looping a list and skip from another list

Question

is there a better way to do this, this code is working but i feel like there is a better way to do it

mainlist = ['a','b','c','d','e','f','i','j','k']
skiplist = [4,6]
avilable=[j for j in range(len(mainlist)+len(skiplist)+1) if j not in skiplist]

i=avilable[0]
for letter in mainlist:
    print (letter," is ",i)
    i= avilable[avilable.index(i)+1]

---

result
a  is  0
b  is  1
c  is  2
d  is  3
e  is  5
f  is  7
i  is  8
j  is  9
k  is  10

Answer 1

Since you've already worked out how to build available you could just zip the two:

mainlist = ['a','b','c','d','e','f','i','j','k']
skiplist = [4,6]
available= [j for j in range(len(mainlist)+len(skiplist)) if j not in skiplist]

for i, j in zip(mainlist, available):
    print(f"{i} is {j}")

Another option might be to use a counter to build the values of j as you go:

mainlist = ['a','b','c','d','e','f','i','j','k']
skiplist = [4,6]

j = 0
for i in mainlist:
   while j in skiplist:
       j += 1
   print(f"{i} is {j}")
   j += 1

Yet another option would be to build a generator using something like itertools.count and filter:

from itertools import count

mainlist = ['a','b','c','d','e','f','i','j','k']
skiplist = [4,6]
available = filter(lambda j: j not in skiplist, count())

for i, j in zip(mainlist, available):
   print(f"{i} is {j}")

Answer 2

You can use enumerate to get index and value same time. Single for loop is enough to solve problem.

mainlist = ['a','b','c','d','e','f','i','j','k']
skiplist = [4,6]

skipAmout = 0
for i, letter in enumerate(mainlist):
    if i  in skiplist:
        skipAmout +=1
    print (letter," is ",i+skipAmout)

Answer 3

mainlist = ['a','b','c','d','e','f','i','j','k']
skiplist = [4,6]
i=0

for letter in mainlist:
    if i in skiplist:
        i+=1
    print (letter," is ",i)
    i+=1

Getting rid of the unnecessary list makes it much simpler