Is there a more efficient way to do this?

Question

I want to make a dict where every key is a letter and the value is the letter with the same place in the reverse alphabet. Is there a way to do it without the for loop?

alph = "abcdefghijklmnopqrstuvwxyz"
alph_r = alph[::-1]
alph_dict = {}

for x in range(0, len(alph)):
alph_dict[alph[x]] = alph_r[x]

Answer 1

very similar to Louis solution:

from string import ascii_lowercase as chars
res = dict(zip(chars, chars[::-1]))

Answer 2

alph = "abcdefghijklmnopqrstuvwxyz"
alph_dict = {i:j for i,j in zip(alph,alph[::-1])}

#for loop is still there, just more compact

Answer 3

Lothar very nice!
Yours is better.
And so we keep learning from one another.

Answer 4

https://code.sololearn.com/cG79e5NKK6OE/?ref=app

Answer 5

Louis I see, thanks!

Answer 6

Steven, what Erick is asking for, is to create a dict for handling the cryption of characters. Therefore he made his try that gives a dict like:

{'a': 'z', 'b': 'y', 'c': 'x', ...', 'y': 'b', 'z': 'a'}

What your code is doing is a bit different. It just creates a dict with characters of alphabeth, and a consecutive number as value.

Answer 7

Lothar Cool!

Answer 8

Since my variant already written by Lothar
The alternative code is follow(acording to c++ inverse string element detection)

k=len(alph)
alph={alph[i]:alph[k-1-i] for i in range(k)}

also which variant is faster?! someone know what faster use comprehension use for loop or slice of list. sinse coping in memory took time in case of huge data

Answer 9

Sherif please share link!
i didn't know.

Answer 10

Probably no