有沒有更好的方法來計算句子中的標點符號？

Question

我想算「端的句子」如句號，感嘆號和問號。

我已經寫了一個小環要做到這一點，但我不知道是否有更好的方法。不允許使用內置函數。

for line in textContent: 
    numberOfFullStops += line.count(".") 
    numberOfQuestionMarks += line.count("?") 
    numberOfQuestionMarks += line.count("!") 

numberOfSentences = numberOfFullStops + numberOfQuestionMarks + numberOfExclamationMarks 

Answer 1

假設你想在一個句子計算終端標點符號，我們可以通過遍歷每個字符串的字符及過濾標點符號產生的字典（字符，計數）對。

演示

這裏有三個選項呈現自上而下與中程對初學者級別的數據結構：

import collections as ct 


sentence = "Here is a sentence, and it has some exclamations!!" 
terminals = ".?!" 

# Option 1 - Counter and Dictionary Comprehension 
cd = {c:val for c, val in ct.Counter(sentence).items() if c in terminals} 
cd 
# Out: {'!': 2} 


# Option 2 - Default Dictionary 
dd = ct.defaultdict(int) 
for c in sentence: 
    if c in terminals: 
     dd[c] += 1 
dd 
# Out: defaultdict(int, {'!': 2}) 


# Option 3 - Regular Dictionary 
d = {} 
for c in sentence: 
    if c in terminals: 
     if c not in d: 
      d[c] = 0 
     d[c] += 1 
d 
# Out: {'!': 2} 

爲了進一步擴展，爲獨立sentences列表，一環周圍後者的選擇。

for sentence in sentences: 
    # add option here 

注：總結每個句子的總標點符號，總的dict.values()，例如sum(cd.values())。

---

更新：假設你要通過終端punctutation拆分句子，使用正則表達式：

import re 


line = "Here is a string of sentences. How do we split them up? Try regular expressions!!!" 


# Option - Regular Expression and List Comprehension 
pattern = r"[.?!]" 
sentences = [sentence for sentence in re.split(pattern, line) if sentence] 
sentences 
# Out: ['Here is a string of sentences', ' How do we split them up', ' Try regular expressions'] 

len(sentences) 
# Out: 3 

通知line有5個終端，但是隻有3句。因此正則表達式是一種更可靠的方法。

參考

• collections.Counter
• collections.defaultdict
• re.split
• List comprehension