题
def fraction?(number)
number - number.truncate
end
def percentile(param_array, percentage)
another_array = param_array.to_a.sort
r = percentage.to_f * (param_array.size.to_f - 1) + 1
if r <= 1 then return another_array[0]
elsif r >= another_array.size then return another_array[another_array.size - 1]
end
ir = r.truncate
another_array[ir] + fraction?((another_array[ir].to_f - another_array[ir - 1].to_f).abs)
end
用法示例:
test_array = [95.1772, 95.1567, 95.1937, 95.1959, 95.1442, 95.061, 95.1591, 95.1195,
95.1065, 95.0925, 95.199, 95.1682]
test_values = [0.0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1.0]
test_values.each do |value|
puts value.to_s + ": " + percentile(test_array, value).to_s
end
输出:
0.0: 95.061
0.1: 95.1205
0.2: 95.1325
0.3: 95.1689
0.4: 95.1692
0.5: 95.1615
0.6: 95.1773
0.7: 95.1862
0.8: 95.2102
0.9: 95.1981
1.0: 95.199
这里的问题是第 80 个百分位数高于第 90 个和第 100 个百分位数。但是,据我所知,我的实现如所描述的那样,并且它返回给定示例的正确答案(0.9)。
我的代码中是否存在我没有看到的错误?或者有更好的方法来做到这一点吗?
解决方案
脚本
这听起来像是一个家庭作业问题。无论如何,这很有趣。
# Score class
class Score
attr_accessor :value, :percentile
def initialize(score)
self.value = score.to_f
end
def <=>(foo)
self.value <=> foo.value
end
end
# load scores
scores = []
DATA.each do |line|
scores << Score.new(line)
end
scores.sort!
scores_count = scores.size
# iterate through scores and calculate percentile
scores.each_with_index do |s, i|
# L/N(100) = P
# L = number of scores beneath this score (score array index)
# N = total number of scores
# P = percentile
s.percentile = (i.to_f/scores_count.to_f*100).ceil
end
# output
puts "What is the precise percentile of each score"
scores.each_with_index do |s,i|
puts "#{s.value} is in the #{s.percentile} percentile"
end
# bonus: what score is in the Xth percentile?
puts "\nWhat score is in the Xth percentile?"
percentiles = [0, 10, 20, 30, 40, 50, 60, 70, 80, 90, 100]
percentiles.each do |p|
# P/100(N) = L
# P = percentile
# N = total number of scores
# L = score array index
l = (p.to_f/100*scores_count).ceil
puts "#{p} percentile? #{scores[l].value}"
end
__END__
95.1772
95.1567
95.1937
95.1959
95.1442
95.061
95.1591
95.1195
95.1065
95.0925
95.199
95.1682
输出
What is the precise percentile of each score
95.061 is in the 0 percentile
95.0925 is in the 9 percentile
95.1065 is in the 17 percentile
95.1195 is in the 25 percentile
95.1442 is in the 34 percentile
95.1567 is in the 42 percentile
95.1591 is in the 50 percentile
95.1682 is in the 59 percentile
95.1772 is in the 67 percentile
95.1937 is in the 75 percentile
95.1959 is in the 84 percentile
95.199 is in the 92 percentile
What score is in the Xth percentile?
0 percentile? 95.061
10 percentile? 95.1065
20 percentile? 95.1195
30 percentile? 95.1442
40 percentile? 95.1567
50 percentile? 95.1591
60 percentile? 95.1772
70 percentile? 95.1937
80 percentile? 95.1959
90 percentile? 95.199
其他提示
成功了。添加 -Infinity
到数组,以便我可以使用范围内的索引 1 - N
. 。我还将最后一行中的值乘以错误的变量。
def percentile(param_array, percentage)
another_array = param_array.to_a.dup
another_array.push(-1.0/0.0) # add -Infinity to be 0th index
another_array.sort!
another_array_size = another_array.size - 1 # disregard -Infinity
r = percentage.to_f * (another_array_size - 1) + 1
if r <= 1 then return another_array[1]
elsif r >= another_array_size then return another_array[another_array_size]
end
ir = r.truncate
fr = fraction? r
another_array[ir] + fr*(another_array[ir+1] - another_array[ir])
end
这 r = ...
线可以替换为 r = percentage.to_f * (another_array_size + 1)
使用第一个链接中的公式而不是 Excel 的公式。
输出:
0.0: 95.061
0.1: 95.0939
0.2: 95.1091
0.3: 95.12691
0.4: 95.1492
0.5: 95.1579
0.6: 95.16456
0.7: 95.1745
0.8: 95.1904
0.9: 95.19568
1.0: 95.199
你也可以用monkeypatch枚举:
module Enumerable
def rank value, n_tiles
count = self.length
raise "You cannot split an array of #{count} elements into #{n_tiles} tiles!" if n_tiles > count
ordered_array = self.sort
split_size = count / n_tiles
boundaries = []
(n_tiles - 1).times do |i|
boundaries << ordered_array[(i + 1) * split_size - 1]
end
boundaries.each_with_index do |boundary, i|
if value > boundaries.last
return n_tiles
elsif value <= boundary
return (i + 1)
end
end
end
end
之后你就可以做这样的事情:
a = [1,4,2,5,3,6]
# Test in which range (rank) the number '1' would be places, if the array is ordered and spit into 3 pieces:
a.rank(1,3)
#=> 1
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