Finding elements not in a list

Question

So heres my code   item    0 1 2 3 4 5 6 7 8 9  z         list of integers  for item in z      if item not in z          print item   z contains a list of integers  I want to compare item to z and print out the numbers that are not in z when compared to item   I can print the elements that are in z when compared not item  but when I try and do the opposite using the code above nothing prints   Any help

User · Answer

gt  gt  items    1 2 3 4   gt  gt  Z    3 4 5 6    gt  gt  print list set items -set Z    1  2

User · Answer

No  z is undefined  item contains a list of integers   I think what you re trying to do is this    z defined elsewhere item    0  1  2  3  4  5  6  7  8  9   for i in item    if i not in z  print i   As has been stated in other answers  you may want to try using sets

User · Answer

In the case where item and z are sorted iterators  we can reduce the complexity from O n 2  to O n m  by doing this  def iexclude sorted iterator  exclude sorted iterator       next val   next exclude sorted iterator      for item in sorted iterator          try              while next val  lt  item                  next val   next exclude sorted iterator                  continue             if item    next val                  continue         except StopIteration              pass         yield item   If the two are iterators  we also have the opportunity to reduce the memory footprint not storing z  exclude sorted iterator  as a list

User · Answer

If you run a loop taking items from z  how do you expect them not to be in z  IMHO it would make more sense comparing items from a different list to z

User · Answer

Your code is a no-op  By the definition of the loop   item  has to be in Z  A  For     in  loop in Python means  Loop though the list called  z   each time you loop  give me the next item in the list   and call it  item    http   docs python org tutorial controlflow html for-statements  I think your confusion arises from the fact that you re using the variable name  item  twice  to mean two different things

User · Answer

You are reassigning item to the values in z as you iterate through z   So the first time in your for loop  item   0  next item   1  etc     You are never checking one list against the other   To do it very explicitly    gt  gt  gt  item    0 1 2 3 4 5 6 7 8 9   gt  gt  gt  z    0 1 2 3 4 5 6 7   gt  gt  gt    gt  gt  gt  for elem in item        if elem not in z          print elem      8 9

User · Answer

Your code is not doing what I think you think it is doing  The line for item in z  will iterate through z  each time making item equal to one single element of z  The original item list is therefore overwritten before you ve done anything with it   I think you want something like this   item    0 1 2 3 4 5 6 7 8 9   for element in item      if element not in z          print element   But you could easily do this like    x for x in item if x not in z    or  if you don t mind losing duplicates of non-unique elements    set item  - set z

User · Answer

list1    1 2 3 4   list2    0 3 3 6   print set list2  - set list1

User · Answer

gt  gt  gt  item   set  0 1 2 3 4 5 6 7 8 9    gt  gt  gt  z   set  2 3 4    gt  gt  gt  print item - z set  0  1  5  6  7  8  9

User · Answer

Using list comprehension   print  x for x in item if x not in Z    or using filter function    filter lambda x  x not in Z  item    Using set in any form may create a bug if the list being checked contains non-unique elements  e g    print item  Out 39    0  1  1  2  3  4  5  6  7  8  9   print Z  Out 40    3  4  5  6   set item  - set Z   Out 41    0  1  2  7  8  9    vs list comprehension as above  print  x for x in item if x not in Z   Out 38    0  1  1  2  7  8  9    or filter function   filter lambda x  x not in Z  item   Out 38    0  1  1  2  7  8  9

[python] Finding elements not in a list

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