How to get a letter range in reversed order in Python?

Question

I do realize this has already been addressed here (e.g., Is it possible to make a letter range in python?). Nevertheless, I hope this question was different.

The below code gives the range between any two alphabets

[chr(i) for i in range(ord('H'),ord('P'))]

Output

['H', 'I', 'J', 'K', 'L', 'M', 'N', 'O']

For the range of alphabets in reversed order, using reversed() returns the empty list

[chr(i) for i in range(ord('P'),ord('H'))]

# or 

[chr(i) for i in reversed(range(ord('P'),ord('H')))]

Output

[]

Expected output

['O', 'N', 'M', 'L', 'K', 'J', 'I', 'H']

How to get a letter range in reversed order in Python?

Your second example (`[chr(i) for i in reversed(range(ord('P'),ord('H')))]`) will fail for the same reason your first one did - you're still telling it to count from P to H. I think you intended to count from H to P and then reverse it. — Kemp, Apr 26 '21 at 09:17
Correct, and you can do that by counting from H to P and then reversing it.. — Kemp, Apr 26 '21 at 15:38

score 2 · Accepted Answer · answered Apr 26 '21 at 09:18

ord('P') is greater than ord('H'), so range(...) makes an empty range.

To get a reverse letter list, do either of the following:

Make a forward list and apply reverse():
```
reversed( range(ord('H'), ord('P')) )
```
Use the 3rd parameter to range():
```
range(ord('P'), ord('H'), -1)
```

score 0 · Answer 2 · answered Apr 26 '21 at 09:22

ord('P') -> 80
ord('H') -> 72

so [chr(i) for i in range(ord('P'),ord('H'))] is simply [chr(i) for i in range(80,72)] and in range defualt increment value is 1.

since start value > end value and default step increment is 1, so this iterator is running resulting in an empty list.

you need to change the increment value from default to -1

[chr(i) for i in range(ord('P')-1,ord('H')-1,-1)]

output

['O', 'N', 'M', 'L', 'K', 'J', 'I', 'H']

2 Answers2