问题描述
字符串列表:
['Georgie Porgie',87,'$$$',['Canadian','Pub Food'],'Queen St. Cafe',82,'$',['Malaysian','Thai'],'Dumplings R Us',71,'Chinese','Mexican Grill',85,'$$','Mexican','Deep Fried Everything',52,'Pub Food']
我正在尝试从上面的列表中创建一个cuine_to_names 字典。菜系在索引 3 处,有时是它们自己的迷你清单。餐厅名称在索引 0 处。它们每四个索引重复一次。美食 - 3::4,名称 - 0::4。
我遇到的问题是从索引 3::4 中提取元素并将它们设为键。我认为麻烦来了,因为有时它们是一个字符串的迷你列表,有时它们只是一个字符串。这使得使用 defaultdict 对我来说很困难,但我是学习该功能的新手。我看到其他一些答案包括 setdefault 之类的东西,但我不知道如何在这种特定情况下使用它。任何指导将不胜感激!
我想要这个输出:
Cuisine to list of restaurant names:
# dict of {str,list of str}
{'Canadian': ['Georgie Porgie'],'Pub Food': ['Georgie Porgie','Deep Fried Everything'],'Malaysian': ['Queen St. Cafe'],'Thai': ['Queen St. Cafe'],'Chinese': ['Dumplings R Us'],'Mexican': ['Mexican Grill']}
我试过了,得到了 TypeError: unhashable type: 'list':
from collections import defaultdict
cuisine_to_name = defaultdict(list)
for cuisine,name in zip(contents_list_2[3::4],contents_list_2[::4]):
cuisine_to_name[cuisine].append(name)
print(cuisine_to_name)
解决方法
问题在于变量 cuisine 是一个列表,而您正试图将其用作字典的键,请改为这样做:
import pprint
from collections import defaultdict
data = ['Georgie Porgie',87,'$$$',['Canadian','Pub Food'],'Queen St. Cafe',82,'$',['Malaysian','Thai'],'Dumplings R Us',71,'Chinese','Mexican Grill',85,'$$','Mexican','Deep Fried Everything',52,'Pub Food']
res = defaultdict(list)
for v,k in zip(data[::4],data[3::4]):
ks = k if isinstance(k,list) else [k]
for ki in ks:
res[ki].append(v)
pprint.pprint(dict(res))
输出
{'Canadian': ['Georgie Porgie'],'Chinese': ['Dumplings R Us'],'Malaysian': ['Queen St. Cafe'],'Mexican': ['Mexican Grill'],'Pub Food': ['Georgie Porgie','Deep Fried Everything'],'Thai': ['Queen St. Cafe']}
,
如果 l 是您的列表,您可以执行以下操作:
l=['Georgie Porgie','Pub Food']
res={l[i]:l[i+3] if type(l[i+3])==list else [l[i+3]] for i in range(0,len(l),4)}
>>>print(res)
{'Georgie Porgie': ['Canadian','Queen St. Cafe': ['Malaysian','Dumplings R Us': ['Chinese'],'Mexican Grill': ['Mexican'],'Deep Fried Everything': ['Pub Food']}