如何计算单词在句子中出现的次数并打印出索引？ （Python）

Question

所以我一直在学习Python几个月，并想知道如何去编写一个程序来计算一个单词在一个句子中出现的次数并打印出索引。

谢谢。

Answer 1

问题和我的答案改变了。这里是最后的建议：

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

有几种方法可以做到这一点，但这里是一个计数的单词的实例数一个平凡的解决方案，但不采取例如punctation考虑：

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]) 

的逻辑是（虽然有点bodgy）基本上会在一定范围内的search_term给定string从而索引的出现次数的单词，将索引添加到列表中;那么它将该词替换为无，然后在下一个词中增加根据当前索引删除的字符数量，并且循环工作得很好。

Answer 4

我们也欢迎学习者。以下可能会让你去;其包括基本治疗标点符号，以及返回的情况下，变化的相应索引处：

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')])