Dynamic substrings on List. 10 elements before variable

Question

I have problem with dynamic substrings. I have list which can have 1000 elements, 100 elements or even 20. I want to make copy of that list, which will have elements from -10 to variable. For example(pseudo-code):

L = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20]
variable = 12
print L[substring:variable]
>>> L = [2,3,4,5,6,7,8,9,10,12]

I can't figure out how make it correct. The point is that variable is always changing by one.

Here is my piece of code:

def Existing(self, Pages):
     if(self.iter <= 10):
         list = self.other_list[:self.iter]
     else:
         list = self.other_list[self.iter-10:self.iter]
     result = 0
     page = Pages[0]

     list.reverse()

     for blocks in Pages:
         if(list.index(blocks) > result):
             result = list.index(blocks)
             page = blocks

     return page

That method is looking for the element which has the farest index. This part can be unclear. So assume that we have

list = [1,2,3,4,1,5,2,1,2,3,4]

Method should return 5, because it is the farest element. List has duplicates and .index() is returning index of the first element so i reverse list. With that code sometimes program returns that some element do not exist in List. The problem (after deep review with debbuger) is with substrings in self.other_list.

Could you help me with that problem? How to make it correct? Thanks for any advice.

EDIT: Because my problem is not clear enough (I was sure that it can be), so here are more examples.

Okay, so list Pages are list which cointains currently pages which are used. Second list "list" are list of all pages which HAS BEEN used. Method is looking for pages which are already used and choose that one which has been not used for the longest time. With word "use" I mean the index of element. What means the farest element? That one which the smallest index (remember about duplicates, the last duplicates means the real index).

So we have:

Pages = [1,3,5,9]

and

list = [1,2,5,3,6,3,5,1,2,9,3,2]

Method should return 5.

To sum up: I'm looking for substring which give result:

With list =[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20] 
For variable 12: [2,3,4,5,6,7,8,9,10,12] 
for 13: [3,4,5,6,7,8,9,10,11,13] 

ect :-) I know that problem can be complicated. So i would aks you to focus only about substrings. :-) Thanks you very much!

Answer 1

If I understood your problem correctly you want to find the index of items from pages that is at minimum position in lst(taking duplicates in consideration).

So, for this you need to first reverse the list and then first the index of each item in pages in lst, if item is not found then return negative Infinity. Out of those indices you can find the max item and you'll get your answer.

from functools import partial

pages = [1,3,5,9]
lst = [1,2,5,3,6,3,5,1,2,9,3,2]

def get_index(seq, i):
    try:
        return seq.index(i)
    except ValueError:
        return float('-inf')


lst.reverse()
print max(pages, key=partial(get_index, lst))
#5

---

Note that the above method will take quadratic time, so it won't perform well for huge lists. If you're not concerned with some additional memory but linear time then you can use set and dict for this:

pages_set = set(pages)
d = {}
for i, k in enumerate(reversed(lst), 1):
    if k not in d and k in pages_set:
        d[k] = len(lst) - i

print min(d, key=d.get)
#5