How do I find an object in a sequence satisfying a particular criterion?
List comprehension and filter go through the entire list. Is the only alternative a handmade loop?
mylist = [10, 2, 20, 5, 50]
find(mylist, lambda x:x>10) # Returns 20
4 Answers
Here's the pattern I use:
mylist = [10, 2, 20, 5, 50]
found = next(i for i in mylist if predicate(i))
Or, in Python 2.4/2.5, next() is a not a builtin:
found = (i for i in mylist if predicate(i)).next()
Do note that next() raises StopIteration if no element was found. In most cases, that's probably good. You asked for the first element, no such element exists, and so the program probably cannot continue.
If, on the other hand, you do know what to do in that case, you can supply a default to next():
conf_files = ['~/.foorc', '/etc/foorc']
conf_file = next((f for f in conf_files if os.path.exists(f)),
'/usr/lib/share/foo.defaults')
answered May 18, 2011 at 3:23
SingleNegationEliminationSingleNegationElimination
155k34 gold badges268 silver badges306 bronze badges
-
Have you come out with it yourself or have you seen it somewhere and if so where? Also, wouldn't it be better to use built-in
nextand pass the default value (empty list in this case probably)? Commented Oct 2, 2013 at 9:56 -
1
next()was introduced in Python 2.6 already - docs.python.org/2.6/library/functions.html?highlight=next#next Commented Oct 2, 2013 at 19:55 -
@Piotr: I think that may depend on the particular use case. If you want the first element of an empty sequence, there is no correct value to return, it has no first element. It might make perfectly good sense to raise an exception.
bool(foo(next(filter(foo, []), [])) != True, that is, unlessfoois something likelambda x: x == [], and[]is certainly not in[], so it's a lie in any case. Commented Oct 2, 2013 at 22:31 -
Your code throws an exception
StopIterationif generator describes empty sequence. Commented Dec 8, 2014 at 16:04
Actually, in Python 3, at least, filter doesn't go through the entire list.
To double check:
def test_it(x):
print(x)
return x>10
var = next(filter(test_it, range(20)))
In Python 3.2, that prints out 0-11, and assigns var to 11.
In 2.x versions of Python you may need to use itertools.ifilter.
-
Good call;
zip,map, andfilterall become lazy in Python3. (Replacements for Python2'simap,izip, andifilter.) Commented May 18, 2011 at 3:26 -
I prefer this to using
foras it is much clearer. Note 1: this will also throwStopIterationif no match. Note 2: if your list is sorted, consider usingbisectif appropriate. Commented Apr 2, 2018 at 9:52
If you only want the first greater than 10 you can use itertools.ifilter:
import itertools
first_gt10 = itertools.ifilter(lambda x: x>10, [10, 2, 20, 5, 50]).next()
If you want all greater than 10, it may be simplest to use a list-comprehension:
all_gt10 = [i for i in mylist if i > 10]
answered May 18, 2011 at 3:13
Too lazy to write:
mylist = [10, 2, 20, 5, 50]
max(mylist, key=lambda x: x>10)
-
6
[i for i in mylist if i > 10]