Reputation:
I have a list of numbers, for example:
list= [1, 2, 4, 1, 2, 4, 1, 2, 4]
You can see the obvious repeating pattern of 1,2,4
I'm trying to find a way to go through that list to determine the cycle and it's length. The numbers vary every time in my program. So for example, the list that I give, I want it to output 3, which is the length of the pattern.
I tried comparing the items together, but ultimately couldn't figure out a way to do it efficiently without going out of the list index every time, because I was just looking for when a number repeated in the sequence. The problem with that was that the pattern could be something like 2 2 1, in which case i would have outputted 2 as the pattern length, when it was actually 3.
Sorry for the lack of information, I'm lost as to how to do this so I don't have much sample code to show.
Upvotes: 1
Views: 1430
Reputation: 879
Returns the length of the pattern it has found or None if it can't find a pattern that repeats at least once.
def GetPattern(lst:list) -> list:
lstLen = len(lst)
i = 1
while i < (lstLen - (lstLen % 2))//2:
tar = lst[:i]
d = lstLen % i
if any([not tar == lst[i*a:i*(a+1)] for a in range(1, (lstLen - d)//i)]):
i += 1
continue
elif not d == 0 and not tar[:d] == lst[-1*d:]:
i += 1
continue
else:
return len(tar)
return None
Upvotes: 0
Reputation: 212895
How about this?
def find_sub(lst):
return next(sub for sub in range(len(lst), 0, -1)
if lst == mylist[:sub] * (len(lst) / sub))
find_sub([1, 2, 4, 1, 2, 4, 1, 2, 4]) # returns 3
find_sub([1, 2, 1, 2, 1, 2, 1, 2]) # returns 2
find_sub([1, 1, 1]) # returns 1
find_sub([1, 2, 3]) # returns 3
find_sub([1, 2, 3, 1, 2, 3, 1]) # returns 7
Upvotes: 0
Reputation: 61225
You can use a generator function to split your list into chunks.
>>> def gen(lst, pat):
... size = len(pat)
... size_l = len(lst)
... for i in range(0, size_l, size):
... yield lst[i:i+size]
...
>>> lst = [1, 2, 4, 1, 2, 4, 1, 2, 4]
>>> pat = [1, 2, 4]
>>> len(list(gen(lst, pat)))
3
Also don't use "list" as variable's name it will shadow the built-in list
class
Upvotes: 1