清理嵌套列表

Question

我有一個嵌套列表的一個巨大的混亂，看起來這樣的事情，只是長：

fruit_mess = [['watermelon,0,1.0\n'], ['apple,0,1.0\n'], ['"pineapple",0,1.0\n'], ['"strawberry, banana",0,1.0\n'], ['peach plum pear,0,1.0\n'], ['"orange, grape",0,1.0\n']] 

最後，我想要的東西，看起來像這樣：

neat_fruit = [['watermelon',0,1.0], ['apple',0,1.0], ['pineapple',0,1.0], ['strawberry, banana',0,1.0], ['peach plum pear',0,1.0], ['orange, grape',0,1.0]] 

但我我不知道如何處理報價中的雙引號，以及如何分割數字中的成果，特別是用分隔一些成果的逗號。我嘗試了一堆東西，但一切似乎都讓它變得更加混亂。任何建議將不勝感激。

Answer 1

使用csv模塊（標準庫）來處理雙引號的水果用逗號在他們的名字：

import csv 
import io 

fruit_mess = [['watermelon,0,1.0\n'], ['apple,0,1.0\n'], ['"pineapple",0,1.0\n'], ['"strawberry, banana",0,1.0\n'], ['peach plum pear,0,1.0\n'], ['"orange, grape",0,1.0\n']] 

# flatten the list of lists into a string: 
data='\n'.join(item[0].strip() for item in fruit_mess)  
reader=csv.reader(io.BytesIO(data)) 
neat_fruit=[[fruit,int(num1),float(num2)] for fruit,num1,num2 in reader] 

print(neat_fruit)  
# [['watermelon', 0, 1.0], ['apple', 0, 1.0], ['pineapple', 0, 1.0], ['strawberry, banana', 0, 1.0], ['peach plum pear', 0, 1.0], ['orange, grape', 0, 1.0]] 

Answer 2

基於正則表達式的解決方案：

>>> import re 
>>> regex = re.compile(r'("[^"]*"|[^,]*),(\d+),([\d.]+)') 
>>> neat_fruit = [] 
>>> for item in fruit_mess: 
...  match = regex.match(item[0]) 
...  result = [match.group(1).strip('"'), int(match.group(2)), float(match.group(3))] 
...  neat_fruit.append(result) 
... 
>>> neat_fruit 
[['watermelon', 0, 1.0], ['apple', 0, 1.0], ['pineapple', 0, 1.0], ['strawberry, 
banana', 0, 1.0], ['peach plum pear', 0, 1.0], ['orange, grape', 0, 1.0]] 

Answer 3

一個更簡單解決方案：

fruit_mess = [['watermelon,0,1.0\n'], ['apple,0,1.0\n'], ['"pineapple",0,1.0\n'], ['"strawberry, banana",0,1.0\n'], ['peach plum pear,0,1.0\n'], ['"orange, grape",0,1.0\n']] 
for i,x in enumerate(fruit_mess): 
    data = x[0].rstrip('\n').rsplit(',', 2) 
    fruit_mess[i] = [data[0], int(data[1]), float(data[2])]