Python 3.9.1 iterate lst to pass to tst2

Question

I want to read all the elements from a list, lst, check the value, change for a number and save in another list, lst2.

lst = ['a','b','c'] 
lst2 = []

for x in range(len(lst)): 
    if lst[x] == 'a': 
        lst2[x] == 1
    if lst[x] == 'b':
        lst2[x] == 2
    if lst[x] == 'c':
        lst2[x] == 3

print(lst2)

Error:

IndexError: list index out of range.

I already tried with while loop and I get the same message. Can I achieve this in other way?

use lst2.append(1) instead of `==`. The `==` operator is to check for True and False — Joe Ferndz, Jan 31 '21 at 19:38
`lst2.append( 1)`, and so on. You can't assign anything to `lst[x]` if there is no item at index `x` in it - note that this would potentially allow you to create lists with 'holes' if it were possible, not a good idea... — Thierry Lathuille, Jan 31 '21 at 19:40
If your intent is to have `lst2` be the same size that `lst1` was, then you can just initialize it to the proper size: `lst2 = [None] * len(lst1)` — Mark Ransom, Jan 31 '21 at 19:42

Red · Accepted Answer · 2021-01-31T20:13:18.443

You need to fill lst2 with some values in order for the subscription assignments to work:

lst = ['a','b','c'] 
lst2 = [None] * len(lst)

for x in range(len(lst)): 
    if lst[x] == 'a': 
        lst2[x] = 1
    if lst[x] == 'b':
        lst2[x] = 2
    if lst[x] == 'c':
        lst2[x] = 3

print(lst2)

Where [None] * len(lst) returns [None, None, None].

Of course, there's the list.append() method, which would be really handy if you can be sure that the values use will be added in consecutive indices.

Python 3.9.1 iterate lst to pass to tst2

1 Answers1