Python中的位置排名和关系处理 [英] Positional Rankings and Dealing with Ties in Python

问题描述

(很抱歉，此问题的先前版本显示了我需要修复的错误功能，已经纠正，希望现在这个问题更有意义.)

我有一个带有分数的对象列表，我试图根据这些分数为它们分配等级.下面基本上是我输出数据的方式.

I have a list of objects with scores, and I'm attempting to assign rank to them based on those scores. Below is basically how I output my data.

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，将位置分配给上方对象的函数是一个简单的for循环，仅将 i 的值分配为对象的最终位置.

I have a tie. Now, the function that assigns positions to the objects above right now is a simple for loop that just assigns the value of i as the object's final position.

positions = {}

i = 1
for key, value in sorted_list:
    # Since in my codebase the strings are IDs, I use the key to fetch the object.
    if value is not None:
        positions[key] = i
        i += 1

这样显然会返回:

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

希望这很有道理.问题的实质是那个循环.更有意义的是，是否像这样返回它们:

Hopefully that makes some sense. The meat of the question is that loop. What makes more sense is if it returns them like so:

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

我要如何编辑上面的函数来做到这一点，请记住我可以在任何给定的时间有任意数量的联系，这取决于我有多少个成员对所述对象进行排名?最高等级应为1，因此可以这样显示:< rank>/<人员总数>

How would I edit the function above to do that, keeping in mind that I could have any number of ties at any given time depending on how many of my members ranked said object? The highest rank should be 1, so it can be displayed as such: <rank>/<total # of people>

先谢谢了.:)

推荐答案

>>> 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}

请记住，字典是无序的，因此要按顺序进行迭代，您需要这样做

Remember that dicts are unordered, so to iterate in order of place, you need to do this

>>> from operator import itemgetter
>>> print sorted(res.items(),key=itemgetter(1))
[('Apolo Ohno', 1), ('Shanie Davis', 2), ('Bodie Miller', 3), ('Lindsay Vohn', 4), ('Shawn White', 4), ('Bryan Veloso', 6)]

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