Python顺序排名

Question

早上好，请给我一张表，我正试图返回职位(排名(学生的分数。我到处找了找，没有任何好转。我以为我可以通过排序来完成，但事实并非如此。这是表格：

Name       Score    Position 
David      89        3rd
James      56        5th
Matt       72        4th
John       91        1st
Iva        56        5th
Anna       91        1st

我试着写这个代码，但它没有处理重复的

score = [89, 56, 72, 91, 56,91]
order = sorted(list(set(score)), reverse=True)
position = []
for n in score:
position.append(order.index(n) + 1)
print(position)

Answer 1

如果您愿意使用外部库，我只能推荐ranking:

import ranking
# Each player and their scores.
score_list = [
('David', 89),
('James', 56),
('Matt', 72),
('John', 91),
('Iva', 56),
('Anna', 91),
]
# Helper for accessing a score record's score, 
# defined here as we need it twice.
get_score = lambda pair: pair[1]
# Sort the scores in place; required by `ranking`.
score_list.sort(key=get_score, reverse=True)
# Define a format string; handy for formatting both the header and the scores.
format_str = "{player:10}   {score:<10}   {rank}"
# Print header.
print(format_str.format(player="Player", score="Score", rank="Position"))
# Rank and print.
for rank, (player, score) in ranking.Ranking(score_list, key=get_score, start=1):
print(format_str.format(rank=rank, player=player, score=score))

输出

Player       Score        Position
John         91           1
Anna         91           1
David        89           3
Matt         72           4
James        56           5
Iva          56           5

Answer 2

不是最漂亮的代码，但我认为这符合您的要求：

from collections import Counter
score = [89, 56, 72, 91, 56, 91]
score_set = sorted(list(set(score)), reverse=True) # unique score
dico = dict((v, i + 1) for i, v in enumerate(score_set)) # position of unique score
count = Counter(score) # occurence of each score
total = 1 # indice to get the current position in the score
# update the dico by counting the number of duplicate.
for k, v in sorted(dico.items(), key=lambda x: x[1]): # sort the dico by the value (rank without duplicate)
count_ = count[k]
dico[k] = total
total += count_ # take into account potential duplicate
position = [dico[e] for e in score]  # apply the dict to our original list
print(position) # et voilà

Answer 3

如果您正在寻找一个不使用任何库的简单解决方案：

score = [89, 56, 72, 91, 56, 91]
sorted_score = sorted(score, reverse=True)
position = [1]
pos = 1
for i in range(1, len(score)):
if sorted_score[i] < sorted_score[i-1]:
pos = i+1
position.append(pos)
print(position)

[1, 1, 3, 4, 5, 5]

Answer 4

您可以使用defaultdict将分数和索引存储在原始列表中，以处理重复项：

from collections import defaultdict                                                   
scores_dict = defaultdict(list)                                                      
for i, s in enumerate(scores): 
scores_dict[s].append(i)
positions = [None] * len(scores)
rank = 1                                                                             
for s in sorted(scores_dict, reverse=True): 
indices = scores_dict[s] 
for i in indices: 
positions[i] = rank 
rank += len(indices)