Python中的位置排名和平局处理

Question

（我很抱歉，这个问题以前的版本显示了我需要修复的错误功能，这个问题已经得到了纠正，我希望这个问题现在变得更有意义。）

我有一个带有分数的对象列表，我试图根据这些分数给它们分配等级。下面是我如何输出我的数据。在

sorted_scores = [
    ('Apolo Ohno', 0),
    ('Shanie Davis', -1),
    ('Bodie Miller', -2),
    ('Lindsay Vohn', -3),  
    ('Shawn White', -3),
    ('Bryan Veloso', -4)
]

我有领带。现在，为上面的对象指定位置的函数是一个简单的for循环，它只将i的值指定为对象的最终位置。在

^{pr2}$

很明显这会回来：

positions = {
    'Apolo Ohno': 1,
    'Shanie Davis': 2,
    'Bodie Miller': 3,
    'Lindsay Vohn': 4,        
    'Shawn White': 5,
    'Bryan Veloso': 6
}

希望这有点道理。问题的关键是这个循环。更有意义的是，如果它像这样返回它们：

positions = {
    'Apolo Ohno': 1,
    'Shanie Davis': 2,
    'Bodie Miller': 3,
    'Lindsay Vohn': 4, # Same value.
    'Shawn White': 4, # Same value.
    'Bryan Veloso': 6
}

我应该如何编辑上面的函数来做到这一点，记住我可以在任何给定的时间有任何数量的领带，这取决于我的会员中有多少人对所说的对象进行排名？最高的等级应该是1，所以可以这样显示：<rank>/<total # of people>

提前谢谢。：）

Answer 1

==规范变更/澄清后更新===

# coding: ascii

def ranks_from_scores(sorted_scores):
    """sorted_scores: a list of tuples (object_id, score), sorted by score DESCENDING
       return a mapping of object IDs to ranks
    """
    ranks = {}
    previous_score = object()
    for index, (obj_id, score) in enumerate(sorted_scores):
        if score != previous_score:
            previous_score = score
            rank = index + 1
        ranks[obj_id] = rank
    return ranks

from operator import itemgetter
import pprint

scores0 = dict([
    ('Apolo Ohno', 0),
    ('Shanie Davis', -1),
    ('Bodie Miller', -2),
    ('Lindsay Vohn', -3),
    ('Shawn White', -3)
    ])

scores1 = {
    'lorem': 100,
    'ipsum': 200,
    'dolor': 300,
    'sit': 300,
    'amet': 300,
    'quia': 400,
    'consectetur': 500,
    'adipiscing': 500,
    'elit': 600,
    }

scores2 = {
    'lorem': 100,
    'ipsum': 200,
    'dolor': 300,
    'sit': 300,
    'amet': 300,
    'quia': 400,
    'consectetur': 500,
    'adipiscing': 500,
    'elit': 6000,
    }

import pprint
funcs = (ranks_from_scores, ) # Watch this space!
tests = (scores0, scores1, scores2)

for test in tests:
    print
    test_list = sorted(test.items(), key=itemgetter(1), reverse=True)
    print "Input:", test_list
    for func in funcs:
        result = func(test_list)
        print "%s ->" % func.__name__
        pprint.pprint(result)

结果：

^{pr2}$

==原始提交===

这段代码假设您真的希望最高的分数获得排名1，而不是最低分数获得排名1（或0！）。在

# coding: ascii

def ranks_from_scores(scores, debug=False):
    """scores (a mapping of object IDs to scores)
       return a mapping of object IDs to ranks
    """
    alist = [(v, k) for k, v in scores.items()]
    alist.sort(reverse=True)
    if debug: print 'alist:', alist
    bdict = {}
    previous_score = object()
    for posn, (score, obj_id) in enumerate(alist):
        if score != previous_score:
            previous_score = score
            rank = posn + 1
        bdict[obj_id] = rank
    if debug:
        print 'bdict:', bdict
        blist = [(v, k) for k, v in bdict.items()]
        print 'blist:', sorted(blist)
    return bdict

ranks_from_scores(
    {'q': 10, 'w': 20, 'e': 20, 'r': 20, 't': 30},
    debug=True,
    )

输出：

alist: [(30, 't'), (20, 'w'), (20, 'r'), (20, 'e'), (10, 'q')]
bdict: {'q': 5, 'r': 2, 'e': 2, 't': 1, 'w': 2}
blist: [(1, 't'), (2, 'e'), (2, 'r'), (2, 'w'), (5, 'q')]

Answer 2

>>> sorted_scores = [
...     ('Apolo Ohno', 0),
...     ('Shanie Davis', -1),
...     ('Bodie Miller', -2),
...     ('Lindsay Vohn', -3),  
...     ('Shawn White', -3),
...     ('Bryan Veloso',-4)
... ]
>>> 
>>> res = {}
>>> prev = None
>>> for i,(k,v) in enumerate(sorted_scores):
...     if v!=prev:
...         place,prev = i+1,v
...     res[k] = place
... 
>>> print res
{'Apolo Ohno': 1, 'Bryan Veloso': 6, 'Shanie Davis': 2, 'Lindsay Vohn': 4, 'Bodie Miller': 3, 'Shawn White': 4}

记住，dicts是无序的，所以要按位置顺序迭代，您需要这样做

^{pr2}$

Answer 3

这样做的方法不是计算元素的位置是任意的序列，而是计算有多少其他元素有更好的分数。在

编辑：

根据大众的需求，O（n）ed和其他一切：

positions = {}
cur_score = None # Score we're examining
cur_count = 0 # Number of others that we've seen with this score

for ix, (name, score) in enumerate(sorted_scores):
  if score == cur_score: # Same score for this player as previous
    cur_count += 1
  else: # Different score from before
    cur_score = score
    cur_count = 0
  positions[name] = ix - cur_count + 1 # Add 1 because ix is 0-based

print positions