How to convert number to its corresponding alphabet

Question

For example, 1 should give me alphabet a ,

2 > b
3 > c
... 
26 > a
And 27 should give an again
28 > b
.. Till 9999

The below code only works with 26 numbers

import string
di=dict(zip(string.letters,[ord(c)%32 for c in string.letters]))
di['c']

Output is 3 .

Can anyone please help?

Answer 1

this would do

def num2char(num):
    return chr(((num-1) % 26) + ord('a'))

Answer 2

you don't need to go through a dictionary to convert numbers to letters that way. You can apply the modulo to positions in a string of letters

letters = "abcdefghijklmnopqrstuvwxyz"
for i in range(1,1000):
  print(i,">",letters[(i-1)%26])

output:

1 > a
2 > b
3 > c
4 > d
5 > e
6 > f
7 > g
8 > h
9 > i
10 > j
11 > k
12 > l
13 > m
14 > n
15 > o
16 > p
17 > q
18 > r
19 > s
20 > t
21 > u
22 > v
23 > w
24 > x
25 > y
26 > z
27 > a
28 > b
29 > c
...

Answer 3

See if either one of these is what you're looking for:

print (ord ('a') - 96)

print (chr (1 + 96))

Answer 4

Your data structure is odd. If your input alphabet contains 9999 symbols, you need a dictionary of that many elements. But actually you don't really need a dictionary at all, just the observation that the input number modulo 26 plus 1 maps to the character codes starting at a .

def strmap(numbers):
    base = ord('a')
    mapped = [chr(base + (n-1)%26) for n in numbers]
    return ''.join(mapped)

Demo: https://ideone.com/FeVkO4