remove None value from a list without removing the 0 value

Question

This was my source I started with   My List   L    0  23  234  89  None  0  35  9    When I run this    L   filter None  L    I get this results    23  234  89  35  9    But this is not what I need  what I really need is     0  23  234  89  0  35  9    Because I m calculating percentile of the data and the 0 make a lot of difference   How to remove the None value from a list without removing 0 value

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Using list comprehension this can be done as follows   l    i for i in my list if i is not None    The value of l is    0  23  234  89  0  35  9

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from operator import is not from functools import partial     filter null   partial filter  partial is not  None      A test case L    1  None  2  None  3  L   list filter null L

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L    0  23  234  89  None  0  35  9   result   list filter lambda x  x    None  L

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Say the list is like below  iterator    None  1  2  0      None  False                This will return only those items whose bool item  is True  print filter lambda item  item  iterator     1  2    This is equivalent to   print  item for item in iterator if item    To just filter None   print filter lambda item  item is not None  iterator     1  2  0      False                Equivalent to   print  item for item in iterator if item is not None    To get all the items that evaluate to False  print filter lambda item  not item  iterator    Will print  None      0  None  False

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A list comprehension is likely the cleanest way   gt  gt  gt  L    0  23  234  89  None  0  35  9  gt  gt  gt   x for x in L if x is not None   0  23  234  89  0  35  9   There is also a functional programming approach but it is more involved   gt  gt  gt  from operator import is not  gt  gt  gt  from functools import partial  gt  gt  gt  L    0  23  234  89  None  0  35  9   gt  gt  gt  list filter partial is not  None   L    0  23  234  89  0  35  9

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My List  L  0  23  234  89  None  0  35  9  Code result   k for k in L if  k is None   False  Result   0  23  234  89  0  35  9

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jamylak answer is quite nice  however if you don t want to import a couple of modules just to do this simple task  write your own lambda in-place    gt  gt  gt  L    0  23  234  89  None  0  35  9   gt  gt  gt  filter lambda v  v is not None  L   0  23  234  89  0  35  9

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For Python 2 7  See Raymond s answer  for Python 3 equivalent    Wanting to know whether something  is not None  is so common in python  and other OO languages   that in my Common py  which I import to each module with  from Common import      I include these lines   def exists it       return  it is not None    Then to remove None elements from a list  simply do   filter exists  L    I find this easier to read  than the corresponding list comprehension  which Raymond shows  as his Python 2 version

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gt  gt  gt  L    0  23  234  89  None  0  35  9   gt  gt  gt   x for x in L if x is not None   0  23  234  89  0  35  9    Just for fun  here s how you can adapt filter to do this without using a lambda   I wouldn t recommend this code - it s just for scientific purposes    gt  gt  gt  from operator import is not  gt  gt  gt  from functools import partial  gt  gt  gt  L    0  23  234  89  None  0  35  9   gt  gt  gt  filter partial is not  None   L   0  23  234  89  0  35  9

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Iteration vs Space  usage could be an issue  In different situations profiling may show either to be  quot faster quot  and or  quot less memory quot  intensive    first  gt  gt  gt  L    0  23  234  89  None  0  35  9        gt  gt  gt   x for x in L if x is not None   0  23  234  89  0  35  9          second  gt  gt  gt  L    0  23  234  89  None  0  35  9   gt  gt  gt  for i in range L count None    L remove None   0  23  234  89  0  35  9        The first approach  as also suggested by  jamylak   Raymond Hettinger  and  Dipto  creates a duplicate list in memory  which could be costly of memory for a large list with few None entries  The second approach goes through the list once  and then again each time until a None is reached  This could be less memory intensive  and the list will get smaller as it goes  The decrease in list size could have a speed up for lots of None entries in the front  but the worst case would be if lots of None entries were in the back  The second approach would likely always be slower than the first approach  That does not make it an invalid consideration  Parallelization and in-place techniques are other approaches  but each have their own complications in Python  Knowing the data and the runtime use-cases  as well profiling the program are where to start for intensive operations or large data  Choosing either approach will probably not matter in common situations  It becomes more of a preference of notation  In fact  in those uncommon circumstances  numpy  example if L is numpy array  L   L L    numpy array None   from here   or cython may be worthwhile alternatives instead of attempting to micromanage Python optimizations

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If it is all  a list of lists  you could modify sir  Raymond s answer  L      None    123    None    151    no none val   list filter None   ne     x 0  for x in L       for python 2 however  no none val    x 0  for x in L if x 0  is not None      Both returns  123  151       lt  lt  list indice 0  for variable in List if variable is not None

[python] remove None value from a list without removing the 0 value

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