a   b

User · Answer

From ser1 remove items present in ser2   Input  ser1   pd Series  1  2  3  4  5   ser2   pd Series  4  5  6  7  8    Solution  ser1  ser1 isin ser2

User · Answer

Use a list comprehension like this   main list    item for item in list 2 if item not in list 1    Output    gt  gt  gt  list 1     a    b    c    d    e    gt  gt  gt  list 2     a    f    c    m     gt  gt  gt    gt  gt  gt  main list    item for item in list 2 if item not in list 1   gt  gt  gt  main list   f    m     Edit   Like mentioned in the comments below  with large lists  the above is not the ideal solution  When that s the case  a better option would be converting list 1 to a set first   set 1   set list 1     this reduces the lookup time from O n  to O 1  main list    item for item in list 2 if item not in set 1

[python] Python find elements in one list that are not in the other

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