Appending an id to a list if not already present in a string

Question

I am trying to check if id is in a list and append the  id only if its not in the list using the below code  however I see that the id is getting appended even though id is already present in the list   can anyone provide inputs on what is wrong here      list     350882 348521 350166 r n       id   348521     if id not in list          list append id      print list  OUTPUT -   350882 348521 350166 r n   348521

User · Answer

Your list just contains a string   Convert it to integer IDs   L     350882 348521 350166 r n    ids    int i  for i in L 0  strip   split    print ids  id   348521 if id not in ids      ids append id  print ids  id   348522 if id not in ids      ids append id  print ids    Turn it back into your odd format L        join str id  for id in ids      r n   print L    Output    350882  348521  350166   350882  348521  350166   350882  348521  350166  348522    350882 348521 350166 348522 r n

User · Answer

What you are trying to do can almost certainly be achieved with a set    gt  gt  gt  x   set  1 2 3    gt  gt  gt  x add 2   gt  gt  gt  x set  1  2  3    gt  gt  gt  x add 4   gt  gt  gt  x add 4   gt  gt  gt  x set  1  2  3  4    gt  gt  gt     using a set s add method you can build your unique set of ids very quickly  Or if you already have a list  unique ids   set id list    as for getting your inputs in numeric form you can do something like   gt  gt  gt  ids    int n  for n in  350882 348521 350166 r n  split     gt  gt  gt  ids  350882  348521  350166

User · Answer

I agree with other answers that you are doing something weird here  You have a list containing a string with multiple entries that are themselves integers that you are comparing to an integer id   This is almost surely not what you should be doing  You probably should be taking input and converting it to integers before storing in your list  You could do that with   input    350882 348521 350166 r n  list append  int x  for x in input split       Then your test will pass  If you really are sure you don t want to do what you re currently doing  the following should do what you want  which is to not add the new id that already exists   list     350882 348521 350166 r n   id   348521 if id not in  int y  for x in list for y in x split         list append id  print list

User · Answer

There are a couple things going on with your example  You have a list containing a string of numbers and newline characters   list     350882 348521 350166 r n     And you are trying to find a number ID within this list   id   348521 if id not in list            Your first conditional is always going to pass  because it will be looking for integer 348521 in list which has one element at index list 0  with the string value of  350882 348521 350166 r n   so integer 348521 will be added to that list  making it a list of two elements  a string and an integer  as your output shows   To reiterate  list is searched for id  not the string in list s first element   If you were trying to find if the string representation of  348521  was contained within the larger string contained within your list  you could do the following  noting that you would need to do this for each element in list   if str id  not in list 0     list 0    350882 348521 350166 r n                                                         However be aware that you would need to wrap str id  with whitespace for the search  otherwise it would also match   2348521999           It is unclear whether you want your  list  to be a  string of integers separated by whitespace  or if you really want a list of integers   If all you are trying to accomplish is to have a list of IDs  and to add IDs to that list only if they are not already contained   and if the order of the elements in the list is not important   then a set would be the best data structure to use   ids   set       int id  for id in  350882 348521 350166 r n  strip   split            Adding an ID already in the set has no effect ids add 348521    If the ordering of the IDs in the string is important then I would keep your IDs in a standard list and use your conditional check   ids    int id  for id in  350882 348521 350166 r n  strip   split        if 348521 not in ids

User · Answer

A more pythonic way  without using set is as follows   lst    1  2  3  4  lst append 3  if 3 not in lst else lst

User · Answer

Your id variable is a number where your list only has one element   It s a string that contains your other IDs   You either need to check if id is in that string  or pull the numbers out of the string and store them in the list separately  list     350882  348521  350166

User · Answer

If you really don t want to change your structure  or at least create a copy of it containing the same data  e g  make a class property with a setter and getter that read from write to that string behind the scenes   then you can use a regular expression to check if an item is in that  list  at any given time  and if not  append it to the  list  as a separate element   if not re match   b   b  format 348521   some list 0    some list append 348521   This is probably faster than converting it to a set every time you want to check if an item is in it  But using set as others have suggested here is a million times better

User · Answer

7 years later  allow me to give a one-liner solution by building on a previous answer  You could do the followwing  numbers    1  2  3   to Add  3  4  5  into numbers without repeating 3  do the following  numbers   list set numbers    3  4  5     This results in 4 and 5 being added to numbers as in  1  2  3  4  5  Explanation  Now let me explain what happens  starting from the inside of the set   instruction  we took numbers and added 3  4  and 5 to it which makes numbers look like  1  2  3  3  4  5   Then  we took that   1  2  3  3  4  5   and transformed it into a set which gets rid of duplicates  resulting in the following  1  2  3  4  5   Now since we wanted a list and not a set  we used the function list   to make that set   1  2  3  4  5   into a list  resulting in  1  2  3  4  5  which we assigned to the variable numbers This  I believe  will work for all types of data in a list  and objects as well if done correctly

[python] Appending an id to a list if not already present in a string

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