How to write a simple python program that prints letters in ascending order?


For example I would like to have:












and so on.

Currently I’m here but I am lost. So feel free to propose a completely different solution.

count = 0
string = ''
for i in range(100):
    count += 1
    if i % 26 == 0:
        count = 0
        string += 'a'
    ch = 'a'
    x = chr(ord(ch) + count)
    string = string[:-1] + x
    print(i + 1, string)

and my output is something like this:

1 a

2 b


26 z

27 za

28 zb


52 zz

53 zza

54 zzb


Asked By: user19025048



Maybe try something like the following:

range(97,123) simply creates a range of numbers from 97 to 122, which converted to ASCII equates to a…z (done using chr())

So all our FUnction does, is it recieves a base string (starts with empty), prints out the base + range of charachters and calls its self with base + every charachter as the new base and depth decremented by 1

def printcharachters(base, depth):
    if depth > 0:
        for a in range(97, 123):
            print(base + chr(a))

        for a in range(97, 123):
            printcharachters(base + chr(a), depth - 1)

printcharachters("", 2)

Replace the depth with your desired depth (for 2 the last string would be zz for 4 it would be zzzz).

Answered By: Noah

Using more of standard library:

import itertools
import string

for i in range(1, 3): # Print items of length 1 and 2
    for prod in itertools.product(string.ascii_lowercase, repeat=i):

What you describe is a sorted output of n-th powers of set {'a'...'z'} in terms of cartesian products in string format (cartesian power n of a set X is, from a simple point of view, a set of all possible tuples (x_1, ..., x_n) where all x_i are contained in X). itertools.product implements exactly cartesian product and outputs in order of presence, so just loop over them. string.ascii_lowercase is a simple string containing all letters a..z in natural order. This is fast (uses C implementation of itertools).

Answered By: SUTerliakov
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