Answer by MarianD for Simplifying a Scrabble Word Finder - Python

Instead of

scores = {"a": 1, "c": 3, "b": 3, "e": 1, "d": 2, "g": 2,
         "f": 4, "i": 1, "h": 4, "k": 5, "j": 8, "m": 3,
         "l": 1, "o": 1, "n": 1, "q": 10, "p": 3, "s": 1,
         "r": 1, "u": 1, "t": 1, "w": 4, "v": 4, "y": 4,
         "x": 8, "z": 10}

you may use more readable

letters = 'a c b e d g f i h k j m l o n q  p s r u t w v y x z '.split()
numbers = '1 3 3 1 2 2 4 1 4 5 8 3 1 1 1 10 3 1 1 1 1 4 4 4 8 10'.split()

scores = dict(zip(letters, map(int, numbers)))

The split() function convert your strings into lists and the last command creates the dictionary scores from the pairs of the corresponding letter/number.

(The map(int, numbers) applies the int() function to every member of the number list, as this list seems like ['1', '3', '3', ...].)