如何将子列表的分数分配给单词并创建新词典

Question

I have a dictionary adict:

a={"good":2, "bad":-2}

a list with sublist of strings b:

b=[["how", "are", "good"],["bad", "BAD", "hello"]]

and a list of integers which is the same length as list b and is a score of each sublist of b:

c=[2, -4] 

I need to assign the score of the sublist to the words in b that do not appear i the keys of a It should create a new dictionary as follows:

{{"how":2, "are":2},{"hello":-4}}

I have tried the following code but it does not work:

for sublst in b:
    for i in sublst:
        if i.lower() not in a.keys():
            newdict=dict(zip(sublst, c))

Answer 1

a={"good":2, "bad":-2} 
b=[["how", "are", "good"],["bad", "BAD", "hello"]]
c=[2, -4]

new_list = []
for i in range(len(b)):
    value = c[i]
    d= {}
    for word in b[i]:
        if(word.lower() not in a.keys()):
            d[word] = value
    new_list.append(d.copy())

print(new_list)

output:

 [{'how': 2, 'are': 2}, {'hello': -4}]

Answer 2

Here's one way using a dictionary comprehension. Note that dictionaries are unhashable, so you cannot have a set of dictionaries. You could have the result be a list of dictionaries instead as bellow:

k = a.keys()
[{w:s for w in l if w.lower() not in k} for l,s in zip(b,c)]
# [{'how': 2, 'are': 2}, {'hello': -4}]

Answer 3

Your code goes wrong at the zip line. Firstly,

sublist = [['how', 'are', 'good']
           ['bad', 'BAD', 'hello']]

while

c = [2, -4]

(sublist, c) works for the first two elements and not the elements satisfying the condition. To make this work, a different list has to be made consisting of

[['how', 'are'], ['hello']]

But this fails to zip the values because zip does not work on list of lists. So the solution to this problem comes out to store the c[i] value for the ith element of b. If any sub-element satisfies the condition, update the dictionary, otherwise keep on iterating and changing the value of c[i]. This method is implemented as follows :-

dic = {}
for i in range(len(b)):
    score = c[i]
    for j in b[i]:
        if j.lower() not in a.keys():
            dic.update({j : score})