リスト内の単語に番号を割り当てるにはどうすればよいですか？

Question

これは私の現在の進歩です。私はPythonに若干の新しさを感じています。私は本当に問題を解決する方法を正確には知らないし、タイトルがやや誤解を招く場合はお詫び申し上げます。私はできる限り最善の方法で問題を説明しようとします。

は

---

出力はリスト形式でこれをする必要があります：

"WHAT", "IS", "MINE", "IS", "YOURS", "AND", "WHAT", "IS", "YOURS", "IS", "MINE" 

"1", "2", "3", "2", "4", "5", "1", "2", "4", "2", "3" 

"1"が"WHAT"で、"2"が"IS"ある...というように。単語が複数回表示される場合は、同じ番号のままになります。これは動作するはず

---

コード

sentence = "WHAT IS MINE IS YOURS AND WHAT IS YOURS IS MINE"; 

sentence = sentence.lower(); 

sentence = sentence.split(); 

uniqueWord = []; 

store = []; 

for i in sentence: 
    if i not in uniqueWord: 
     uniqueWord.append(i); 

lengthOfUniqueWord = len(uniqueWord); 

print(sentence); 

print(uniqueWord); 

for i in range(lengthOfUniqueWord): 
    i = str(i+1); 
    store.append(i); 

print(store); 

for positions in enumerate(uniqueWord, 1): 
    print(positions); 

---

出力

['what', 'is', 'mine', 'is', 'yours', 'and', 'what', 'is', 'yours', 'is', 'mine'] 
['what', 'is', 'mine', 'yours', 'and'] 
['1', '2', '3', '4', '5'] 
(1, 'what') 
(2, 'is') 
(3, 'mine') 
(4, 'yours') 
(5, 'and') 

Answer 1

：

sentence = "WHAT IS MINE IS YOURS AND WHAT IS YOURS IS MINE";  
sentence = sentence.lower(); 
sentence = sentence.split(); 

uniqueWord = []; 

for i in sentence: 
    if i not in uniqueWord: 
     uniqueWord.append(i); 

for word in sentence: 
    print uniqueWord.index(word) + 1 

ここでは、へのリンクですindex

Answer 2

のドキュメント他の人も私が思いついた;-)回答型付けながら - それはまた、Pythonのコーディングに関する更なるヒントが含まれている願っています：Pythonの2.7.11で（これは私のシステムで提供します

#! /usr/bin/env python 
from __future__ import print_function 


def word_indexer(text): 
    """Split the text in words maintaining order and return two 
    aligned lists: 1) the words all in sequence, and 2) the matching 
    unique 1-based case insensitive index (insert based rank).""" 

    words_in_order = text.split() 
    word_index = [] 
    unique_word_rank = {} 
    rank = 1 
    for word in words_in_order: 
     normalized_word = word.lower() 
     if normalized_word not in unique_word_rank: 
      unique_word_rank[normalized_word] = rank 
      rank += 1 
     word_index.append(unique_word_rank[normalized_word]) 

    return words_in_order, word_index 


def main(): 
    """Do the word indexing.""" 
    sentence = "WHAT IS MINE IS YOURS AND WHAT IS YOURS IS MINE" 
    words_in_order, word_index = word_indexer(sentence) 
    # print as in question in two lines: 
    print(words_in_order) 
    print(word_index) 
    # to display like a table: 
    for ndx, word in zip(word_index, words_in_order): 
     print(ndx, word) 

if __name__ == '__main__': 
    main() 

が、 Python 3.5.1でも動作します）：

['WHAT', 'IS', 'MINE', 'IS', 'YOURS', 'AND', 'WHAT', 'IS', 'YOURS', 'IS', 'MINE'] 
[1, 2, 3, 2, 4, 5, 1, 2, 4, 2, 3] 
1 WHAT 
2 IS 
3 MINE 
2 IS 
4 YOURS 
5 AND 
1 WHAT 
2 IS 
4 YOURS 
2 IS 
3 MINE 

これが役立ちます - そしてハッピーハッキング！

Answer 3

新しい単語がdict dに挿入された場合にのみインクリメントするcountイテレータを設定します。各単語の一致数から最終的なリストをビルドします。ここでは

from itertools import count 

sentence = "WHAT IS MINE IS YOURS AND WHAT IS YOURS IS MINE" 
s = sentence.split() 
c = count(1) 
d = {} 

for word in s: 
    if word.lower() not in d: 
     d[word.lower()] = next(c) 

result = [str(d[word.lower()]) for word in s] 
# ['1', '2', '3', '2', '4', '5', '1', '2', '4', '2', '3'] 

Answer 4

は、同じ目標を達成するための別の方法である：

sentence = "WHAT IS MINE IS YOURS AND WHAT IS YOURS IS MINE".lower().split() 
uniqueWord = list(set(sentence)) 

print(sentence); 
print(uniqueWord); 

store = [uniqueWord.index(x) for x in sentence] 

print(store);