What is the pythonic way to detect the last element in a for loop

Question

I d like to know the best way  more compact and  pythonic  way  to do a special treatment for the last element in a for loop  There is a piece of code that should be called only between elements  being suppressed in the last one   Here is how I currently do it   for i  data in enumerate data list       code that is done for every element     if i    len data list  - 1          code that is done between elements   Is there any better way   Note  I don t want to make it with hacks such as using reduce

User · Answer

Google brought me to this old question and I think I could add a different approach to this problem.

Most of the answers here would deal with a proper treatment of a for loop control as it was asked, but if the data_list is destructible, I would suggest that you pop the items from the list until you end up with an empty list:

while True:
    element = element_list.pop(0)
    do_this_for_all_elements()
    if not element:
        do_this_only_for_last_element()
        break
    do_this_for_all_elements_but_last()

you could even use while len(element_list) if you don't need to do anything with the last element. I find this solution more elegant then dealing with next().

User · Answer

One simple solution that comes to mind would be   for i in MyList        Check if  i  is the last element in the list     if i    MyList -1             Do something different for the last     else            Do something for all other elements   A second equally simple solution could be achieved by using a counter     Count the no  of elements in the list ListLength   len MyList    Initialize a counter count   0  for i in MyList        increment counter     count    1       Check if  i  is the last element in the list       by using the counter     if count    ListLength            Do something different for the last     else            Do something for all other elements

User · Answer

This is similar to Ants Aasma s approach but without using the itertools module  It s also a lagging iterator which looks-ahead a single element in the iterator stream   def last iter it         Ensure it s an iterator and get the first field     it   iter it      prev   next it      for item in it            Lag by one item so I know I m not at the end         yield 0  prev         prev   item       Last item     yield 1  prev  def test data       result   list last iter data       if not result          return     if len result   gt  1          assert set x 0  for x in result  -1      set  0    result     assert result -1  0     1  test     test  1   test  1  2   test range 5   test xrange 4    for is last  item in last iter  Hi         print is last  item

User · Answer

you can determine the last element with this code   for i element in enumerate list       if  i  len list -1           print  quot last element is quot    element

User · Answer

Count the items once and keep up with the number of items remaining   remaining   len data list  for data in data list      code that is done for every element      remaining -  1     if remaining          code that is done between elements   This way you only evaluate the length of the list once  Many of the solutions on this page seem to assume the length is unavailable in advance  but that is not part of your question  If you have the length  use it

User · Answer

I like the approach of  ethan-t  but while True is dangerous from my point of view   data list    1  2  3  2  1     sample data L   list data list     destroy L instead of data list while L      e   L pop 0      if L          print f process element  e        else          print f process last element  e    del L   Here  data list is so that last element is equal by value to the first one of the list  L can be exchanged with data list but in this case it results empty after the loop  while True is also possible to use if you check that list is not empty before the processing or the check is not needed  ouch     data list    1  2  3  2  1  if data list      while True          e   data list pop 0          if data list              print f process element  e            else              print f process last element  e                break else      print  list is empty     The good part is that it is fast  The bad - it is destructible  data list becomes empty    Most intuitive solution   data list    1  2  3  2  1     sample data for i  e in enumerate data list       if i    len data list  - 1          print f process element  e        else          print f process last element  e      Oh yes  you have already proposed it

User · Answer

if the items are unique   for x in list       code     if x    list -1            code   other options   pos   -1 for x in list      pos    1      code     if pos    len list  - 1           code   for x in list       code  code - e g  print x   if len list   gt  0      for x in list  -1           code     for x in list -1            code

User · Answer

The  code between  is an example of the Head-Tail pattern  You have an item  which is followed by a sequence of   between  item   pairs   You can also view this as a sequence of  item  between  pairs followed by an item   It s generally simpler to take the first element as special and all the others as the  quot standard quot  case  Further  to avoid repeating code  you have to provide a function or other object to contain the code you don t want to repeat   Embedding an if statement in a loop which is always false except one time is kind of silly  def item processing  item           the common processing   head tail iter   iter  someSequence   head   next head tail iter  item processing  head   for item in head tail iter         the between processing      item processing  item    This is more reliable because it s slightly easier to prove   It doesn t create an extra data structure  i e   a copy of a list  and doesn t require a lot of wasted execution of an if condition which is always false except once

User · Answer

You can use a sliding window over the input data to get a peek at the next value and use a sentinel to detect the last value  This works on any iterable  so you don t need to know the length beforehand  The pairwise implementation is from itertools recipes   from itertools import tee  izip  chain  def pairwise seq       a b   tee seq      next b  None      return izip a b   def annotated last seq          Returns an iterable of pairs of input item and a boolean that show if     the current item is the last item in the sequence         MISSING   object       for current item  next item in pairwise chain seq   MISSING             yield current item  next item is MISSING   for item  is last item in annotated last data list       if is last item            current item is the last item

User · Answer

Is there no possibility to iterate over all-but the last element  and treat the last one outside of the loop   After all  a loop is created to do something similar to all elements you loop over  if one element needs something special  it shouldn t be in the loop    see also this question  does-the-last-element-in-a-loop-deserve-a-separate-treatment   EDIT  since the question is more about the  in between   either the first element is the special one in that it has no predecessor  or the last element is special in that it has no successor

User · Answer

The most simple solution coming to my mind is   for item in data list      try          print new      except NameError  pass     new   item print  The last item      str new     So we always look ahead one item by delaying the the processing one iteration  To skip doing something during the first iteration I simply catch the error   Of course you need to think a bit  in order for the NameError to be raised when you want it   Also keep the  counstruct  try      new except NameError  pass else        continue here if no error was raised   This relies that the name new wasn t previously defined  If you are paranoid you can ensure that new doesn t exist using   try      del new except NameError      pass   Alternatively you can of course also use an if statement  if notfirst  print new  else  notfirst   True   But as far as I know the overhead is bigger     Using  timeit  yields            try  new    test            except NameError  pass           100000000 loops  best of 3  16 2 ns per loop   so I expect the overhead to be unelectable

User · Answer

There can be multiple ways  slicing will be fastest  Adding one more which uses  index   method    gt  gt  gt  l1    1 5 2 3 5 1 7 43                                                    gt  gt  gt   i for i in l1 if l1 index i  1  len l1                                   43

User · Answer

Use slicing and is to check for the last element   for data in data list       lt code that is done for every element gt      if not data is data list -1            lt code that is done between elements gt    Caveat emptor  This only works if all elements in the list are actually different  have different locations in memory   Under the hood  Python may detect equal elements and reuse the same objects for them  For instance  for strings of the same value and common integers

User · Answer

Better late than never   Your original code used enumerate    but you only used the i index to check if it s the last item in a list   Here s an simpler alternative  if you don t need enumerate    using negative indexing  for data in data list      code that is done for every element     if data    data list -1           code that is done between elements  if data    data list -1  checks if the current item in the iteration is NOT the last item in the list  Hope this helps  even nearly 11 years later

User · Answer

Delay the special handling of the last item until after the loop    gt  gt  gt  for i in  1  2  3           pass      gt  gt  gt  i 3

User · Answer

There is nothing wrong with your way  unless you will have 100 000 loops and wants save 100 000  if  statements  In that case  you can go that way    iterable    1 2 3    Your date iterator   iter iterable    get the data iterator  try       wrap all in a try   except     while 1            item   iterator next            print item   put the  for loop  code here except StopIteration  e     make the process on the last element here     print item   Outputs    1 2 3 3   But really  in your case I feel like it s overkill   In any case  you will probably be luckier with slicing    for item in iterable  -1        print item print  last     iterable -1    outputs 1 2 last   3   or just    for item in iterable       print item print iterable -1    outputs 1 2 3 last   3   Eventually  a KISS way to do you stuff  and that would work with any iterable  including the ones without   len      item      for item in iterable       print item print item   Ouputs   1 2 3 3   If feel like I would do it that way  seems simple to me

User · Answer

Most of the times it is easier  and cheaper  to make the first iteration the special case instead of the last one   first   True for data in data list      if first          first   False     else          between items        item     This will work for any iterable  even for those that have no len     file   open   path to file   for line in file      process line line         No way of telling if this is the last line    Apart from that  I don t think there is a generally superior solution as it depends on what you are trying to do  For example  if you are building a string from a list  it s naturally better to use str join   than using a for loop    with special case        Using the same principle but more compact   for i  line in enumerate data list       if i  gt  0          between items       item     Looks familiar  doesn t it        For  ofko  and others who really need to find out if the current value of an iterable without len   is the last one  you will need to look ahead   def lookahead iterable          Pass through all values from the given iterable  augmented by the     information if there are more values to come after the current one      True   or if it is the last value  False                 Get an iterator and pull the first value      it   iter iterable      last   next it        Run the iterator to exhaustion  starting from the second value       for val in it            Report the  previous  value  more to come           yield last  True         last   val       Report the last value      yield last  False   Then you can use it like this    gt  gt  gt  for i  has more in lookahead range 3            print i  has more  0 True 1 True 2 False

User · Answer

So  this is definitely not the  quot shorter quot  version - and one might digress if  quot shortest quot  and  quot Pythonic quot  are actually compatible  But if one needs this pattern often  just put the logic in to a 10-liner generator - and get any meta-data related to an element s position directly on the for call  Another advantage here is that it will work wit an arbitrary iterable  not only Sequences   sentinel   object    def iter check last iterable       iterable   iter iterable      current element   next iterable   sentinel      while current element is not  sentinel          next element   next iterable   sentinel          yield  next element is  sentinel  current element          current element   next element  In  107   for is last  el in iter check last range 3                  print is last  el                        False 0 False 1 True 2

User · Answer

if you are going through the list  for me this worked too   for j in range 0  len Array        if len Array  - j  gt  1          notLast

User · Answer

Assuming input as an iterator  here s a way using tee and izip from itertools   from itertools import tee  izip items  between   tee input iterator  2     Input must be an iterator  first   items next   do to every item first     All  do to every  operations done to first item go here  for i  b in izip items  between       do between items b     All  between  operations go here      do to every item i     All  do to every  operations go here    Demo    gt  gt  gt  def do every x   print  E   x      gt  gt  gt  def do between x   print  B   x      gt  gt  gt  test input   iter range 5    gt  gt  gt   gt  gt  gt  from itertools import tee  izip  gt  gt  gt   gt  gt  gt  items  between   tee test input  2   gt  gt  gt  first   items next    gt  gt  gt  do every first  E 0  gt  gt  gt  for i b in izip items  between           do between b          do every i      B 0 E 1 B 1 E 2 B 2 E 3 B 3 E 4  gt  gt  gt

User · Answer

Just check if data is not the same as the last data in data list  data list -1    for data in data list      code that is done for every element     if data    data list - 1           code that is done between elements

User · Answer

For me the most simple and pythonic way to handle a special case at the end of a list is   for data in data list  -1       handle element data  handle special element data list -1     Of course this can also be used to treat the first element in a special way

User · Answer

Instead of counting up  you can also count down     nrToProcess   len list    for s in list      s doStuff       nrToProcess -  1     if nrToProcess  0     this is the last one       s doSpecialStuff

User · Answer

If you re simply looking to modify the last element in data list then you can simply use the notation   L -1    However  it looks like you re doing more than that  There is nothing really wrong with your way  I even took a quick glance at some Django code for their template tags and they do basically what you re doing

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I will provide with a more elegant and robust way as follows  using unpacking  def mark last iterable       try           init  last   iterable     except ValueError     if iterable is empty         return      for e in init          yield e  True     yield last  False  Test  for a  b in mark last  1  2  3        print a  b   The result is   1 True 2 True 3 False

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Although that question is pretty old  I came here via google and I found a quite simple way  List slicing  Let s say you want to put an   amp   between all list entries   s      l    1  2  3  for i in l  -1       s   s   str i       amp    s   s   str l -1     This returns  1  amp  2  amp  3

[for-loop] What is the pythonic way to detect the last element in a 'for' loop?

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