how to get the index of element of the iterated list directly, without using index()? How to count the instances and not values?

Question

In the "for" loop, does the program "know" the position of the iterated element in the list? I can't use index() method for getting the index because I have a bunch of identical ones and zeros, so I need the way of counting the instances of iterated elements, without checking their values. I think that maybe the "for" loop is already counting the instances or checking the next() element of the list, so I need to get to there.
P.S. No, I don't wanna use i=0 ........ i+=1, I know it will work but for some reasons I will not
P.P.S. No, dictionary won't work either

Answer 1

you dont need to repost. can you please provide desired input and output for your code if this doesnt help

https://code.sololearn.com/caKtuDpJGjWf/?ref=app

Answer 2

yeah. in that case you could that example of mine and use if list[i] == 1:
and i instead of index

Answer 3

objects = [o1, o2, o3, o4, o5]
#each object has its method, and I need to run those depending on the list I have here
list = [1,0,1,1,1,0,1,0,1,1,1,0,1]
for var in list:
if var == 1:
object_used = objects[list.index(var)]
object_used.method()
#obviously, this will not work because index() function returns the index of first "1" in the list, and I need to get the real position of var I'm checking

Answer 4

how? tell me how to use that, cos that doesn't sound like it will work
and please don't tell me it's
for I in range(list):
if list[i] == 1:
object = objects[i]

Answer 5

i need to understand how the "for" is coded