查找列表中 -1、1 和 0 的多数票 - python

Question

如何找到可以包含 -1、1 和 0 的列表的多数票？

例如，给定列表：

x = [-1, -1, -1, -1, 0]

大多数是 -1 ，所以输出应该 return -1

另一个例子，给定一个列表：

x = [1, 1, 1, 0, 0, -1]

多数票将是1

当我们平局时，多数票应该 return 0，例如：

x = [1, 1, 1, -1, -1, -1]

这也应该return零：

x = [1, 1, 0, 0, -1, -1]

获得多数票的最简单情况似乎是对列表求和并检查它是否定的、肯定的还是 0。

>>> x = [-1, -1, -1, -1, 0]
>>> sum(x) # So majority -> 0
-4
>>> x = [-1, 1, 1, 1, 0]
>>> sum(x) # So majority -> 1
2
>>> x = [-1, -1, 1, 1, 0]
>>> sum(x) # So majority is tied, i.e. -> 0
0

求和后，我可以做这个检查以获得多数票，即：

>>> x = [-1, 1, 1, 1, 0]
>>> majority = -1 if sum(x) < 0 else 1 if sum(x)!=0 else 0
>>> majority
1
>>> x = [-1, -1, 1, 1, 0]
>>> majority = -1 if sum(x) < 0 else 1 if sum(x)!=0 else 0
>>> majority
0

但如前所述，它很丑陋：Python putting an if-elif-else statement on one line 而不是 pythonic。

所以解决方案似乎是

>>> x = [-1, -1, 1, 1, 0]
>>> if sum(x) == 0:
...     majority = 0
... else:
...     majority = -1 if sum(x) < 0 else 1
... 
>>> majority
0

已编辑

但有些情况下 sum() 不起作用，例如@RobertB

>>> x = [-1, -1, 0, 0, 0, 0]
>>> sum(x) 
-2

但在这种情况下，多数票应该是 0!!

Answer 1

我假设投 0 票算作投票。所以 sum 不是一个合理的选择。

试试计数器：

>>> from collections import Counter
>>> x = Counter([-1,-1,-1, 1,1,1,1,0,0,0,0,0,0,0,0])
>>> x
Counter({0: 8, 1: 4, -1: 3})
>>> x.most_common(1)
[(0, 8)]
>>> x.most_common(1)[0][0]
0

所以你可以这样写代码：

from collections import Counter

def find_majority(votes):
    vote_count = Counter(votes)
    top_two = vote_count.most_common(2)
    if len(top_two)>1 and top_two[0][1] == top_two[1][1]:
        # It is a tie
        return 0
    return top_two[0][0]

>>> find_majority([1,1,-1,-1,0]) # It is a tie
0
>>> find_majority([1,1,1,1, -1,-1,-1,0])
1
>>> find_majority([-1,-1,0,0,0]) # Votes for zero win
0
>>> find_majority(['a','a','b',]) # Totally not asked for, but would work
'a'

Answer 2

from collections import Counter

result = Counter(votes).most_common(2)

result = 0 if result[0][1] == result[1][1] else result[0][0]

空 votes 列表或基数为 1 的 votes 列表的错误处理是微不足道的，留作 reader.

的练习

Answer 3

我相信这适用于所有提供的测试用例。如果我做错了什么，请告诉我。

from collections import Counter

def fn(x):
    counts = Counter(x)
    num_n1 = counts.get(-1, 0)
    num_p1 = counts.get(1, 0)
    num_z = counts.get(0, 0)
    if num_n1 > num_p1:
        return -1 if num_n1 > num_z else 0
    elif num_p1 > num_n1:
        return 1 if num_p1 > num_z else 0
    else:
        return 0

Answer 4

import numpy as np

def fn(vote):
   n=vote[np.where(vote<0)].size
   p=vote[np.where(vote>0)].size
   ret=np.sign(p-n)
   z=vote.size-p-n
   if z>=max(p,n):
      ret=0
   return ret

# some test cases
print fn(np.array([-1,-1, 1,1,1,1,0,0,0,0,0,0,0,0]))
print fn(np.array([-1, -1, -1, 1,1,1,0,0]))
print fn(np.array([0,0,0,1,1,1]))
print fn(np.array([1,1,1,1, -1,-1,-1,0]))
print fn(np.array([-1, -1, -1, -1, 1, 0]))

Answer 5

如果你使用 python >= 3.4 ，你可以使用 statistics.mode ，当你没有唯一模式时捕获 StatisticsError ：

from statistics import mode, StatisticsError

def majority(l):
    try:
        return mode(l)
    except StatisticsError:
        return 0

statistics 实现本身使用了计数器命令：

import  collections
def _counts(data):
    # Generate a table of sorted (value, frequency) pairs.
    table = collections.Counter(iter(data)).most_common()
    if not table:
        return table
    # Extract the values with the highest frequency.
    maxfreq = table[0][1]
    for i in range(1, len(table)):
        if table[i][1] != maxfreq:
            table = table[:i]
            break
    return table

def mode(data):
    """Return the most common data point from discrete or nominal data.

    ``mode`` assumes discrete data, and returns a single value. This is the
    standard treatment of the mode as commonly taught in schools:

    >>> mode([1, 1, 2, 3, 3, 3, 3, 4])
    3

    This also works with nominal (non-numeric) data:

    >>> mode(["red", "blue", "blue", "red", "green", "red", "red"])
    'red'

    If there is not exactly one most common value, ``mode`` will raise
    StatisticsError.
    """
    # Generate a table of sorted (value, frequency) pairs.
    table = _counts(data)
    if len(table) == 1:
        return table[0][0]
    elif table:
        raise StatisticsError(
                'no unique mode; found %d equally common values' % len(table)
                )
    else:
        raise StatisticsError('no mode for empty data')

另一种使用计数器并捕获空列表的方法：

def majority(l):
    cn = Counter(l).most_common(2)
    return 0 if len(cn) > 1 and cn[0][1] == cn[1][1] else next(iter(cn),[0])[0]

Answer 6

您可以 count occurences 个 0 并测试它们是否占多数。

>>> x = [1, 1, 0, 0, 0]
>>> if sum(x) == 0 or x.count(0) >= len(x) / 2.0:
...     majority = 0
... else:
...     majority = -1 if (sum(x) < 0) else 1
... majority
0

Answer 7

此解决方案基于计算出现次数和排序：

import operator
def determineMajority(x):
    '''
    >>> determineMajority([-1, -1, -1, -1, 0])
    -1

    >>> determineMajority([1, 1, 1, 0, 0, -1])
    1

    >>> determineMajority([1, 1, 1, -1, -1, -1])
    0

    >>> determineMajority([1, 1, 1, 0, 0, 0])
    0

    >>> determineMajority([1, 1, 0, 0, -1, -1])
    0

    >>> determineMajority([-1, -1, 0, 0, 0, 0])
    0
    '''

    # Count three times
    # sort on counts
    xs = sorted(
        [(i, x.count(i)) for i in range(-1,2)],
        key=operator.itemgetter(1),
        reverse=True
    )

    if xs[0][1] > xs[1][1]:
        return xs[0][0]
    else:
        # tie
        return 0


if __name__ == '__main__':
    import doctest
    doctest.testmod()

此外，还有一个 if 语句。正如评论中提到的那样，

会发生什么是未定义的

x = [1, 1, 0, 0, -1]

Answer 8

除了内置的列表运算符和其他东西，您不需要任何东西，不需要导入任何东西。

  votes = [ -1,-1,0,1,0,1,-1,-1] # note that we don't care about ordering

    counts = [ votes.count(-1),votes.count(0),votes.count(1)] 

    if (counts[0]>0 and counts.count(counts[0]) > 1) or (counts[1]>0 and counts.count(counts[1])>1):
         majority=0
    else:
         majority=counts.index(max(counts))-1 # subtract 1 as indexes start with 0

    print majority

3d 行将各个选票的计数放入新列表中，counts.index() 向我们显示我们找到最大选票的列表位置。

我敢说这应该尽可能地像 pythonic，而不会让人眼花缭乱。

Upd：在没有文本字符串的情况下重写并更新为 return 0，以防出现多个相同的结果（在原始 post 中没有注意到这一点），添加了一个 IF 以防只有一票，例如投票=[-1]

Answer 9

from collections import Counter

def find_majority_vote(votes):
  counter = Counter(votes)
  most_common = counter.most_common(2)
  if len(most_common)==2:
    return 0 if most_common[0][1] == most_common[1][1] else most_common[0][0]
  else:
    return most_common[0][0]

Answer 10

# These are your actual votes
votes = [-1, -1, -1, -1, 0]

# These are the options on the ballot
ballot = (-1, 0, 1)

# This is to initialize your counters
counters = {x: 0 for x in ballot}

# Count the number of votes
for vote in votes:
    counters[vote] += 1

results = counters.values().sort()

if len(set(values)) < len(ballot) and values[-1] == values [-2]:
    # Return 0 if there's a tie
    return 0
else:
    # Return your winning vote if there isn't a tie
    return max(counters, key=counters.get)

Answer 11

这适用于任意数量的候选人。如果两位候选人之间有平局，则 return 为零，否则 return 为得票最多的候选人。

from collections import Counter
x = [-1, -1, 0, 0, 0, 0]
counts = list((Counter(x).most_common())) ## Array in descending order by votes
if len(counts)>1 and (counts[0][1] == counts[1][1]): ## Comparing top two candidates 
   print 0
else:
   print counts[0][0]

我们只比较两个候选值，因为如果两个候选值之间有平局，它应该 return 0 并且它不依赖于第三个候选值

Answer 12

一个非常简单的方法。

a = [-1, -1, -1, -1, 0]   # Example
count = {}
for i in a:
    if i not in count:
        count[i] = 1
    else:
        count[i] += 1
m_count = max(count.values())
for key in count:
    if count[key] == m_count:
        print key

在上面的示例中，输出将为 -1，但是，如果出现平局，则将打印两个键。

Answer 13

一个明显的方法是制作一个计数器并根据数据列表更新它x。然后你可以得到最频繁的数字列表（从-1、0、1）。如果有1个这样的数字，这就是你想要的，否则选择0（按照你的要求）。

counter = {-1: 0, 0: 0, 1: 0}
for number in x:
    counter[number] += 1
best_values = [i for i in (-1, 0, 1) if counter[i] == max(counter.values())]
if len(best_values) == 1:
    majority = best_values[0]
else:
    majority = 0

查找列表中 -1、1 和 0 的多数票 - python

Finding majority votes on -1s, 1s and 0s in list - python

python

binary

if-statement

voting

list

已编辑