如何計算單詞在句子中出現的次數並打印出索引？ （Python）

Question

所以我一直在學習Python幾個月，並想知道如何去編寫一個程序來計算一個單詞在一個句子中出現的次數並打印出索引。

謝謝。

Answer 1

問題和我的答案改變了。這裏是最後的建議：

string = "wordetcetcetcetcetcetcetcword" 

import re 
find = "word" 
p = re.compile(find) 
matches = [m.start() for m in p.finditer(string)] 
print(matches) 

返回：

[0, 25] 

Answer 2

有幾種方法可以做到這一點，但這裏是一個計數的單詞的實例數一個平凡的解決方案，但不採取例如punctation考慮：

from collections import Counter 
s = "This is true. This is false." 
c = Counter(s.split(' ')) 
print(c['This']) # Prints "2" 

Answer 3

def count_index(string, search_term): 
    return (search_term, 
     string.count(search_term), 
     [string.replace(search_term, '', i).index(search_term) + (len(search_term)*i) for i in range(string.count(search_term))] 
    ) 

返回

>>> a = test.count_index("python is a very good language, i like python because python is good", "python") 
>>> a 
('python', 3, [0, 39, 54]) 

的邏輯是（雖然有點bodgy）基本上會在一定範圍內的search_term給定string從而索引的出現次數的單詞，將索引添加到列表中;那麼它將該詞替換爲無，然後在下一個詞中增加根據當前索引刪除的字符數量，並且循環工作得很好。

Answer 4

我們也歡迎學習者。以下可能會讓你去;其包括基本治療標點符號，以及返回的情況下，變化的相應索引處：

import string 
# 
mask = str.maketrans('', '', string.punctuation) # Punctuation mask. 
# 
def aFunc(sentence, word): 
    words = sentence.translate(mask).split(' ') # Remove punctuation. 
    indices = [(e,w) for (e,w) in enumerate(words) if w.lower() == word.lower()] # Collect (index,word) pairs. 
    return (len(indices), indices) 

---

s = 'The cat fell out of the hat. Then thE cAt fell asleep against the haT=:)' 

aFunc(s, 'HAT') 
(2, [(6, 'hat'), (14, 'haT')]) 

aFunc(s, 'the') 
(4, [(0, 'The'), (5, 'the'), (8, 'thE'), (13, 'the')]) 

aFunc(s, 'Cat') 
(2, [(1, 'cat'), (9, 'cAt')])