python list index out of range error while iteratively popping elements

Question

I have written a simple python program   l  1 2 3 0 0 1  for i in range 0 len l           if l i   0             l pop i    This gives me error  list index out of range  on line if l i   0   After debugging I could figure out that i is getting incremented and list is getting reduced  However  I have loop termination condition i  lt  len l   Then why I am getting such error

User · Answer

I am using python 3 3 5  The above solution of using while loop did not work for me  Even if i put print  i  after len l  it gave me an error  I ran the same code in command line  shell   window that pops up when we run a function  it runs without error  What i did was calculated len l  outside the function in main program and passed the length as a parameter  It worked  Python is weird sometimes

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Code   while True          n    1         try            DATA n   message    text           except            key   DATA n-1   message    text             break   Console    Traceback  most recent call last     File  botnet py   line 82  in  lt module gt      key  DATA n-1   message    text   IndexError  list index out of range

User · Answer

You are reducing the length of your list l as you iterate over it  so as you approach the end of your indices in the range statement  some of those indices are no longer valid   It looks like what you want to do is   l    x for x in l if x    0    which will return a copy of l without any of the elements that were zero  that operation is called a list comprehension  by the way    You could even shorten that last part to just if x  since non-zero numbers evaluate to True   There is no such thing as a loop termination condition of i  lt  len l   in the way you ve written the code  because len l  is precalculated before the loop  not re-evaluated on each iteration   You could write it in such a way  however   i   0 while i  lt  len l      if l i     0         l pop i     else         i    1

User · Answer

I think the best way to solve this problem is   l    1  2  3  0  0  1  while 0 in l      l remove 0    Instead of iterating over list I remove 0 until there aren t any 0 in list

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The problem was that you attempted to modify the list you were referencing within the loop that used the list len    When you remove the item from the list  then the new len   is calculated on the next loop   For example  after the first run  when you removed  i  using l pop i   that happened successfully but on the next loop the length of the list has changed so all index numbers have been shifted  To a certain point the loop attempts to run over a shorted list throwing the error   Doing this outside the loop works  however it would be better to build and new list by first declaring and empty list before the loop  and later within the loop append everything you want to keep to the new list   For those of you who may have come to the same problem

User · Answer

x    x    int i  for i in input   split    i   0     while i  lt  len x           print x i           if x i  5   0              del x i          else              i    1 print  x

User · Answer

The expression len l  is evaluated only one time  at the moment the range   builtin is evaluated  The range object constructed at that time does not change  it can t possibly know anything about the object l   P S  l is a lousy name for a value  It looks like the numeral 1  or the capital letter I

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You re changing the size of the list while iterating over it  which is probably not what you want and is the cause of your error     Edit  As others have answered and commented  list comprehensions are better as a first choice and especially so in response to this question  I offered this as an alternative for that reason  and while not the best answer  it still solves the problem   So on that note  you could also use filter  which allows you to call a function to evaluate the items in the list you don t want     Example    gt  gt  gt  l    1 2 3 0 0 1   gt  gt  gt  filter lambda x  x  gt  0  l   1  2  3    Live and learn   Simple is better  except when you need things to be complex

User · Answer

I think most solutions talk here about List Comprehension  but if you d like to perform in place deletion and keep the space complexity to O 1   The solution is   i   0 for j in range len arr    if  arr j     0       arr i    arr j      i   1 arr   arr  i

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What Mark Rushakoff said is true  but if you iterate in the opposite direction  it is possible to remove elements from the list in the for-loop as well  E g    x    1 2 3 0 0 1  for i in range len x -1  -1  -1       if x i     0          x pop i    It s like a tall building that falls from top to bottom  even if it is in the middle of collapse  you still can  enter  into it and visit yet-to-be-collapsed floors

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List comprehension will lead you to a solution   But the right way to copy a object in python is using python module copy - Shallow and deep copy operations    l  1 2 3 0 0 1  for i in range 0 len l       if l i   0         l pop i    If instead of this   import copy l  1 2 3 0 0 1  duplicate l   copy copy l  for i in range 0 len l       if l i   0         m remove i  l   m   Then  your own code would have worked  But for optimization  list comprehension is a good solution

User · Answer

I recently had a similar problem and I found that I need to decrease the list index by one   So instead of   if l i   0    You can try   if l i-1   0    Because the list indices start at 0 and your range will go just one above that

[python] python : list index out of range error while iteratively popping elements

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