如何根据特定标准对列表进行排序

Question

for example I have a list

list = [['word1', 0.234], ['word2', 0.2], ['word3', 0.5], ['word4', 0,67]]

and I want to sort it by importance. Importance means how far is the value from 0.5. The list after sorting should be:

list = [['word2', 0.2], ['word1', 0.234], ['word4', 0.67], ['word3', 0,5]]

First I solved it by writing simple function based on bubble sorting but it's very slow.

def most_interesting_words(seznam):
"""sort list i.e. [['word', spamicity], ...], first tokens whose value was furthest from the neutral value of 
for i in range(len(seznam)):
    for j in range(len(seznam)):
        #if dist_from_05(seznam[i][1]) > dist_from_05(seznam[j][1]):
        if abs(0.5 - seznam[i][1]) > abs(0.5 - seznam[j][1]):
            tmp = seznam[i]
            seznam[i] = seznam[j]
            seznam[j] = tmp
return seznam

How it can be solved with sorted() function?

Answer 1

To sort by distance from 0.5 you can use the following sorted() function:

my_list = [['word1', 0.234], ['word2', 0.2], ['word3', 0.5], ['word4', 0.67]]
print(sorted(my_list, key=lambda x: abs(0.5 - x[1]), reverse=True))

This would give you the following output:

[['word2', 0.2], ['word1', 0.234], ['word4', 0.67], ['word3', 0.5]]

Answer 2

Using the parameter key . Example

myList = [['word1', 0.234], ['word2', 0.2], ['word3', 0.5], ['word4', 0.67]]
print(sorted(myList,key= lambda x: abs(0.5-x[1]),reverse=True))

Output:

[[word2, 0.2], [word1, 0.234], [word4, 0.67], [word3, 0.5]]

If you want to actually sort you list and not create a new one use myList.sort(key= lambda x: abs(0.5-x),reverse=True) instead