Trouver les positions des caractères et le nombre de fois qu'ils sont utilisés dans un texte avec Python

Question

Je veux faire un tableau des positions des caractères dans un texte. Exemple :

Texte = " une pomme et une banane ".

Char:Positions (de 0 à 20) et combien de fois utilisé dans ces positions.

Characters--> A:4101010...0-B:100000...0-C:000000...0-D:001000...0-E:000010...0-...-Z:000000...0

Qu'est-ce qui ne va pas ?

position_list = []

i = 0
for char in range(29):
    position_list.append([])
    for position in range(20):
        position_list[i].append(0)
    i += 1

alphabet = ["a", "b", "c", "ç", "d", "e", "f", "g", "", "h", "", "i", "j", "k", "l", "m", "n", "o", "ö", "p", "r", "s", "", "t", "u", "ü", "v", "y", "z"]
alphabet_index = 0
text = ["sample", "text"]

for word in text:
    x = 0
    for char in alphabet:
        start = 0
        while len(word) > start:
            char_pos = word.find(char, start)
            start += 1
            if char_pos == -1:
                break
            else:
                position_list[x][char_pos] += 1
        x += 1

print(position_list)

Answer 1

Cela résout votre problème :

alphabet = ["a", "b", "c", "ç", "d", "e", "f", "g", "", "h", "", "i", "j", "k", "l",
            "m", "n", "o", "ö", "p", "r", "s", "", "t", "u", "ü", "v", "y", "z"]

position_list = [[0]*20 for i in range(len(alphabet))]

text = "an apple and a banana"

for word in text.split():
    for i, c in enumerate(word):
        position_list[alphabet.index(c)][i] += 1

Answer 2

J'espère avoir bien compris votre problème. Ce script va créer un dictionnaire à partir de votre alphabet où les valeurs sont les positions du caractère dans les mots du texte :

from pprint import pprint
from itertools import zip_longest

Text = "an apple and a banana"
alphabet = ["a", "b", "c", "ç", "d", "e", "f", "g", "", "h", "", "i", "j", "k", "l", "m", "n", "o", "ö", "p", "r", "s", "", "t", "u", "ü", "v", "y", "z"]
word_max_len = max(map(len, Text.split()))
position_dict = {a: [0 for _ in range(word_max_len)] for a in alphabet}

for i, v in enumerate(zip_longest(*Text.split())):
    for a in v:
        if a is None:
            continue
        position_dict[a][i] += 1

pprint(position_dict)

Imprimés :

{'a': [4, 1, 0, 1, 0, 1],
 'b': [1, 0, 0, 0, 0, 0],
 'c': [0, 0, 0, 0, 0, 0],
 'd': [0, 0, 1, 0, 0, 0],
 'e': [0, 0, 0, 0, 1, 0],
 'f': [0, 0, 0, 0, 0, 0],
 'g': [0, 0, 0, 0, 0, 0],
 'h': [0, 0, 0, 0, 0, 0],
 'i': [0, 0, 0, 0, 0, 0],
 'j': [0, 0, 0, 0, 0, 0],
 'k': [0, 0, 0, 0, 0, 0],
 'l': [0, 0, 0, 1, 0, 0],
 'm': [0, 0, 0, 0, 0, 0],
 'n': [0, 2, 1, 0, 1, 0],
 'o': [0, 0, 0, 0, 0, 0],
 'p': [0, 1, 1, 0, 0, 0],
 'r': [0, 0, 0, 0, 0, 0],
 's': [0, 0, 0, 0, 0, 0],
 't': [0, 0, 0, 0, 0, 0],
 'u': [0, 0, 0, 0, 0, 0],
 'v': [0, 0, 0, 0, 0, 0],
 'y': [0, 0, 0, 0, 0, 0],
 'z': [0, 0, 0, 0, 0, 0],
 'ç': [0, 0, 0, 0, 0, 0],
 'ö': [0, 0, 0, 0, 0, 0],
 'ü': [0, 0, 0, 0, 0, 0],
 '': [0, 0, 0, 0, 0, 0],
 '': [0, 0, 0, 0, 0, 0],
 '': [0, 0, 0, 0, 0, 0]}

Answer 3

Je pense que c'est la solution

import string
text = "an apple and a banana"
CharList = string.ascii_lowercase + string.ascii_uppercase
long_word = max(map(len, text.split()))
final_result = [[0] * long_word for i in range(len(CharList))]

for word in text.split():
    for pos, letter in enumerate(word):
        final_result[CharList.index(letter)][pos] += 1

for i in range(len(final_result)):
    letter = CharList[i]
    li = final_result[i]
    print(f'{letter}:{li}')