looping a list and skip from another list-CodePudding

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

CodePudding user response：

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}")

CodePudding user response：

Without that extra list:

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

i = 0
for letter in mainlist:
    while i in skiplist:
        i  = 1
    print(letter, ' is ', i)
    i  = 1

Or a fancy itertools solution, also using a set instead of the list, which would be more efficient if the skiplist were large:

from itertools import count, filterfalse

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

numbers = filterfalse(set(skiplist).__contains__, count())

for letter, number in zip(mainlist, numbers):
    print(letter, ' is ', number)

CodePudding user response：

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)

CodePudding user response：

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