Binary search bisection in Python

Question

Is there a library function that performs binary search on a list tuple and return the position of the item if found and  False   -1  None  etc   if not   I found the functions bisect left right in the bisect module  but they still return a position even if the item is not in the list  That s perfectly fine for their intended usage  but I just want to know if an item is in the list or not  don t want to insert anything    I thought of using bisect left and then checking if the item at that position is equal to what I m searching  but that seems cumbersome  and I also need to do bounds checking if the number can be larger than the largest number in my list   If there is a nicer method I d like to know about it   Edit To clarify what I need this for  I m aware that a dictionary would be very well suited for this  but I m trying to keep the memory consumption as low as possible  My intended usage would be a sort of double-way look-up table  I have in the table a list of values and I need to be able to access the values based on their index  And also I want to be able to find the index of a particular value or None if the value is not in the list   Using a dictionary for this would be the fastest way  but would  approximately  double the memory requirements   I was asking this question thinking that I may have overlooked something in the Python libraries  It seems I ll have to write my own code  as Moe suggested

User · Answer

It might be worth mentioning that the bisect docs now provide searching examples: http://docs.python.org/library/bisect.html#searching-sorted-lists

(Raising ValueError instead of returning -1 or None is more pythonic – list.index() does it, for example. But of course you can adapt the examples to your needs.)

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def binary search length of a list single method list       index   0     first   0     last   1      while True          mid     first   last     2          if not single method list get index               break         index   mid   1         first   index         last   index   1     return mid

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If you just want to see if it s present  try turning the list into a dict     Generate a list l    n n for n in range 1000      Convert to dict - doesn t matter what you map values to d   dict  x  1  for x in l   count   0 for n in range 1000000         Compare with  if n in l      if n in d          count    1   On my machine   if n in l  took 37 seconds  while  if n in d  took 0 4 seconds

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Using a dict wouldn t like double your memory usage unless the objects you re storing are really tiny  since the values are only pointers to the actual objects    gt  gt  gt  a    foo   gt  gt  gt  b    a   gt  gt  gt  c    a   gt  gt  gt  b 0  is c 0  True   In that example   foo  is only stored once   Does that make a difference for you   And exactly how many items are we talking about anyway

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I agree that  DaveAbrahams s answer using the bisect module is the correct approach   He did not mention one important detail in his answer   From the docs  bisect bisect left a  x  lo 0  hi len a    The bisection module does not require the search array to be precomputed ahead of time   You can just present the endpoints to the bisect bisect left instead of it using the defaults of 0 and len a    Even more important for my use  looking for a value X such that the error of a given function is minimized   To do that  I needed a way to have the bisect left s algorithm call my computation instead   This is really simple   Just provide an object that defines   getitem   as a  For example  we could use the bisect algorithm to find a square root with arbitrary precision   import bisect  class sqrt array object       def   init   self  digits           self precision   float 10   digits       def   getitem   self  key           return  key self precision   2 0  sa   sqrt array 4      search  in the range of 0 to 10 with a  precision  of 0 0001 index   bisect bisect left sa  7  0  10 10  4  print 7  0 5 print index  10  4 0

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If you just want to see if it s present  try turning the list into a dict     Generate a list l    n n for n in range 1000      Convert to dict - doesn t matter what you map values to d   dict  x  1  for x in l   count   0 for n in range 1000000         Compare with  if n in l      if n in d          count    1   On my machine   if n in l  took 37 seconds  while  if n in d  took 0 4 seconds

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This one is    not recursive  which makes it more memory-efficient than most recursive approaches  actually working fast since it runs without any unnecessary if s and conditions based on a mathematical assertion that the floor of  low   high  2 is always smaller than high where low is the lower limit and high is the upper limit      def binsearch t  key  low   0  high   len t  - 1         bisecting the range     while low  lt  high          mid    low   high   2         if t mid   lt  key              low   mid   1         else              high   mid       at this point  low  should point at the place       where the value of  key  is possibly stored      return low if t low     key else -1

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def binary search length of a list single method list       index   0     first   0     last   1      while True          mid     first   last     2          if not single method list get index               break         index   mid   1         first   index         last   index   1     return mid

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Only used if set your position as global     position  set global   def bst array taget     just pass the array and target         global position         low   0         high   len array      while low  lt   high          mid    lo hi   2         if a mid     target              position   mid             return -1         elif a mid   lt  target               high   mid 1         else              low   mid-1     return -1   I guess this is much better and effective  please correct me      Thank you

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s is a list    binary s  0  len s  - 1  find  is the initial call  Function returns an index of the queried item  If there is no such item it returns -1   def binary s p q find       if find  s  p q  2           return  p q  2     elif p  q-1 or p  q          if find  s q               return q         else              return -1     elif find  lt  s  p q  2           return binary s p  p q  2 find      elif find  gt  s  p q  2           return binary s  p q  2 1 q find

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This is a little off-topic  since Moe s answer seems complete to the OP s question   but it might be worth looking at the complexity for your whole procedure from end to end   If you re storing thing in a sorted lists  which is where a binary search would help   and then just checking for existence  you re incurring  worst-case  unless specified    Sorted Lists   O  n log n  to initially create the list  if it s unsorted data   O n   if it s sorted   O  log n  lookups  this is the binary search part  O  n   insert   delete   might be O 1  or O log n  average case  depending on your pattern    Whereas with a set    you re incurring   O n  to create O 1  lookup O 1  insert   delete   The thing a sorted list really gets you are  next    previous   and  ranges   including inserting or deleting ranges   which are O 1  or O  range    given a starting index   If you aren t using those sorts of operations often  then storing as sets  and sorting for display might be a better deal overall   set   incurs very little additional overhead in python

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This is right from the manual   http   docs python org 2 library bisect html  8 5 1  Searching Sorted Lists  The above bisect   functions are useful for finding insertion points but can be tricky or awkward to use for common searching tasks  The following five functions show how to transform them into the standard lookups for sorted lists   def index a  x        Locate the leftmost value exactly equal to x      i   bisect left a  x      if i    len a  and a i     x          return i     raise ValueError   So with the slight modification your code should be   def index a  x        Locate the leftmost value exactly equal to x      i   bisect left a  x      if i    len a  and a i     x          return i     return -1

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It might be worth mentioning that the bisect docs now provide searching examples: http://docs.python.org/library/bisect.html#searching-sorted-lists

(Raising ValueError instead of returning -1 or None is more pythonic – list.index() does it, for example. But of course you can adapt the examples to your needs.)

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Using a dict wouldn t like double your memory usage unless the objects you re storing are really tiny  since the values are only pointers to the actual objects    gt  gt  gt  a    foo   gt  gt  gt  b    a   gt  gt  gt  c    a   gt  gt  gt  b 0  is c 0  True   In that example   foo  is only stored once   Does that make a difference for you   And exactly how many items are we talking about anyway

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I needed binary search in python and generic for Django models  In Django models  one model can have foreign key to another model and I wanted to perform some search on the retrieved models objects  I wrote following function you can use this   def binary search values  key  lo 0  hi None  length None  cmp None               This is a binary search function which search for given key in values      This is very generic since values and key can be of different type      If they are of different type then caller must specify  cmp  function to     perform a comparison between key and values  item       param values   List of items in which key has to be search      param key  search key      param lo  start index to begin search      param hi  end index where search will be performed      param length  length of values      param cmp  a comparator function which can be used to compare key and values      return  -1 if key is not found else index             assert type values 0      type key  or cmp   can t be compared      assert not  hi and length     hi    length  both can t be specified at the same time       lo   lo     if not lo          lo   0     if hi          hi   hi     elif length          hi   length - 1     else          hi   len values  - 1      while lo  lt   hi          mid   lo    hi - lo     2         if not cmp              if values mid     key                  return mid             if values mid   lt  key                  lo   mid   1             else                  hi   mid - 1         else              val   cmp values mid   key                0 - gt  a    b                gt  0 - gt  a  gt  b                lt  0 - gt  a  lt  b             if val    0                  return mid             if val  lt  0                  lo   mid   1             else                  hi   mid - 1     return -1

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I agree that  DaveAbrahams s answer using the bisect module is the correct approach   He did not mention one important detail in his answer   From the docs  bisect bisect left a  x  lo 0  hi len a    The bisection module does not require the search array to be precomputed ahead of time   You can just present the endpoints to the bisect bisect left instead of it using the defaults of 0 and len a    Even more important for my use  looking for a value X such that the error of a given function is minimized   To do that  I needed a way to have the bisect left s algorithm call my computation instead   This is really simple   Just provide an object that defines   getitem   as a  For example  we could use the bisect algorithm to find a square root with arbitrary precision   import bisect  class sqrt array object       def   init   self  digits           self precision   float 10   digits       def   getitem   self  key           return  key self precision   2 0  sa   sqrt array 4      search  in the range of 0 to 10 with a  precision  of 0 0001 index   bisect bisect left sa  7  0  10 10  4  print 7  0 5 print index  10  4 0

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Using a dict wouldn t like double your memory usage unless the objects you re storing are really tiny  since the values are only pointers to the actual objects    gt  gt  gt  a    foo   gt  gt  gt  b    a   gt  gt  gt  c    a   gt  gt  gt  b 0  is c 0  True   In that example   foo  is only stored once   Does that make a difference for you   And exactly how many items are we talking about anyway

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Dave Abrahams  solution is good  Although I have would have done it minimalistic   def binary search L  x       i   bisect bisect left L  x      if i    len L  or L i     x          return -1     return i

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While there s no explicit binary search algorithm in Python  there is a module - bisect - designed to find the insertion point for an element in a sorted list using a binary search  This can be  tricked  into performing a binary search  The biggest advantage of this is the same advantage most library code has - it s high-performing  well-tested and just works  binary searches in particular can be quite difficult to implement successfully - particularly if edge cases aren t carefully considered    Basic Types  For basic types like Strings or ints it s pretty easy - all you need is the bisect module and a sorted list    gt  gt  gt  import bisect  gt  gt  gt  names     bender    fry    leela    nibbler    zoidberg    gt  gt  gt  bisect bisect left names   fry   1  gt  gt  gt  keyword    fry   gt  gt  gt  x   bisect bisect left names  keyword   gt  gt  gt  names x     keyword True  gt  gt  gt  keyword    arnie   gt  gt  gt  x   bisect bisect left names  keyword   gt  gt  gt  names x     keyword False   You can also use this to find duplicates        gt  gt  gt  names     bender    fry    fry    fry    leela    nibbler    zoidberg    gt  gt  gt  keyword    fry   gt  gt  gt  leftIndex   bisect bisect left names  keyword   gt  gt  gt  rightIndex   bisect bisect right names  keyword   gt  gt  gt  names leftIndex rightIndex    fry    fry    fry     Obviously you could just return the index rather than the value at that index if desired   Objects  For custom types or objects  things are a little bit trickier  you have to make sure to implement rich comparison methods to get bisect to compare correctly    gt  gt  gt  import bisect  gt  gt  gt  class Tag object      a simple wrapper around strings         def   init   self  tag               self tag   tag         def   lt   self  other               return self tag  lt  other tag         def   gt   self  other               return self tag  gt  other tag      gt  gt  gt  tags    Tag  bender    Tag  fry    Tag  leela    Tag  nibbler    Tag  zoidbe rg     gt  gt  gt  key   Tag  fry    gt  gt  gt  leftIndex   bisect bisect left tags  key   gt  gt  gt  rightIndex   bisect bisect right tags  key   gt  gt  gt  print  tag tag for tag in tags leftIndex rightIndex      fry     This should work in at least Python 2 7 -  3 3

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This code works with integer lists in a recursive way  Looks for the simplest case scenario  which is  list length less than 2  It means the answer is already there and  a test is performed to check for the correct answer  If not  a middle value is set and tested to be the correct  if not bisection is performed by calling again the function  but setting middle value as the upper or lower limit  by shifting it to the left or right   def binary search intList  intValue  lowValue  highValue       if highValue - lowValue   lt 2          return intList lowValue     intValue or intList highValue     intValue     middleValue   lowValue     highValue - lowValue  2      if intList middleValue     intValue          return True     if intList middleValue    intValue          return binary search intList  intValue  lowValue  middleValue - 1     return binary search intList  intValue  middleValue   1  highValue

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bisect left finds the first position p at which an element could be inserted in a given sorted range while maintaining the sorted order   That will be the position of x if x exists in the range  If p is the past-the-end position  x wasn t found   Otherwise  we can test to see if x is there to see if x was found   from bisect import bisect left  def binary search a  x  lo 0  hi None       if hi is None  hi   len a      pos   bisect left a  x  lo  hi                     find insertion position     return pos if pos    hi and a pos     x else -1    don t walk off the end

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I needed binary search in python and generic for Django models  In Django models  one model can have foreign key to another model and I wanted to perform some search on the retrieved models objects  I wrote following function you can use this   def binary search values  key  lo 0  hi None  length None  cmp None               This is a binary search function which search for given key in values      This is very generic since values and key can be of different type      If they are of different type then caller must specify  cmp  function to     perform a comparison between key and values  item       param values   List of items in which key has to be search      param key  search key      param lo  start index to begin search      param hi  end index where search will be performed      param length  length of values      param cmp  a comparator function which can be used to compare key and values      return  -1 if key is not found else index             assert type values 0      type key  or cmp   can t be compared      assert not  hi and length     hi    length  both can t be specified at the same time       lo   lo     if not lo          lo   0     if hi          hi   hi     elif length          hi   length - 1     else          hi   len values  - 1      while lo  lt   hi          mid   lo    hi - lo     2         if not cmp              if values mid     key                  return mid             if values mid   lt  key                  lo   mid   1             else                  hi   mid - 1         else              val   cmp values mid   key                0 - gt  a    b                gt  0 - gt  a  gt  b                lt  0 - gt  a  lt  b             if val    0                  return mid             if val  lt  0                  lo   mid   1             else                  hi   mid - 1     return -1

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If you just want to see if it s present  try turning the list into a dict     Generate a list l    n n for n in range 1000      Convert to dict - doesn t matter what you map values to d   dict  x  1  for x in l   count   0 for n in range 1000000         Compare with  if n in l      if n in d          count    1   On my machine   if n in l  took 37 seconds  while  if n in d  took 0 4 seconds

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While there s no explicit binary search algorithm in Python  there is a module - bisect - designed to find the insertion point for an element in a sorted list using a binary search  This can be  tricked  into performing a binary search  The biggest advantage of this is the same advantage most library code has - it s high-performing  well-tested and just works  binary searches in particular can be quite difficult to implement successfully - particularly if edge cases aren t carefully considered    Basic Types  For basic types like Strings or ints it s pretty easy - all you need is the bisect module and a sorted list    gt  gt  gt  import bisect  gt  gt  gt  names     bender    fry    leela    nibbler    zoidberg    gt  gt  gt  bisect bisect left names   fry   1  gt  gt  gt  keyword    fry   gt  gt  gt  x   bisect bisect left names  keyword   gt  gt  gt  names x     keyword True  gt  gt  gt  keyword    arnie   gt  gt  gt  x   bisect bisect left names  keyword   gt  gt  gt  names x     keyword False   You can also use this to find duplicates        gt  gt  gt  names     bender    fry    fry    fry    leela    nibbler    zoidberg    gt  gt  gt  keyword    fry   gt  gt  gt  leftIndex   bisect bisect left names  keyword   gt  gt  gt  rightIndex   bisect bisect right names  keyword   gt  gt  gt  names leftIndex rightIndex    fry    fry    fry     Obviously you could just return the index rather than the value at that index if desired   Objects  For custom types or objects  things are a little bit trickier  you have to make sure to implement rich comparison methods to get bisect to compare correctly    gt  gt  gt  import bisect  gt  gt  gt  class Tag object      a simple wrapper around strings         def   init   self  tag               self tag   tag         def   lt   self  other               return self tag  lt  other tag         def   gt   self  other               return self tag  gt  other tag      gt  gt  gt  tags    Tag  bender    Tag  fry    Tag  leela    Tag  nibbler    Tag  zoidbe rg     gt  gt  gt  key   Tag  fry    gt  gt  gt  leftIndex   bisect bisect left tags  key   gt  gt  gt  rightIndex   bisect bisect right tags  key   gt  gt  gt  print  tag tag for tag in tags leftIndex rightIndex      fry     This should work in at least Python 2 7 -  3 3

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Dave Abrahams  solution is good  Although I have would have done it minimalistic   def binary search L  x       i   bisect bisect left L  x      if i    len L  or L i     x          return -1     return i

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Check out the examples on Wikipedia http   en wikipedia org wiki Binary search algorithm  def binary search a  key  imin 0  imax None       if imax is None            if max amount not set  get the total         imax   len a  - 1      while imin  lt   imax            calculate the midpoint         mid    imin   imax   2         midval   a mid             determine which subarray to search         if midval  lt  key                change min index to search upper subarray             imin   mid   1         elif midval  gt  key                change max index to search lower subarray             imax   mid - 1         else                return index number              return mid     raise ValueError

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bisect left finds the first position p at which an element could be inserted in a given sorted range while maintaining the sorted order   That will be the position of x if x exists in the range  If p is the past-the-end position  x wasn t found   Otherwise  we can test to see if x is there to see if x was found   from bisect import bisect left  def binary search a  x  lo 0  hi None       if hi is None  hi   len a      pos   bisect left a  x  lo  hi                     find insertion position     return pos if pos    hi and a pos     x else -1    don t walk off the end

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If you just want to see if it s present  try turning the list into a dict     Generate a list l    n n for n in range 1000      Convert to dict - doesn t matter what you map values to d   dict  x  1  for x in l   count   0 for n in range 1000000         Compare with  if n in l      if n in d          count    1   On my machine   if n in l  took 37 seconds  while  if n in d  took 0 4 seconds

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Why not look at the code for bisect left right and adapt it to suit your purpose   like this   def binary search a  x  lo 0  hi None       if hi is None          hi   len a      while lo  lt  hi          mid    lo hi   2         midval   a mid          if midval  lt  x              lo   mid 1         elif midval  gt  x               hi   mid         else              return mid     return -1

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Check out the examples on Wikipedia http   en wikipedia org wiki Binary search algorithm  def binary search a  key  imin 0  imax None       if imax is None            if max amount not set  get the total         imax   len a  - 1      while imin  lt   imax            calculate the midpoint         mid    imin   imax   2         midval   a mid             determine which subarray to search         if midval  lt  key                change min index to search upper subarray             imin   mid   1         elif midval  gt  key                change max index to search lower subarray             imax   mid - 1         else                return index number              return mid     raise ValueError

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Why not look at the code for bisect left right and adapt it to suit your purpose   like this   def binary search a  x  lo 0  hi None       if hi is None          hi   len a      while lo  lt  hi          mid    lo hi   2         midval   a mid          if midval  lt  x              lo   mid 1         elif midval  gt  x               hi   mid         else              return mid     return -1

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Many good solutions above but I haven t seen a simple  KISS keep it simple  cause  I m  stupid use of the Python built in generic bisect function to do a binary search   With a bit of code around the bisect function  I think I have an example below where I have tested all cases for a small string array of names  Some of the above solutions allude to say this  but hopefully the simple code below will help anyone confused like I was     Python bisect is used to indicate where to insert an a new value search item into a sorted list  The below code which uses bisect left which will return the index of the hit if the search item in the list array is found  Note bisect and bisect right will return the index of the element after the hit or match as the insertion point  If not found  bisect left will return an index to the next item in the sorted list which will not    the search value   The only other case is where the search item would go at the end of the list where the index returned would be beyond the end of the list array  and which in the code below the early exit by Python with  and  logic handles   first condition False Python does not check subsequent conditions    Code from bisect import bisect left names   Adam   Donny   Jalan   Zach   Zayed   search    lenNames   len names  while search    none       search  input  Enter name to search for or  none  to terminate program        if search     none           break     i   bisect left names search      print i    show index returned by Python bisect left     if i  lt   lenNames  and names i     search          print names i   found    return True - if function     else          print search  not found    return False     if function   Exhaustive test cases    Enter name to search for or  none  to terminate program Zayed   4   Zayed found   Enter name to search for or  none  to terminate program Zach   3   Zach found   Enter name to search for or  none  to terminate program Jalan   2   Jalan found   Enter name to search for or  none  to terminate program Donny   1   Donny found   Enter name to search for or  none  to terminate program Adam   0   Adam found   Enter name to search for or  none  to terminate program Abie   0   Abie not found   Enter name to search for or  none  to terminate program Carla   1   Carla not found   Enter name to search for or  none  to terminate program Ed   2   Ed not found   Enter name to search for or  none  to terminate program Roger   3   Roger not found   Enter name to search for or  none  to terminate program Zap   4   Zap not found   Enter name to search for or  none  to terminate program Zyss   5   Zyss not found

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Only used if set your position as global     position  set global   def bst array taget     just pass the array and target         global position         low   0         high   len array      while low  lt   high          mid    lo hi   2         if a mid     target              position   mid             return -1         elif a mid   lt  target               high   mid 1         else              low   mid-1     return -1   I guess this is much better and effective  please correct me      Thank you

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This code works with integer lists in a recursive way  Looks for the simplest case scenario  which is  list length less than 2  It means the answer is already there and  a test is performed to check for the correct answer  If not  a middle value is set and tested to be the correct  if not bisection is performed by calling again the function  but setting middle value as the upper or lower limit  by shifting it to the left or right   def binary search intList  intValue  lowValue  highValue       if highValue - lowValue   lt 2          return intList lowValue     intValue or intList highValue     intValue     middleValue   lowValue     highValue - lowValue  2      if intList middleValue     intValue          return True     if intList middleValue    intValue          return binary search intList  intValue  lowValue  middleValue - 1     return binary search intList  intValue  middleValue   1  highValue

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This is right from the manual   http   docs python org 2 library bisect html  8 5 1  Searching Sorted Lists  The above bisect   functions are useful for finding insertion points but can be tricky or awkward to use for common searching tasks  The following five functions show how to transform them into the standard lookups for sorted lists   def index a  x        Locate the leftmost value exactly equal to x      i   bisect left a  x      if i    len a  and a i     x          return i     raise ValueError   So with the slight modification your code should be   def index a  x        Locate the leftmost value exactly equal to x      i   bisect left a  x      if i    len a  and a i     x          return i     return -1

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Simplest is to use bisect and check one position back to see if the item is there   def binary search a x lo 0 hi -1       i   bisect a x lo hi      if i    0          return -1     elif a i-1     x          return i-1     else          return -1

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s is a list    binary s  0  len s  - 1  find  is the initial call  Function returns an index of the queried item  If there is no such item it returns -1   def binary s p q find       if find  s  p q  2           return  p q  2     elif p  q-1 or p  q          if find  s q               return q         else              return -1     elif find  lt  s  p q  2           return binary s p  p q  2 find      elif find  gt  s  p q  2           return binary s  p q  2 1 q find

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This one is    not recursive  which makes it more memory-efficient than most recursive approaches  actually working fast since it runs without any unnecessary if s and conditions based on a mathematical assertion that the floor of  low   high  2 is always smaller than high where low is the lower limit and high is the upper limit      def binsearch t  key  low   0  high   len t  - 1         bisecting the range     while low  lt  high          mid    low   high   2         if t mid   lt  key              low   mid   1         else              high   mid       at this point  low  should point at the place       where the value of  key  is possibly stored      return low if t low     key else -1

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Using a dict wouldn t like double your memory usage unless the objects you re storing are really tiny  since the values are only pointers to the actual objects    gt  gt  gt  a    foo   gt  gt  gt  b    a   gt  gt  gt  c    a   gt  gt  gt  b 0  is c 0  True   In that example   foo  is only stored once   Does that make a difference for you   And exactly how many items are we talking about anyway

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This is a little off-topic  since Moe s answer seems complete to the OP s question   but it might be worth looking at the complexity for your whole procedure from end to end   If you re storing thing in a sorted lists  which is where a binary search would help   and then just checking for existence  you re incurring  worst-case  unless specified    Sorted Lists   O  n log n  to initially create the list  if it s unsorted data   O n   if it s sorted   O  log n  lookups  this is the binary search part  O  n   insert   delete   might be O 1  or O log n  average case  depending on your pattern    Whereas with a set    you re incurring   O n  to create O 1  lookup O 1  insert   delete   The thing a sorted list really gets you are  next    previous   and  ranges   including inserting or deleting ranges   which are O 1  or O  range    given a starting index   If you aren t using those sorts of operations often  then storing as sets  and sorting for display might be a better deal overall   set   incurs very little additional overhead in python

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Simplest is to use bisect and check one position back to see if the item is there   def binary search a x lo 0 hi -1       i   bisect a x lo hi      if i    0          return -1     elif a i-1     x          return i-1     else          return -1

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Why not look at the code for bisect left right and adapt it to suit your purpose   like this   def binary search a  x  lo 0  hi None       if hi is None          hi   len a      while lo  lt  hi          mid    lo hi   2         midval   a mid          if midval  lt  x              lo   mid 1         elif midval  gt  x               hi   mid         else              return mid     return -1

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Binary Search           List - values inside list    searchItem - Item to search    size - Size of list    upperBound - higher index of list    lowerBound - lower index of list def binarySearch list  searchItem  size  upperBound  lowerBound           print list          print upperBound          print lowerBound          mid     upperBound   lowerBound      2         print mid          if int list int mid       value                 return  value exist          elif int list int mid     lt  value               return searchItem list  value  size  upperBound  mid   1          elif int list int mid     gt  value                 return searchItem list  value  size  mid - 1  lowerBound       To call above function use    list    0  1  2  3  4  5  6  7  8  9  10  searchItem   1         print searchItem list 0   item  len list 0   -1  len list 0   - 1  0

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This is a little off-topic  since Moe s answer seems complete to the OP s question   but it might be worth looking at the complexity for your whole procedure from end to end   If you re storing thing in a sorted lists  which is where a binary search would help   and then just checking for existence  you re incurring  worst-case  unless specified    Sorted Lists   O  n log n  to initially create the list  if it s unsorted data   O n   if it s sorted   O  log n  lookups  this is the binary search part  O  n   insert   delete   might be O 1  or O log n  average case  depending on your pattern    Whereas with a set    you re incurring   O n  to create O 1  lookup O 1  insert   delete   The thing a sorted list really gets you are  next    previous   and  ranges   including inserting or deleting ranges   which are O 1  or O  range    given a starting index   If you aren t using those sorts of operations often  then storing as sets  and sorting for display might be a better deal overall   set   incurs very little additional overhead in python

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Binary Search           List - values inside list    searchItem - Item to search    size - Size of list    upperBound - higher index of list    lowerBound - lower index of list def binarySearch list  searchItem  size  upperBound  lowerBound           print list          print upperBound          print lowerBound          mid     upperBound   lowerBound      2         print mid          if int list int mid       value                 return  value exist          elif int list int mid     lt  value               return searchItem list  value  size  upperBound  mid   1          elif int list int mid     gt  value                 return searchItem list  value  size  mid - 1  lowerBound       To call above function use    list    0  1  2  3  4  5  6  7  8  9  10  searchItem   1         print searchItem list 0   item  len list 0   -1  len list 0   - 1  0

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Why not look at the code for bisect left right and adapt it to suit your purpose   like this   def binary search a  x  lo 0  hi None       if hi is None          hi   len a      while lo  lt  hi          mid    lo hi   2         midval   a mid          if midval  lt  x              lo   mid 1         elif midval  gt  x               hi   mid         else              return mid     return -1

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Many good solutions above but I haven t seen a simple  KISS keep it simple  cause  I m  stupid use of the Python built in generic bisect function to do a binary search   With a bit of code around the bisect function  I think I have an example below where I have tested all cases for a small string array of names  Some of the above solutions allude to say this  but hopefully the simple code below will help anyone confused like I was     Python bisect is used to indicate where to insert an a new value search item into a sorted list  The below code which uses bisect left which will return the index of the hit if the search item in the list array is found  Note bisect and bisect right will return the index of the element after the hit or match as the insertion point  If not found  bisect left will return an index to the next item in the sorted list which will not    the search value   The only other case is where the search item would go at the end of the list where the index returned would be beyond the end of the list array  and which in the code below the early exit by Python with  and  logic handles   first condition False Python does not check subsequent conditions    Code from bisect import bisect left names   Adam   Donny   Jalan   Zach   Zayed   search    lenNames   len names  while search    none       search  input  Enter name to search for or  none  to terminate program        if search     none           break     i   bisect left names search      print i    show index returned by Python bisect left     if i  lt   lenNames  and names i     search          print names i   found    return True - if function     else          print search  not found    return False     if function   Exhaustive test cases    Enter name to search for or  none  to terminate program Zayed   4   Zayed found   Enter name to search for or  none  to terminate program Zach   3   Zach found   Enter name to search for or  none  to terminate program Jalan   2   Jalan found   Enter name to search for or  none  to terminate program Donny   1   Donny found   Enter name to search for or  none  to terminate program Adam   0   Adam found   Enter name to search for or  none  to terminate program Abie   0   Abie not found   Enter name to search for or  none  to terminate program Carla   1   Carla not found   Enter name to search for or  none  to terminate program Ed   2   Ed not found   Enter name to search for or  none  to terminate program Roger   3   Roger not found   Enter name to search for or  none  to terminate program Zap   4   Zap not found   Enter name to search for or  none  to terminate program Zyss   5   Zyss not found

User · Answer

This is a little off-topic  since Moe s answer seems complete to the OP s question   but it might be worth looking at the complexity for your whole procedure from end to end   If you re storing thing in a sorted lists  which is where a binary search would help   and then just checking for existence  you re incurring  worst-case  unless specified    Sorted Lists   O  n log n  to initially create the list  if it s unsorted data   O n   if it s sorted   O  log n  lookups  this is the binary search part  O  n   insert   delete   might be O 1  or O log n  average case  depending on your pattern    Whereas with a set    you re incurring   O n  to create O 1  lookup O 1  insert   delete   The thing a sorted list really gets you are  next    previous   and  ranges   including inserting or deleting ranges   which are O 1  or O  range    given a starting index   If you aren t using those sorts of operations often  then storing as sets  and sorting for display might be a better deal overall   set   incurs very little additional overhead in python

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