返回一个列表

Question

我有一个关于返回以下功能 我正在给一个食物列表中的文件，看起来像这样的问题：

''' 
bread 
bun 

milk 
soya milk 
''' 

，我一定要退回的食品名单列表例如[['bread','bun'], ['milk','soya milk']]

我对Python和编程非常新，因此我被困在for循环中以创建我的列表。任何输入将不胜感激 - kev

Answer 1

它是非常接近。而不是使用while len(next_food) > 0:，您应该使用if和next_food为空且非空白时处理这两种情况。就像你评论表明的那样，在返回之前你应该包含最后一个子列表。

另一个需要检查的问题是next_food是否在最后包含换行符。如果它在那里，你应该去掉换行符。最后有一个替代if len(next_food):检查的快捷方式。只要写if next_food:就可以了。

Answer 2

的作品......

grocery_list_file = open('foods.txt','r').read() 
foods = grocery_list_file.split("\n\n") #split on blank lines 

result = [] 
for food in foods: 
    newFood = food.split("\n") # split the lines, creating the output... 
    result += [newFood] 
return result 

在单行：

print [f.strip().split("\n") for f in open('foods.txt','r').read().split("\n\n")] 

Answer 3

你想，当你达到新类别追加的子表，然后开始一个新的SUB_LIST。当您到达文件末尾时，将剩余的子列表追加到最后非常重要。

new_list.append("\n") #to make sure it appends the last category 
for next_food in new_list: 
     if next_food = "\n": 
      result.append(sub_list) 
      sub_list = [] 
     else: 
      sub_list.append(next_food) 

Answer 4

这不是一个很好的解决方案......但它的一些有趣的小窍门..

>>> s = ''' 
... bread 
... bun 
... 
... milk 
... soya milk 
... ''' 
>>> import re 
>>> parts = re.sub("[\[\]']","",str(s.strip().splitlines())).split(", ,") 
>>> import string 
>>> print [map(string.strip,p.split(",")) for p in parts] 
[['bread', 'bun'], ['milk', 'soya milk']] 

Answer 5

使用itertools.groupby：

from itertools import groupby 


def build_grocery_list(): 
    # using "with" to open the file - recommended way 
    with open("foods.txt") as f: 
     # lines will contain all the lines in the file, without "\n" characters 
     lines = f.read().splitlines() 

     # initialize result as an empty list 
     result = [] 

     # Now for the fun part: group file lines basing on whether they are empty 
     # (bool(string) is analogous to as using "if string:" -- will be True if 
     # the string is not empty) 
     # 
     # groupby works in such a way that it appends stuff to the group as long 
     # as "key" condition is the same, returning (key, group) pairs. 
     # 
     # So, we get pairs: (bool(string), string-group) where: 
     # - bool(string) is the group "key", delimiting empty and non-empty 
     # strings 
     # - string-group is a lazy *generator*, hence the "list(group)" 
     for nonblank, group in groupby(lines, bool): 
      if nonblank: 
       result.append(list(group)) 

    return result 

如果你正在学习Python的，我真的建议您熟悉优秀的itertools模块 - 非常方便！

Answer 6

最简单且相当可读的方式是：

>>> [el.strip().split('\n') for el in text.split('\n\n')] 
[['bread', 'bun'], ['milk', 'soya milk']] 

1. 的开裂\n\n导致了线，紧跟着一个空行

2. 的.strip()消除了前端和后端换行符所以只有 元素之间的换行符

3. split那么打破这些元素是到一个列表，从而你的列表清单

或者，你可以使用itertools.groupby：

>>> [groups for groups in (list(g) for k, g in groupby(text.splitlines(), bool)) if groups[0]] 
[['bread', 'bun'], ['milk', 'soya milk']] 

Answer 7

如果输入文件是足够小，可以完全读入内存，我会做这样的：

with open('grocery_list.txt', 'rt') as grocery_list_file: 
    data = grocery_list_file.read() 

sublist = [item.strip().split('\n') for item in data.split('\n\n')] 

输出：

sublist: [['bread', 'bun'], ['milk', 'soya milk']]