itertools.ifilter Vs. filter Vs. list comprehensions
Solution 1
The example below includes a number generator that prints a message immediately before yielding the number, shows up how filter()
first builds the list, then runs through that and filters it. Whereas itertools.ifilter
filters as it goes, never building a list. If you're filtering 500,000 significant things, you want ifilter
, so you're not building a list.
import itertools
def number_generator():
for i in range(0, 3):
print "yield", i
yield i
print "stopping"
function = lambda x: x > 0
numbers = number_generator()
print "itertools.ifilter:"
for n in itertools.ifilter(function, numbers):
print n
print "\nfilter:"
numbers = number_generator()
for n in filter(function, numbers):
print n
Output:
itertools.ifilter: yield 0 yield 1 1 yield 2 2 stopping filter: yield 0 yield 1 yield 2 stopping 1 2
Solution 2
Your understanding is corret: the only difference is that ifilter
returns an iterator, while using filter
is like calling:
list(ifilter(...))
You may also be interested in what PEP 289 says about filter and ifilter:
List comprehensions greatly reduced the need for
filter()
andmap()
. Likewise, generator expressions are expected to minimize the need foritertools.ifilter()
anditertools.imap()
. [...]
Also note that ifilter
became filter
in Python-3 (hence removed from itertools).
Solution 3
ifilter
returns a generator, not a list.
Generators create their items on the fly when needed, instead of allocating the entire list first. That's the only difference between ifilter
and filter
Solution 4
Here, you can see the diference:
filter(function, iterable): Construct a list from those elements of iterable for which function returns true.
itertools.ifilter(predicate, iterable): Make an iterator that filters elements from iterable returning only those for which the predicate is True.
This means that to obtain 'ifiltered' items you should iterate with returned iterator, but 'filter' returns all elements in a list with out iteration needed.
inspectorG4dget
I recently graduated with a PhD in Computer Science from the University of Ottawa. Most of my Saturday nights are spent writing hobby code (right now, my fascination is with Evolutionary Algorithms) - which speaks to the lack of my social life. I love to learn new things. Paradoxically, don't enjoy reading more than 10 pages per new topic (much to my disappointment). I have therefore developed an ability to speed read - something that the readability and speed read it bookmarklets have helped with. Author of Pyvolytion in the python package index (which I talked about at PyConCA, 2012 and published it in the IEEE) and Pystitia, co-author of cfg in the python package index. I can be found on Twitter, Careers 2.0, and LinkedIn
Updated on August 09, 2020Comments
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inspectorG4dget over 3 years
I am trying to become more familiar with the
itertools
module and have found a function calledifilter
.From what I understand, it filters and iterable based on the given function and returns an iterator over a list containing the elements of the iterable on which the function evaluates to
True
.Question 1: is my understanding thus far correct?
Question 2: aside from the fact that this returns and iterator, how is it different from the built-in
filter
function?Question 3 Which is faster?
From what I can tell, it is not. Am I missing something? (I ran the following test)
>>> itertools.ifilter(lambda x: x%2, range(5)) <itertools.ifilter object at 0x7fb1a101b210> >>> for i in itertools.ifilter(lambda x: x%2, range(5)): print i ... 1 3 >>> filter(lambda x: x%2, range(5)) [1, 3] >>> function = lambda x: x%2 >>> [item for item in range(5) if function(item)] [1,3]
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Rik Poggi over 12 yearsA generator is an iterator, and to be strict it returns a itertools.ifilter object.