finding and replacing elements in a list

Question

I have to search through a list and replace all occurrences of one element with another  So far my attempts in code are getting me nowhere  what is the best way to do this   For example  suppose my list has the following integers   gt  gt  gt  a    1 2 3 4 5 1 2 3 4 5 1    and I need to replace all occurrences of the number 1 with the value 10 so the output I need is    gt  gt  gt  a    10  2  3  4  5  10  2  3  4  5  10    Thus my goal is to replace all instances of the number 1 with the number 10

User · Answer

On long lists and rare occurrences its about 3x faster using list index   - compared to single step iteration methods presented in the other answers   def list replace lst  old 1  new 10          replace list elements  inplace         i   -1     try          while 1              i   lst index old  i   1              lst i    new     except ValueError          pass

User · Answer

You can simply use list comprehension in python   def replace element YOUR LIST  set to NEW VALUE       return  i             if SOME CONDITION             else NEW VALUE             for i in YOUR LIST    for your case  where you want to replace all occurrences of 1 with 10  the code snippet will be like this   def replace element YOUR LIST  set to 10       return  i             if i    1    keeps all elements not equal to one             else set to    replaces 1 with 10             for i in YOUR LIST

User · Answer

I know this is a very old question and there s a myriad of ways to do it  The simpler one I found is using numpy package   import numpy  arr   numpy asarray  1  6  1  9  8   arr  arr    8     0   change all occurrences of 8 by 0 print arr

User · Answer

This can be easily done by using enumerate function code- lst  1 2 3 4 1 6 7 9 10 1 2  for index item in enumerate lst       if item  1          lst index  10  Replaces the item  1  in list with  10   print lst

User · Answer

Here s a cool and scalable design pattern that runs in O n  time     a    1 2 3 4 5 6 7 6 5 4 3 2 1   replacements         1  10      2  20      3  30     a    replacements get x  x  for x in a   print a    Returns  10  20  30  4  5  6  7  6  5  4  30  20  10

User · Answer

a    1 2 3 4 5 1 2 3 4 5 1 12  for i in range  len a        if a i   2          a i  123   You can use a for and or while loop  however if u know the builtin Enumerate function  then it is recommended to use Enumerate 1

User · Answer

List comprehension works well  and looping through with enumerate can save you some memory  b c the operation s essentially being done in place    There s also functional programming  See usage of map    gt  gt  gt  a    1 2 3 2 3 4 3 5 6 6 5 4 5 4 3 4 3 2 1   gt  gt  gt  map lambda x  x if x    4 else  sss   a   1  2  3  2  3   sss   3  5  6  6  5   sss   5   sss   3   sss   3  2  1

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gt  gt  gt  a   1  2  3  4  5  1  2  3  4  5  1   gt  gt  gt  for n  i in enumerate a         if i    1           a n    10      gt  gt  gt  a  10  2  3  4  5  10  2  3  4  5  10

User · Answer

If you have several values to replace  you can also use a dictionary   a    1  2  3  4  1  5  3  2  6  1  1  dic    1 10  2 20  3  foo    print  dic get n  n  for n in a     gt   10  20   foo   4  10  5   foo   20  6  10  10

User · Answer

Try using a list comprehension and the ternary operator    gt  gt  gt  a  1 2 3 1 3 2 1 1   gt  gt  gt   4 if x  1 else x for x in a   4  2  3  4  3  2  4  4

User · Answer

To replace easily all 1 with 10 in  a    1 2 3 4 5 1 2 3 4 5 1 one could use the following one-line lambda map combination  and  Look  Ma  no IFs or FORs        This substitutes all  1  with  10  in list  a  and places result in list  c    c   list map lambda b  b replace  1   10    a

User · Answer

The following is a very straightforward method in Python 3 x  a    1 2 3 4 5 1 2 3 4 5 1          Replacing every 1 with 10  for i in range len a       if a i     1       a i    10    print a   This method works  Comments are welcome  Hope it helps    Also try understanding how outis s and damzam s solutions work  List compressions and lambda function are useful tools

User · Answer

My usecase was replacing None with some default value    I ve timed approaches to this problem that were presented here  including the one by  kxr - using str count   Test code in ipython with Python 3 8 1   def rep1 lst  replacer   0           List comprehension  new list          return  item if item is not None else replacer for item in lst    def rep2 lst  replacer   0           List comprehension  in-place             lst        item if item is not None else replacer for item in lst       return lst   def rep3 lst  replacer   0           enumerate   with comparison - in-place         for idx  item in enumerate lst           if item is None              lst idx    replacer      return lst   def rep4 lst  replacer   0           Using str index   Exception  in-place          idx   -1       none amount   lst count None      while True          try              idx   lst index None  idx 1          except ValueError              break         else              lst idx    replacer      return lst   def rep5 lst  replacer   0           Using str index   str count  in-place          idx   -1     for   in range lst count None            idx   lst index None  idx 1          lst idx    replacer      return lst   def rep6 lst  replacer   0           Using map  return map iterator          return map lambda item  item if item is not None else replacer  lst    def rep7 lst  replacer   0           Using map  return new list          return list map lambda item  item if item is not None else replacer  lst     lst    5  10  6   lst    None  10  6   timeit rep1 lst       timeit rep2 lst       timeit rep3 lst       timeit rep4 lst       timeit rep5 lst       timeit rep6 lst       timeit rep7 lst        I get   26 3 ms    163   s per loop  mean    std  dev  of 7 runs  10 loops each  29 3 ms    206   s per loop  mean    std  dev  of 7 runs  10 loops each  33 8 ms    191   s per loop  mean    std  dev  of 7 runs  10 loops each  11 9 ms    37 8   s per loop  mean    std  dev  of 7 runs  100 loops each  11 9 ms    60 2   s per loop  mean    std  dev  of 7 runs  100 loops each  260 ns    1 84 ns per loop  mean    std  dev  of 7 runs  1000000 loops each  56 5 ms    204   s per loop  mean    std  dev  of 7 runs  10 loops each    Using the internal str index is in fact faster than any manual comparison   I didn t know if the exception in test 4 would be more laborious than using str count  the difference  seems negligible   Note that map    test 6  returns an iterator and not an actual list  thus test 7

User · Answer

gt  gt  gt  a  1 2 3 4 5 1 2 3 4 5 1   gt  gt  gt  item to replace   1  gt  gt  gt  replacement value   6  gt  gt  gt  indices to replace    i for i x in enumerate a  if x  item to replace   gt  gt  gt  indices to replace  0  5  10   gt  gt  gt  for i in indices to replace          a i    replacement value       gt  gt  gt  a  6  2  3  4  5  6  2  3  4  5  6   gt  gt  gt

User · Answer

The answers for this old but relevant question are wildly variable in speed  The fastest of the solution posted by kxr  However  this is even faster and otherwise not here  def f1 arr  find  replace         fast and readable     base 0     for cnt in range arr count find            offset arr index find  base          arr offset  replace         base offset 1  Here is timing for the various solutions  The faster ones are 3X faster than accepted answer and 5X faster than the slowest answer here  To be fair  all methods needed to do inlace replacement of the array sent to the function  Please see timing code below  def f1 arr  find  replace         fast and readable     base 0     for cnt in range arr count find            offset arr index find  base          arr offset  replace         base offset 1          def f2 arr find replace         accepted answer     for i e in enumerate arr           if e  find               arr i  replace          def f3 arr find replace         in place list comprehension     arr     replace if e  find else e for e in arr       def f4 arr find replace         in place map and lambda -- SLOW     arr    list map lambda x  x if x    find else replace  arr        def f5 arr find replace         find index with comprehension     for i in  i for i  e in enumerate arr  if e  find           arr i  replace          def f6 arr find replace         FASTEST but a little les clear     try          while True              arr arr index find   replace     except ValueError          pass      def f7 lst  old  new        quot  quot  quot replace list elements  inplace  quot  quot  quot      i   -1     try          while 1              i   lst index old  i   1              lst i    new     except ValueError          pass           import time       def cmpthese funcs  args     cnt 1000  rate True  micro True        quot  quot  quot Generate a Perl style function benchmark quot  quot  quot                         def pprint table table            quot  quot  quot Perl style table output quot  quot  quot          def format field field  fmt      0f                 if type field  is str  return field             if type field  is tuple  return field 1  format field 0               return fmt format field                def get max col w table  index               return max  len format field row index    for row in table                     col paddings  get max col w table  i  for i in range len table 0             for i row in enumerate table                 left col             row tab  row 0  ljust col paddings 0                  rest of the cols             row tab   format field row j   rjust col paddings j   for j in range 1 len row                print     join row tab                        results        for i in range cnt           for f in funcs              start time perf counter ns               f  args              stop time perf counter ns               results setdefault f   name        append stop-start      results  k float sum v   len v  for k v in results items             fastest sorted results key results get  reverse True      table            if rate  table 0  append  rate sec       if micro  table 0  append   u03bcsec pass       table 0  extend fastest      for e in fastest          tmp  e          if rate              tmp append        format int round float cnt  1000000 0 results e               if micro              tmp append      1f   format results e  float cnt             for x in fastest              if x  e  tmp append  --               else  tmp append     1    format  results x -results e   results e            table append tmp        pprint table table                         if   name       main         import sys     import time      print sys version      cases             small  found   9  100             small  not found   99  100             large  found   9  1000             large  not found   99  1000            for txt  tgt  mul in cases          print f  n txt             arr  1 2 3 4 5 6 7 8 9 0  mul          args  arr tgt  X           cmpthese  f1 f2 f3  f4  f5  f6  f7  args      And the results  3 9 1  default  Feb  3 2021  07 38 02    Clang 12 0 0  clang-1200 0 32 29    small  found     rate sec   sec pass     f4     f3     f5     f2     f6     f7     f1 f4  133 982       7 5     -- -38 8  -49 0  -52 5  -78 5  -78 6  -82 9  f3  219 090       4 6  63 5      -- -16 6  -22 4  -64 8  -65 0  -72 0  f5  262 801       3 8  96 1   20 0      --  -6 9  -57 8  -58 0  -66 4  f2  282 259       3 5 110 7   28 8    7 4      -- -54 6  -54 9  -63 9  f6  622 122       1 6 364 3  184 0  136 7  120 4      --  -0 7  -20 5  f7  626 367       1 6 367 5  185 9  138 3  121 9    0 7      -- -19 9  f1  782 307       1 3 483 9  257 1  197 7  177 2   25 7   24 9      --  small  not found     rate sec   sec pass     f4     f5     f2     f3     f6     f7     f1 f4   13 846      72 2     -- -40 3  -41 4  -47 8  -85 2  -85 4  -86 2  f5   23 186      43 1  67 5      --  -1 9  -12 5  -75 2  -75 5  -76 9  f2   23 646      42 3  70 8    2 0      -- -10 8  -74 8  -75 0  -76 4  f3   26 512      37 7  91 5   14 3   12 1      -- -71 7  -72 0  -73 5  f6   93 656      10 7 576 4  303 9  296 1  253 3      --  -1 0   -6 5  f7   94 594      10 6 583 2  308 0  300 0  256 8    1 0      --  -5 6  f1  100 206      10 0 623 7  332 2  323 8  278 0    7 0    5 9      --  large  found     rate sec   sec pass     f4     f2     f5     f3     f6     f7     f1 f4      145   6 889 4     -- -33 3  -34 8  -48 6  -85 3  -85 4  -85 8  f2      218   4 593 5  50 0      --  -2 2  -22 8  -78 0  -78 1  -78 6  f5      223   4 492 4  53 4    2 3      -- -21 1  -77 5  -77 6  -78 2  f3      282   3 544 0  94 4   29 6   26 8      -- -71 5  -71 6  -72 3  f6      991   1 009 5 582 4  355 0  345 0  251 1      --  -0 4   -2 8  f7      995   1 005 4 585 2  356 9  346 8  252 5    0 4      --  -2 4  f1    1 019     981 3 602 1  368 1  357 8  261 2    2 9    2 5      --  large  not found     rate sec   sec pass     f4     f5     f2     f3     f6     f7     f1 f4      147   6 812 0     -- -35 0  -36 4  -48 9  -85 7  -85 8  -86 1  f5      226   4 424 8  54 0      --  -2 0  -21 3  -78 0  -78 1  -78 6  f2      231   4 334 9  57 1    2 1      -- -19 6  -77 6  -77 7  -78 2  f3      287   3 484 0  95 5   27 0   24 4      -- -72 1  -72 2  -72 8  f6    1 028     972 3 600 6  355 1  345 8  258 3      --  -0 4   -2 7  f7    1 033     968 2 603 6  357 0  347 7  259 8    0 4      --  -2 3  f1    1 057     946 2 619 9  367 6  358 1  268 2    2 8    2 3      --

[python] finding and replacing elements in a list

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