Iterating over every two elements in a list

Question

How do I make a for loop or a list comprehension so that every iteration gives me two elements   l    1 2 3 4 5 6   for i k in          print str i        str k        str i k    Output   1 2 3 3 4 7 5 6 11

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I hope this will be even more elegant way of doing it   a    1 2 3 4 5 6  zip a   2   a 1  2      1  2    3  4    5  6

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I need to divide a list by a number and fixed like this    l    1 2 3 4 5 6   def divideByN data  n           return  data i n    i 1  n  for i in range len data   n       gt  gt  gt  print divideByN l 2     1  2    3  4    5  6     gt  gt  gt  print divideByN l 3     1  2  3    4  5  6

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Well you need tuple of 2 elements  so  data    1 2 3 4 5 6  for i k in zip data 0  2   data 1  2        print str i        str k        str i k    Where    data 0  2  means create subset collection of elements that  index   2    0  zip x y  creates a tuple collection from x and y collections same index elements

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A simple solution    l    1  2  3  4  5  6   for i in range 0  len l   2       print str l i         str l i   1         str l i    l i   1

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You need a pairwise    or grouped    implementation   For Python 2   from itertools import izip  def pairwise iterable        s - gt   s0  s1    s2  s3    s4  s5            a   iter iterable      return izip a  a   for x  y in pairwise l      print   d    d    d     x  y  x   y    Or  more generally   from itertools import izip  def grouped iterable  n        s - gt   s0 s1 s2    sn-1    sn sn 1 sn 2    s2n-1    s2n s2n 1 s2n 2    s3n-1            return izip   iter iterable   n   for x  y in grouped l  2      print   d    d    d     x  y  x   y    In Python 3  you can replace izip with the built-in zip   function  and drop the import   All credit to martineau for his answer to my question  I have found this to be very efficient as it only iterates once over the list and does not create any unnecessary lists in the process    N B  This should not be confused with the pairwise recipe in Python s own itertools documentation  which yields s - gt   s0  s1    s1  s2    s2  s3        as pointed out by  lazyr in the comments   Little addition for those who would like to do type checking with mypy on Python 3   from typing import Iterable  Tuple  TypeVar  T   TypeVar  T    def grouped iterable  Iterable T   n 2  - gt  Iterable Tuple T                s - gt   s0 s1 s2    sn-1    sn sn 1 sn 2    s2n-1              return zip   iter iterable     n

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This is simple solution  which is using range function to pick alternative elements from a list of elements  Note  This is only valid for an even numbered list  a list    1  2  3  4  5  6  empty list       for i in range 0  len a list   2       empty list append a list i    a list i   1      print empty list     3  7  11

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With unpacking  l    1 2 3 4 5 6  while l      i  k   l   l     print f  i   k   i k      Note  this will consume l  leaving it empty afterward

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Thought that this is a good place to share my generalization of this for n 2  which is just a sliding window over an iterable   def sliding window iterable  n       its     itertools islice iter  i  None               for i  iter             in enumerate itertools tee iterable  n                                         return itertools izip  its

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A simplistic approach     a i  a i 1   for i in range 0 len a  2     this is useful if your array is a and you want to iterate on it by pairs   To iterate on triplets or more just change the  range  step command  for example     a i  a i 1  a i 2   for i in range 0 len a  3      you have to deal with excess values if your array length and the step do not fit

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Here we can have alt elem method which can fit in your for loop   def alt elem list  index 2       for i  elem in enumerate list  start 1           if not i   index             yield tuple list i-index i     a   range 10  for index in  2  3  4       print  With index   0   format index       for i in alt elem a  index          print i    Output   With index  2  0  1   2  3   4  5   6  7   8  9  With index  3  0  1  2   3  4  5   6  7  8  With index  4  0  1  2  3   4  5  6  7    Note  Above solution might not be efficient considering operations performed in func

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for  i  k  in zip l   2   l 1  2        print i       k       i k   zip  iterable  returns a tuple with the next element of each iterable   l   2  returns the 1st  the 3rd  the 5th  etc  element of the list  the first colon indicates that the slice starts at the beginning because there s no number behind it  the second colon is only needed if you want a  step in the slice   in this case 2    l 1  2  does the same thing but starts in the second element of the lists so it returns the 2nd  the 4th  6th  etc  element of the original list

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The polished Python3 solution is given in one of the itertools recipes  import itertools  def grouper iterable  n  fillvalue None        quot Collect data into fixed-length chunks or blocks quot        grouper  ABCDEFG   3   x   -- gt  ABC DEF Gxx quot      args    iter iterable     n     return itertools zip longest  args  fillvalue fillvalue

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There are many ways to do that  For example   lst    1 2 3 4 5 6    lst i   lst i 1   for i   in enumerate lst  -1         gt  gt  gt   1  2    2  3    3  4    4  5    5  6     i for i in zip   iter lst   2        gt  gt  gt   1  2    3  4    5  6

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Using typing so you can verify data using mypy static analysis tool   from typing import Iterator  Any  Iterable  TypeVar  Tuple  T    TypeVar  T    Pairs Iter   Iterator Tuple T   T     def legs iterable  Iterator T    - gt  Pairs Iter      begin   next iterable      for end in iterable          yield begin  end         begin   end

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you can use more itertools package   import more itertools  lst   range 1  7  for i  j in more itertools chunked lst  2       print f  i     j     i j

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The title of this question is misleading  you seem to be looking for consecutive pairs  but if you want to iterate over the set of all possible pairs than this will work    for i v in enumerate items  -1            for u in items i 1

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For anyone it might help  here is a solution to a similar problem but with overlapping pairs  instead of mutually exclusive pairs    From the Python itertools documentation   from itertools import izip  def pairwise iterable        s - gt   s0 s1    s1 s2    s2  s3            a  b   tee iterable      next b  None      return izip a  b    Or  more generally   from itertools import izip  def groupwise iterable  n 2        s - gt   s0 s1     sn-1    s1 s2     sn    s2 s3     sn 1            t   tee iterable  n      for i in range 1  n           for j in range 0  i               next t i   None      return izip  t

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While all the answers using zip are correct  I find that implementing the functionality yourself leads to more readable code   def pairwise it       it   iter it      while True          try              yield next it   next it          except StopIteration                no more elements in the iterator             return   The it   iter it  part ensures that it is actually an iterator  not just an iterable  If it already is an iterator  this line is a no-op   Usage   for a  b in pairwise  0  1  2  3  4  5        print a   b

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Use the zip and iter commands together   I find this solution using iter to be quite elegant   it   iter l  list zip it  it       1  2    3  4    5  6     Which I found in the Python 3 zip documentation   it   iter l  print   f  u     v     u v   for u  v in zip it  it    sep   n      1   2   3   3   4   7   5   6   11     To generalise to N elements at a time   N   2 list zip    iter l     N        1  2    3  4    5  6

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Another try at cleaner solution def grouped itr  n 2       itr   iter itr      end   object       while True          vals   tuple next itr  end  for   in range n           if vals -1  is end              return         yield vals

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In case you re interested in the performance  I did a small benchmark  using my library simple benchmark  to compare the performance of the solutions and I included a function from one of my packages  iteration utilities grouper  from iteration utilities import grouper import matplotlib as mpl from simple benchmark import BenchmarkBuilder  bench   BenchmarkBuilder     bench add function   def Johnsyweb l       def pairwise iterable            s - gt   s0  s1    s2  s3    s4  s5                a   iter iterable          return zip a  a       for x  y in pairwise l           pass   bench add function   def Margus data       for i  k in zip data 0  2   data 1  2            pass   bench add function   def pyanon l       list zip l l 1       2    bench add function   def taskinoor l       for i in range 0  len l   2           l i   l i 1    bench add function   def mic e it       def pairwise it           it   iter it          while True              try                  yield next it   next it              except StopIteration                  return      for a  b in pairwise it           pass   bench add function   def MSeifert it       for item1  item2 in grouper it  2           pass  bench use random lists as arguments sizes  2  i for i in range 1  20    benchmark result   bench run   mpl rcParams  figure figsize      8  10  benchmark result plot both relative to MSeifert      So if you want the fastest solution without external dependencies you probably should just use the approach given by Johnysweb  at the time of writing it s the most upvoted and accepted answer    If you don t mind the additional dependency then the grouper from iteration utilities will probably be a bit faster   Additional thoughts  Some of the approaches have some restrictions  that haven t been discussed here   For example a few solutions only work for sequences  that is lists  strings  etc    for example Margus pyanon taskinoor solutions which uses indexing while other solutions work on any iterable  that is sequences and generators  iterators  like Johnysweb mic e my solutions   Then Johnysweb also provided a solution that works for other sizes than 2 while the other answers don t  okay  the iteration utilities grouper also allows setting the number of elements to  group     Then there is also the question about what should happen if there is an odd number of elements in the list  Should the remaining item be dismissed  Should the list be padded to make it even sized  Should the remaining item be returned as single  The other answer don t address this point directly  however if I haven t overlooked anything they all follow the approach that the remaining item should be dismissed  except for taskinoors answer - that will actually raise an Exception    With grouper you can decide what you want to do    gt  gt  gt  from iteration utilities import grouper   gt  gt  gt  list grouper  1  2  3   2      as single   1  2    3      gt  gt  gt  list grouper  1  2  3   2  truncate True      ignored   1  2     gt  gt  gt  list grouper  1  2  3   2  fillvalue None      padded   1  2    3  None

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gt  gt  gt  l    1 2 3 4 5 6    gt  gt  gt  zip l l 1      1  2    2  3    3  4    4  5    5  6     gt  gt  gt  zip l l 1      2    1  2    3  4    5  6     gt  gt  gt   a b for a b in zip l l 1      2    3  7  11    gt  gt  gt     d    d    d     a b a b  for a b in zip l l 1      2     1   2   3    3   4   7    5   6   11

[python] Iterating over every two elements in a list

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