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?
>>> L = [0, 23, 234, 89, None, 0, 35, 9]
>>> [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)
>>> from operator import is_not
>>> from functools import partial
>>> L = [0, 23, 234, 89, None, 0, 35, 9]
>>> filter(partial(is_not, None), L)
[0, 23, 234, 89, 0, 35, 9]
A list comprehension is likely the cleanest way:
>>> L = [0, 23, 234, 89, None, 0, 35, 9
>>> [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:
>>> from operator import is_not
>>> from functools import partial
>>> L = [0, 23, 234, 89, None, 0, 35, 9]
>>> list(filter(partial(is_not, None), L))
[0, 23, 234, 89, 0, 35, 9]
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).
Iteration vs Space, usage could be an issue. In different situations profiling may show either to be "faster" and/or "less memory" intensive.
# first
>>> L = [0, 23, 234, 89, None, 0, 35, 9, ...]
>>> [x for x in L if x is not None]
[0, 23, 234, 89, 0, 35, 9, ...]
# second
>>> L = [0, 23, 234, 89, None, 0, 35, 9]
>>> 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.
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, {}, (), []]
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]"""
<< list_indice[0] for variable in List if variable is not None >>
