如何从python中的列表列表中创建字典？

Question

Let's suppose I have the following list made out of lists

list1 = [['a','b'],['a'],['b','c'],['c','d'],['b'], ['a','d']]

I am wondering if there is a way to convert every element of list1 in a dictionary where all the new dictionaries will use the same key. Eg: if ['a'] gets to be {'a':1} , and ['b'] gets to be {'b':2} , I would like for all keys a the value of 1 and for all keys b the value of 2 . Therefore, when creating the dictionary of ['a','b'] , I would like to turn into {'a':1, 'b':2} .

What I have found so far are ways to create a dictionary out of lists of lists but using the first element as the key and the rest of the list as the value:

Please note that's not what I am interested in. The result I would want to obtain from list1 is something like:

dict_list1 = [{'a':1,'b':2}, {'a':1}, {'b':2,'c':3}, {'c':3,'d':4}, {'b':2}, {'a':1,'d':4}]

I am not that interested in the items being that numbers but in the numbers being the same for each different key.

Answer 1

You need to declare your mapping first:

mapping = dict(a=1, b=2, c=3, d=4)

Then, you can just use dict comprehension:

[{e: mapping[e] for e in li} for li in list1]
# [{'a': 1, 'b': 2}, {'a': 1}, {'b': 2, 'c': 3}, {'c': 3, 'd': 4}, {'b': 2}, {'a': 1, 'd': 4}]

Answer 2

Using chain and OrderedDict you can do auto mapping

from itertools import chain
from collections import OrderedDict

list1 = [['a','b'],['a'],['b','c'],['c','d'],['b'], ['a','d']]
# do flat list for auto index
flat_list = list(chain(*list1))
# remove duplicates
flat_list = list(OrderedDict.fromkeys(flat_list))
mapping = {x:flat_list.index(x)+1 for x in set(flat_list)}

[{e: mapping[e] for e in li} for li in list1]

Answer 3

在这里，尝试一下ord()同样适用于大写字母和小写字母：

[{e: ord(e)%32 for e in li} for li in list1]