Python start line with a letter, combining two for loops?

Question

I'm trying to have my program output:

a. am {1: '2'}
b. at {1: '2'}
c. far {1: '1'}
...
ff. glad {1: '2'}
gg. i {1: '2'}
hh. it {1: '1'}
...
ooo. lets {1: '0'}
ppp. outcome {1: '2'}

The first part, the letters, is the line number but represented with a letter, looping back to the start of the alphabet when it reaches "z."

The second part, the words, is the alphabetized word that it scans from a string.

The third part, is {the number of times a word appears:sentence number(s) it appears in}.

For some reason, everything works except I cannot get the enumeration loop to work correctly on the lettering. It currently outputs this:

a. am {1: '2'}
a. at {1: '2'}
a. far {1: '1'}
a. glad {1: '2'}

My code is:

import string
from itertools import cycle
x=range(1)

...

letters = list(string.ascii_lowercase)
m = len(letters)
for key in sorted(word_counter):
    order = {len(word_counter[key]): str(sent_num[key]).replace('[','').replace(']','')}
        #creates a dictionary with the number of times word appears on the left
        #and the sentence it appears in on the right
    for i, (let, e) in enumerate(zip(cycle(letters), x)):
        testing = (let*(i//m+1))
        print("{}. {} {}".format(testing, key, order)) 

If I change the 'x=range(1)' to anything else it just repeats the previous line that many times then moves onto the next word. Can anyone help me?

Answer 1

You don't need the inner loop, zip your cycle object with the main list and let the counting go on from there:

...
for i, (let, key) in enumerate(zip(cycle(letters), sorted(word_counter))):
    order = {len(word_counter[key]): str(sent_num[key]).replace('[','').replace(']','')}
    testing = let*(i//m+1)
    print("{}. {} {}".format(testing, key, order))