'''
[[[[[[[[[[[[[[[[[[[[[[[[[[[[1]]]]]]]]]]]]]]]]]]]]]]]]]]]]
find 1
'''
key = 1
x = [[[[[[[[[[[[[[[[[[[[[[[[[[[[key]]]]]]]]]]]]]]]]]]]]]]]]]]]]
#len(x) -> 1
Actually my question is simple, but i couldn't solve this problem... What should i use to solve it, i mean should i use recrusive functions or for loop ???
for i in range(len(x)):
for j in range(len(x)):
for k in range(len(x)):
for _ in range(len(x)):
for ...
You can do that as follow using more_itertools
import more_itertools
key = 1
x = [[[[[[[[[[[[[[[[[[[[[[[[[[[[key]]]]]]]]]]]]]]]]]]]]]]]]]]]]
flattened_list = list(more_itertools.collapse(x))
print(flattened_list) # [1]
Here is a non-recursive approach for your current structure:
def extract_list(l: list):
while True:
l = l.pop()
if not isinstance(l, list):
return l
x = [[[[[[[[[[[[[[[[[[[[[[[[[[[[1]]]]]]]]]]]]]]]]]]]]]]]]]]]]
print(extract_list(x)) # 1
You can also do it with type check
key = 1
x = [[[[[[[[[[[[[[[[[[[[[[[[[[[[key]]]]]]]]]]]]]]]]]]]]]]]]]]]]
def find_key(array):
if type(array) is list:
return find_key(array[0])
return array
print(find_key(x))
You could do this a couple ways but here's a solution using recursion:
x = [[[[[[[[[[[[[[[[[[[[[[[[[[[[1]]]]]]]]]]]]]]]]]]]]]]]]]]]]
def find_num(l):
if isinstance(inner := l[0], list):
return find_num(inner)
return inner
print(find_num(x))
1
