Python loop that also accesses previous and next values

Question

How can I iterate over a list of objects  accessing the previous  current  and next items  Like this C C   code  in Python   foo   somevalue  previous   next   0   for  i 1  i lt objects length    i          if  objects i   foo            previous   objects i-1           next   objects i 1

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For anyone looking for a solution to this with also wanting to cycle the elements  below might work -  from collections import deque    foo     A    B    C    D    def prev and next input list       CURRENT   input list     PREV   deque input list      PREV rotate -1      PREV   list PREV      NEXT   deque input list      NEXT rotate 1      NEXT   list NEXT      return zip PREV  CURRENT  NEXT   for previous   current   next  in prev and next foo       print previous   current   next

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I think this works and not complicated  array   1 5 6 6 3 2  for i in range 0 len array        Current   array i      Next   array i 1      Prev   array i-1

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using conditional expressions for conciseness for python    2 5  def prenext l v        i l index v     return l i-1  if i gt 0 else None l i 1  if i lt len l -1 else None     example x range 10  prenext x 3   gt  gt  gt   2 4  prenext x 0   gt  gt  gt   None 2  prenext x 9   gt  gt  gt   8 None

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Pythonic and elegant way   objects    1  2  3  4  5  value   3 if value in objects     index   objects index value     previous value   objects index-1     next value   objects index 1  if index   1  lt  len objects  else None

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I don t know how this hasn t come up yet since it uses only built-in functions and is easily extendable to other offsets   values    1  2  3  4  offsets    None    values  -1   values  values 1      None  for value in list zip  offsets        print value     previous  current  next    None  1  2   1  2  3   2  3  4   3  4  None

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Using generators  it is quite simple   signal      Signal value    def pniter  iter  signal signal        iA   iB   signal     for iC in iter          if iB is signal              iB   iC             continue         else              yield iA  iB  iC         iA   iB         iB   iC     iC   signal     yield iA  iB  iC  if   name         main         print  test 1        for a  b  c in pniter  range  10             print  a  b  c       print   ntest 2        for a  b  c in pniter   20  30  40  50  60  70  80             print  a  b  c       print   ntest 3        cam     1  30  2  40  10  9  -5  36       for a  b  c in pniter  cam            print  a  b  c       for a  b  c in pniter  cam            print  a  a if a is signal else cam  a    b  b if b is signal else cam  b    c  c if c is signal else cam  c        print   ntest 4        for a  b  c in pniter   20  30  None  50  60  70  80             print  a  b  c       print   ntest 5        for a  b  c in pniter   20  30  None  50  60  70  80      sig             print  a  b  c       print   ntest 6        for a  b  c in pniter   20     Signal value     None    Signal value    60  70  80    signal            print  a  b  c     Note that tests that include None and the same value as the signal value still work  because the check for the signal value uses  is  and the signal is a value that Python doesn t intern  Any singleton marker value can be used as a signal  though  which might simplify user code in some circumstances

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Solutions until now only deal with lists  and most are copying the list  In my experience a lot of times that isn t possible   Also  they don t deal with the fact that you can have repeated elements in the list   The title of your question says  Previous and next values inside a loop   but if you run most answers here inside a loop  you ll end up iterating over the entire list again on each element to find it   So I ve just created a function that  using the itertools module  splits and slices the iterable  and generates tuples with the previous and next elements together  Not exactly what your code does  but it is worth taking a look  because it can probably solve your problem   from itertools import tee  islice  chain  izip  def previous and next some iterable       prevs  items  nexts   tee some iterable  3      prevs   chain  None   prevs      nexts   chain islice nexts  1  None    None       return izip prevs  items  nexts    Then use it in a loop  and you ll have previous and next items in it   mylist     banana    orange    apple    kiwi    tomato    for previous  item  nxt in previous and next mylist       print  Item is now   item   next is   nxt   previous is   previous   The results   Item is now banana next is orange previous is None Item is now orange next is apple previous is banana Item is now apple next is kiwi previous is orange Item is now kiwi next is tomato previous is apple Item is now tomato next is None previous is kiwi   It ll work with any size list  because it doesn t copy the list   and with any iterable  files  sets  etc   This way you can just iterate over the sequence  and have the previous and next items available inside the loop  No need to search again for the item in the sequence   A short explanation of the code    tee is used to efficiently create 3 independent iterators over the input sequence chain links two sequences into one  it s used here to append a single-element sequence  None  to prevs islice is used to make a sequence of all elements except the first  then chain is used to append a None to its end There are now 3 independent sequences based on some iterable that look like    prevs  None  A   B   C   D   E items  A     B   C   D   E nexts  B     C   D   E   None  finally izip is used to change 3 sequences into one sequence of triplets    Note that izip stops when any input sequence gets exhausted  so the last element of prevs will be ignored  which is correct - there s no such element that the last element would be its prev  We could try to strip off the last elements from prevs but izip s behaviour makes that redundant   Also note that tee  izip  islice and chain come from the itertools module  they operate on their input sequences on-the-fly  lazily   which makes them efficient and doesn t introduce the need of having the whole sequence in memory at once at any time   In python 3  it will show an error while importing izip you can use zip instead of izip  No need to import zip  it is predefined in python 3 -  source

User · Answer

You could just use index on the list to find where somevalue is and then get the previous and next as needed     def find prev next elem  elements       previous  next   None  None     index   elements index elem      if index  gt  0          previous   elements index -1      if index  lt   len elements -1           next   elements index  1      return previous  next   foo    three  list     one   two   three    four    five    previous  next   find prev next foo  list   print previous   should print  two  print next   should print  four

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Using a list comprehension  return a 3-tuple with current  previous and next elements   three tuple     current                   my list idx - 1  if idx  gt   1 else None                   my list idx   1  if idx  lt  len my list  - 1 else None  for idx  current in enumerate my list

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Two simple solutions   If variables for both previous and next values have to be defined   alist     Zero    One    Two    Three    Four    Five    prev   alist 0  curr   alist 1   for nxt in alist 2        print f prev   prev   curr   curr   next   nxt        prev   curr     curr   nxt  Output 1   prev  Zero  curr  One  next  Two prev  One  curr  Two  next  Three prev  Two  curr  Three  next  Four prev  Three  curr  Four  next  Five   If all values in the list have to be traversed by the current value variable   alist     Zero    One    Two    Three    Four    Five    prev   None curr   alist 0   for nxt in alist 1      None       print f prev   prev   curr   curr   next   nxt        prev   curr     curr   nxt  Output 2   prev  None  curr  Zero  next  One prev  Zero  curr  One  next  Two prev  One  curr  Two  next  Three prev  Two  curr  Three  next  Four prev  Three  curr  Four  next  Five prev  Four  curr  Five  next  None

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Very C C   style solution      foo   5     objectsList    3  6  5  9  10      prev   nex   0          currentIndex   0     indexHigher   len objectsList -1  control the higher limit of list          found   False     prevFound   False     nexFound   False           main logic      for currentValue in objectsList   getting each value of list         if currentValue    foo              found   True             if currentIndex  gt  0   check if target value is in the first position                    prevFound   True                 prev   objectsList currentIndex-1              if currentIndex  lt  indexHigher   check if target value is in the last position                 nexFound   True                 nex   objectsList currentIndex 1              break  I am considering that target value only exist 1 time in the list         currentIndex  1          if found          print  quot Value  s found quot    foo          if prevFound              print  quot Previous Value   quot   prev          else              print  quot Previous Value  Target value is in the first position of list  quot           if nexFound              print  quot Next Value   quot   nex          else              print  quot Next Value  Target value is in the last position of list  quot       else          print  quot Target value does not exist in the list  quot

User · Answer

AFAIK this should be pretty fast  but I didn t test it   def iterate prv nxt my list       prv  cur  nxt   None  iter my list   iter my list      next nxt  None       while True          try              if prv                  yield next prv   next cur   next nxt  None              else                  yield None  next cur   next nxt  None                  prv   iter my list          except StopIteration              break   Example usage    gt  gt  gt  my list     a    b    c    gt  gt  gt  for prv  cur  nxt in iterate prv nxt my list          print prv  cur  nxt      None a b a b c b c None

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This should do the trick   foo   somevalue previous   next    None l   len objects  for index  obj in enumerate objects       if obj    foo          if index  gt  0              previous   objects index - 1          if index  lt   l - 1               next    objects index   1    Here s the docs on the enumerate function

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Here s a version using generators with no boundary errors   def trios iterable       it   iter iterable      try          prev  current   next it   next it      except StopIteration          return     for next in it          yield prev  current  next         prev  current   current  next  def find prev next objects  foo       prev  next   0  0     for temp prev  current  temp next in trios objects           if current    foo              prev  next   temp prev  temp next     return prev  next  print find prev next range 10   1   print find prev next range 10   0   print find prev next range 10   10   print find prev next range 0   10   print find prev next range 1   10   print find prev next range 2   10     Please notice that the boundary behavior is that we never look for  foo  in the first or last element  unlike your code  Again  the boundary semantics are strange   and are hard to fathom from your code

[python] Python loop that also accesses previous and next values

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