Question

我有一个我要排名的数字列表，从1（最高）开始，然后打印出来。对于有关系的数字，我只想按顺序编号，以先到者为准。

lst = [[0.0] , [0.0] , [0.0] , [0.1] , [0.2]]

#expected output
rank_lst = [3,4,5,2,1]

我只想要一个简单的功能，类似于我所做的事情：

rank_lst = [sorted(lst).index(values) for values in lst]

但是此值从0开始，将最低的数字列为0，并按如下所示对具有相同数字的并列值进行排名：

#output
rank_list = [0,0,0,1,2]

Answer 1

您可以使用numpy.argsort()

由于要从高到低排序，因此可以先将值设为负。

import numpy as np

lst = [[0.0] , [0.0] , [0.0] , [0.1] , [0.2]]
np_lst = np.array([v[0] for v in lst])
rank_lst = list(np.argsort(np.argsort(-np_lst)) + 1)

编辑：再输入np.argsort()即可获得所需的内容。

Answer 2

您可以尝试使用collections.defaultdict反向收集列表的索引，然后使用collections.deque一次将索引从左侧索引中弹出。

from collections import defaultdict
from collections import deque

lst = [0, 1, 0, 2, 0]

d = defaultdict(deque)
for i, x in enumerate(sorted(lst, reverse=True), start=1):
    d[x].append(i)

result = [d[x].popleft() for x in lst]

print(result)
# [3, 2, 4, 1, 5]

或者使用嵌套列表，如问题中所示：

from collections import defaultdict
from collections import deque
from operator import itemgetter

lst = [[0.0], [0.0], [0.0], [0.1], [0.2]]

single = list(map(itemgetter(0), lst))

d = defaultdict(deque)
for i, x in enumerate(sorted(single, reverse=True), start=1):
    d[x].append(i)

result = [d[x].popleft() for x in single]

print(result)
# [3, 4, 5, 2, 1]

Answer 3

您可以这样做

lst = [[0.0], [0.0], [0.0], [0.1], [0.2]]
# wanted: rank_lst == [3, 4, 5, 2, 1]

# add original position
data = list(enumerate(lst, start=1))
# -> [(1, [0.0]), (2, [0.0]), (3, [0.0]), (4, [0.1]), (5, [0.2])]

# sort by value
data = list(sorted(data, key=lambda x: x[1], reverse=True))
# -> [(5, [0.2]), (4, [0.1]), (1, [0.0]), (2, [0.0]), (3, [0.0])]

# add sorted position
data = list(enumerate(data, start=1))
# -> [(1, (5, [0.2])), (2, (4, [0.1])), (3, (1, [0.0])), (4, (2, [0.0])), (5, (3, [0.0]))]

# resort to original order
data = list(sorted(data, key=lambda x: x[1][0]))
# -> [(3, (1, [0.0])), (4, (2, [0.0])), (5, (3, [0.0])), (2, (4, [0.1])), (1, (5, [0.2]))]

# extract sorted order number
rank_lst = [x[0] for x in data]
# -> [3, 4, 5, 2, 1]

当然，您可以排除所有对list的呼叫。我只用它们来创建输出。

Answer 4

那呢？

input = [[0.0], [0.0], [0.0], [0.1], [0.2]]
ranked_list = [item[0] for item in sorted([[index, value] for index, value in enumerate(a)], key=lambda x: x[1], reverse=True)]
final_list = [None] * len(input)
i = 1
for rank in ranked_list:
    final_list[rank] = i
    i += 1

输出：[3, 4, 5, 2, 1]

它在列表上使用enumerate，然后按值（相反）对其进行排序，然后使用列表理解将索引放在新列表中。最后，创建一个len(input)位置的空列表，并用输入的每个值的等级填充它。

编辑：现在实际上已经解决了所提出的问题，对不起您的误解

Answer 5

>>> lst = [[0.0] , [0.0] , [0.0] , [0.1] , [0.2]]
>>> lst1 = [i[0] for i in lst] # you only need numbers, not lists for each value
>>> xx = list(enumerate(lst1)) # add an "index" to each value
>>> xx.sort(key=lambda i: -i[1]) # sort on values
>>> xxx = [(i[0], i[1], j+1) for j, i in enumerate(xx)] # add the rank of each value
>>> xxx.sort(key=lambda i : i[0]) # put them back in the original order using the index
>>> res = [i[2] for i in xxx] # get just the ranks
>>> res
[3, 4, 5, 2, 1]

如何对数字列表进行排名并打印出排名？

5 个答案: