如何找到包含-1s，1s和0s的列表的多数票？
例如，给出一个列表：
x = [-1, -1, -1, -1, 0]
多数为-1，因此输出应返回-1
另一个例子，给出一个列表：
x = [1, 1, 1, 0, 0, -1]
多数票将是1
当我们有平局时，多数投票应该返回0，例如：
x = [1, 1, 1, -1, -1, -1]
这也应该返回零：
x = [1, 1, 0, 0, -1, -1]
获得多数投票的最简单案例似乎是对列表进行求和，并检查它是否为负，正或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
总和之后，我可以做这个检查以获得多数投票，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，它很难看：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
但有些情况sum()不会起作用，@ RobertB＆amp; s
>>> x = [-1, -1, 0, 0, 0, 0]
>>> sum(x)
-2
但在这种情况下，多数投票应为0 !!
答案 0 :(得分：10)
我假设0票作为票数。所以sum不是一个合理的选择。
尝试一个计数器：
>>> 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
所以你可以写代码如下：
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'
答案 1 :(得分：5)
如果您使用的是python＆gt; = 3.4，则可以使用statistics.mode，当您没有唯一模式时，可以使用StatisticsError：
from statistics import mode, StatisticsError
def majority(l):
try:
return mode(l)
except StatisticsError:
return 0
statistics实现本身使用Counter dict：
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')
使用Counter并捕获空列表的另一种方法：
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]
答案 2 :(得分：1)
你可以count occurences为0并测试他们是否占多数。
>>> 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
答案 3 :(得分：1)
此解决方案基于计算事件和排序：
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。如评论中所述，未定义
会发生什么x = [1,1,0,0，-1]
答案 4 :(得分：0)
from collections import Counter
result = Counter(votes).most_common(2)
result = 0 if result[0][1] == result[1][1] else result[0][0]
设置基数为1的空votes列表或votes列表的错误处理是微不足道的，并留给读者练习。
答案 5 :(得分：0)
我相信这适用于所有提供的测试用例。如果我做错了，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
答案 6 :(得分：0)
# 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)
答案 7 :(得分：0)
这适用于任何数量的候选人。如果两个候选人之间存在平局，则返回零，否则返回的选票最多。
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]
我们仅比较两位候选人，因为如果两位候选人之间存在平局，则应该返回0并且不依赖于第三候选人值
答案 8 :(得分：0)
A very simple approach.
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
In the above example the output will be -1, however if there is a tie, both the keys will be printed.
答案 9 :(得分：0)
一种显而易见的方法是制作计数器并根据数据列表x进行更新。然后你可以得到最常见的数字列表（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
答案 10 :(得分：-1)
除了内置列表操作符和内容之外，您不需要任何其他内容，无需导入任何内容。
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 line将相应投票的计数放入新列表中，counts.index（）显示我们找到最多投票的列表位置。
我敢说，这应该尽可能地像pythonic一样，而不会陷入眼睛凿眼的oneliners。
更新：重写没有文本字符串并更新为在几个相同结果的情况下返回0（在原始帖子中没有注意到这一点），如果只有一个投票添加了IF案例，例如votes = [ - 1]
答案 11 :(得分：-1)
<div id="postMessage" style="height:30px;"></div>
<div style="height:1500px;background:#333;"></div>
<script src="jquery.min.js"></script>
<script>
$(document).ready(function() {
var eventMethod = window.addEventListener ? "addEventListener" : "attachEvent";
var eventer = window[eventMethod];
var messageEvent = eventMethod == "attachEvent" ? "onmessage" : "message";
eventer(messageEvent,function(e) {
var key = e.message ? "message" : "data";
var data = e[key];
window.scrollTo(0, data);
},false);
});
</script>
答案 12 :(得分：-2)
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]))