How do I check if an element is in a nested list?
I am trying to define a function nested(x, ys) that tests if a value x appears inside of a nested list of integers ys. The result has to have the value True of False.
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3Can we see your effort?squiguy– squiguy2013-11-30 00:29:17 +00:00Commented Nov 30, 2013 at 0:29
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2Is ys a known depth (ie a 2d array) or variable depth (ie a tree)?Hugh Bothwell– Hugh Bothwell2013-11-30 00:29:37 +00:00Commented Nov 30, 2013 at 0:29
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def nested(x,ys): if type(ys[0]) == type([]): return (nested(x,ys[0]) or nested(x,ys[1:])) else: if x == ys[0]: return True else: return FalseFatmir Alijaj– Fatmir Alijaj2013-11-30 00:53:43 +00:00Commented Nov 30, 2013 at 0:53
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This is my effort until now...Fatmir Alijaj– Fatmir Alijaj2013-11-30 00:54:25 +00:00Commented Nov 30, 2013 at 0:54
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for instance, ys = [[2,1,],3]Fatmir Alijaj– Fatmir Alijaj2013-11-30 00:55:45 +00:00Commented Nov 30, 2013 at 0:55
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2 Answers
Loop over the nested lists and test those; the any() function makes this efficient:
def nested(x, ys):
return any(x in nested for nested in ys)
This assumes ys is nested to one level only.
If recursion is required, you could use:
def flatten(lst):
for elem in lst:
if isinstance(elem, (list, tuple)):
for nested in flatten(elem):
yield nested
else:
yield elem
def nested(x, ys):
return any(x == nested for nested in flatten(ys))
I used a simplified test for list and tuple only to avoid 'flattening' strings.
answered Nov 30, 2013 at 0:30
Martijn Pieters
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3 Comments
BlackVegetable
I take it there is no recursive nested check in the standard library then?
Martijn Pieters
There is not; arbitrarily-nested lists are not that uniform to fit a standard function, I'd say.
Fatmir Alijaj
Dank je wel, Martij Pieters!
This will work for any level of depth:
import collections
def nested(x, ys):
if x in ys: return True
# only keep iterables
nests = [y for y in ys if isinstance(y, collections.Iterable)]
# call nested() on each iterable
return any(nested(x, nest) for nest in nests)
# returns True, will go into multiple nests to find 5
print nested(5, [[1,2],[3,4],[1,2,[3,5]]])
3 Comments
Fatmir Alijaj
Thanks bcorso. But this is too advanced for me. It works, of course. Thanks.
Fatmir Alijaj
How can I use the function nested as a recursion in order to solve the problem?
bcorso
This does use recursion. any(nested(x, nest) for nest in nests) recursively calls nested on any iterable element.