Check if an item is in a nested list

Question

in a simple list following check is trivial:

x = [1, 2, 3]

2 in x  -> True

but if it is a list of list, such as:

x = [[         


        

Answer 1

TL;DR

x = [0, [1, 2, 3], [2, 3, [4, 5, [6], []], [7, 8]]]
def find_n(input_list, n):
    for el in input_list:
        if el == n or (isinstance(el, list) and find_n(el, n)):
            return True
    return False
print(find_n(x, 6))

Note that, somewhat interestingly:

def find_n(input_list, n):
    return any([el == n or (isinstance(el, list) and find_n(el, n)) for el in input_list])
return (find_n(x, 6))

Is over 50% slower to execute.

Original Answer(s)

What if you have a depth greater than 2? Here's one approach to the generic case:

x = [0, [1, 2, 3], [2, 3, [4, 5, [6], []], [7, 8]]]

def flatten(input_list):
    flat_list = []
    for sublist_or_el in input_list:
        if isinstance(sublist_or_el, list):
            for sublist_or_el2 in flatten(sublist_or_el):
                flat_list.append(sublist_or_el2)
        else:
            flat_list.append(sublist_or_el)
    return flat_list

print(6 in flatten(x))

Not sure about speed though, but as I said, it's one approach that may be useful to someone!

EDIT - BETTER (FASTER) ANSWER:

This reduces the time taken (if n is found, actually even if it is not found, about half the time actually...) by returning early. This is slightly faster than @Curt F.'s answer, and slower than creating a function that assumes a maximum depth of 2 (the accepted answer).

x = [0, [1, 2, 3], [2, 3, [4, 5, [6], []], [7, 8]]]
def find_n(input_list, n):
    flat_list = []
    for sublist_or_el in input_list:
        if isinstance(sublist_or_el, list):
            if find_n(sublist_or_el, n) == True:
                return True
        elif sublist_or_el == n:
            return True
    return False
print(find_n(x, 6))

---

Quick timing (very hacky, sorry, busy today!):

import time

x = [0, [1, 2, 3], [2, 3, [4, 5, [6], []], [7, 8]]]

def a():
    def flatten(input_list):
        flat_list = []
        for sublist_or_el in input_list:
            if isinstance(sublist_or_el, list):
                for sublist_or_el2 in flatten(sublist_or_el):
                    flat_list.append(sublist_or_el2)
            else:
                flat_list.append(sublist_or_el)
        return flat_list
    return (6 in flatten(x))

def b():
    def find_n(input_list, n):
        flat_list = []
        for sublist_or_el in input_list:
            if isinstance(sublist_or_el, list):
                if find_n(sublist_or_el, n) == True:
                    return True
            elif sublist_or_el == n:
                return True
        return False
    return (find_n(x, 6))


zz = 0
for i in range(100000):
    start_time = time.clock()
    res = a()
    zz += time.clock() - start_time
print(a())
print((zz)/100, "seconds")

zz = 0
for i in range(100000):
    start_time = time.clock()
    res = b()
    zz += time.clock() - start_time
print(b())
print((zz)/100, "seconds")