Python - 根据列表中的反向位置计算分数

Question

I can do this is very long-winded, hacky code. Am looking for pointers as to how to do this in a clean, pythonic way. I am using Python 3.5.2

I have 2 ordered lists of words thus....

ting_word_bag = ['word', 'tree', 'dependency', 'number', 'pattern']

ted_word_bag = ['dependency', 'verb', 'grammar', 'word', 'parser'] 

I want to iterate over the words in ted_word_bag and assign a value to each, based on the word's position in the ting_word_bag . This is straightforward.

What is less so is that I want to reverse the values so the first word in the ting_word_bag list is worth 5 points, and the last word in the list is worth 1 point (based on a list of 5 elements)

For this example, the total score for the ted_word_bag would be 8 points. 5 for 'word' and 3 for 'dependency' .

Any pointers on how to do this simply and quickly would be much appreciated.

Cheers.

Answer 1

Using comprehension, you can build a dict to store the "score" for each word:

>>> some_dict = {j:i for i,j in enumerate(ting_word_bag)}
>>> some_dict
{'information': 5, 'word': 0, 'pattern': 4, 'tree': 1, 'number': 3, 'dependency': 2}

Using reversed or another method, you can obtain what you want.

Next, with the .get(index, default-value) of the dictionary, you sum each value of your second list:

>>> sum(some_dict.get(i,0) for i in ted_word_bag)
8 #with a correct value of some_dict

Answer 2

ting_word_bag = ['word', 'tree', 'dependency', 'number', 'pattern']

ted_word_bag = ['dependency', 'verb', 'grammar', 'word', 'parser'] 

Given that the following:

from itertools import count

scores = dict(zip(reversed(ting_word_bag), count(1)))

total = sum(scores.get(i,0) for i in ted_word_bag)

Properly assigns the scores:

>>> scores
{'pattern': 1, 'tree': 4, 'number': 2, 'word': 5, 'dependency': 3}

Even so the resulting total is the same:

>>> total
8

Answer 3

Make a list comprehension to enumerate ting_word_bag in reversed order:

points = [(i,w) for i,w in enumerate(ting_word_bag[::-1])]

Another comprehension for the score:

score =  sum([x[0] for x in points if x[1] in ted_word_bag]) 

Answer 4

this code does this (assuming for not found items the score is 0 ):

ting_word_bag = ['word', 'tree', 'dependency', 'number', 'pattern']
ted_word_bag = ['dependency', 'verb', 'grammar', 'word', 'parser']

score = {x: i + 1 for i, x in enumerate(reversed(ting_word_bag))}
values = [(score.get(x, 0), x) for x in ted_word_bag]

print(values)
print(sum(i[0] for i in values))

output:

[(3, 'dependency'), (0, 'verb'), (0, 'grammar'), (5, 'word'), (0, 'parser')]
8