Replace values in list using Python

Question

I have a list where I want to replace values with None where condition   returns True    0  1  2  3  4  5  6  7  8  9  10    For example  if condition checks bool item 2  should return    None  1  None  3  None  5  None  7  None  9  None    What is the most efficient way to do this

User · Answer

ls    x if  condition  else None for x in ls

User · Answer

This might help     test list    5  8  test list 0    None print test list  prints  None  8

User · Answer

Riffing on a side question asked by the OP in a comment  i e       what if I had a generator that yields   the values from range 11  instead of a   list  Would it be possible to replace   values in the generator    Sure  it s trivially easy      def replaceiniter it  predicate  replacement None     for item in it      if predicate item   yield replacement     else  yield item   Just pass any iterable  including the result of calling a generator  as the first arg  the predicate to decide if a value must be replaced as the second arg  and let  er rip   For example    gt  gt  gt  list replaceiniter xrange 11   lambda x  x 2    0  None  2  None  4  None  6  None  8  None  10

User · Answer

gt  gt  gt  L   range  11   gt  gt  gt    x if x 2    1 else None for x in L    None  1  None  3  None  5  None  7  None  9  None

User · Answer

In case you want to replace values in place  you can update your original list with values from a list comprehension by assigning to the whole slice of the original   data     range 11      0  1  2  3  4  5  6  7  8  9  10  id before   id data  data       x if x   2 else None for x in data  data   Out   None  1  None  3  None  5  None  7  None  9  None  id before    id data     check if list is still the same   Out  True   If you have multiple names pointing to the original list  for example you wrote data2 data before changing the list and you skip the slice notation for assigning to data  data will rebind to point to the newly created list while data2 still points to the original unchanged list   data     range 11      0  1  2  3  4  5  6  7  8  9  10  data2   data id before   id data  data    x if x   2 else None for x in data     no     here data   Out   None  1  None  3  None  5  None  7  None  9  None  id before    id data     check if list is still the same   Out  False data2   Out   0  1  2  3  4  5  6  7  8  9  10    Note  This is no recommendation for generally preferring one over the other  changing list in place or not   but behavior you should be aware of

User · Answer

Here s another way    gt  gt  gt  L   range  11   gt  gt  gt  map lambda x  x if x 2 else None  L   None  1  None  3  None  5  None  7  None  9  None

User · Answer

Build a new list with a list comprehension   new items    x if x   2 else None for x in items    You can modify the original list in-place if you want  but it doesn t actually save time   items    0  1  2  3  4  5  6  7  8  9  10  for index  item in enumerate items       if not  item   2           items index    None   Here are  Python 3 6 3  timings demonstrating the non-timesave   In  1     timeit         items    0  1  2  3  4  5  6  7  8  9  10          for index  item in enumerate items               if not  item   2                   items index    None         1 06   s    33 7 ns per loop  mean    std  dev  of 7 runs  1000000 loops each   In  2     timeit         items    0  1  2  3  4  5  6  7  8  9  10          new items    x if x   2 else None for x in items          891 ns    13 6 ns per loop  mean    std  dev  of 7 runs  1000000 loops each    And Python 2 7 6 timings   In  1     timeit         items    0  1  2  3  4  5  6  7  8  9  10          for index  item in enumerate items               if not  item   2                   items index    None          1000000 loops  best of 3  1 27   s per loop In  2     timeit         items    0  1  2  3  4  5  6  7  8  9  10          new items    x if x   2 else None for x in items           1000000 loops  best of 3  1 14   s per loop

[python] Replace values in list using Python

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