How to generate a dictionary with nth number of keys and letters in sequential order?

user_9208714 0
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I need to generate a dictionary in the following format:

x = 3 #number of keys
y = 2 #number of key values (letters in sequential order)

The desired output is:

{1:[a, b],2:[c, d],3:[e,f]}

As a note, I am able to generate a list in range of letters in the following way:

list(map(chr, range(97, 98 + 3)))
['a', 'b', 'c', 'd']

However, I am struggling to figure out the rest.

dictionary python
2 Answers

One way is to use an iterator:

x, y = 3, 2

letters = 'abcdefghijklmnopqrstuvwxyz'
# or whatever method you prefer, like what you proposed:
# letters = map(chr, range(97, 97 + 26))

letters_iter = iter(letters)
output = {k+1: [next(letters_iter) for _ in range(y)] for k in range(x)}
print(output) # {1: ['a', 'b'], 2: ['c', 'd'], 3: ['e', 'f']}

This assumes that x times y does not exceed 26. If you want to make the letters to cycle, then you can use itertools.cycle instead of iter:

from itertools import islice, cycle
from string import ascii_lowercase

x, y = 3, 2

letters_iter = cycle(ascii_lowercase)
output = {k+1: [*islice(letters_iter, y)] for k in range(x)}
print(output) # {1: ['a', 'b'], 2: ['c', 'd'], 3: ['e', 'f']}

(Here, I used islice to shorten the inner list comprehension further. The star * in [*...] is (generalized) unpacking. Also I used string.ascii_lowercase, but any other ways would work.)

user_114508200

There are many ways to do this. So let's have a discrete function like this:

from string import ascii_lowercase as LC

def make_dict(x, y):
    return {k+1:list(LC[k*y:(k+1)*y]) for k in range(x)}

print(make_dict(3, 2))

Output:

{1: ['a', 'b'], 2: ['c', 'd'], 3: ['e', 'f']}
user_175803810
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