找到红宝石中偶数长度数组的中位数

问题描述

我无法弄清楚为什么我的偶数长度部分不正确。

I cannot figure out why I cannot get the even length portion correct.

def median(array)
  array.sort!
  if array.length % 2 == 0                                            #if amount of array members is even
    (array[(array.length/2) + 1] + array[array.length/2]) / 2.to_f    #return average of the 2 middle array members
  else                                                                #if amount of array members is odd
    array[array.length/2.ceil]                                        #return middle number
  end   
end

我的尝试是例如一个长度为6的数组，其第3和第4个索引值为7和9.

My attempt is for example, an array whose length is 6, and whose 3rd and 4th index value are 7 and 9.

array[6/3+1] + array [6/3]
(array[4] + array[3]) /2
9 + 7 / 2

我收到此错误

Error!
median returns the correct median of an even-length array
expected: 5.5 got: 6.0 (compared using ==)

我见过一个较短的解决方案，但是如果我能理解我想要遵循的逻辑路径，那就太好奇了，谢谢你们一起玩吧！

I have seen a shorter solution, but am most curious if I can make sense of the logic path I am trying to follow, thanks for playing along!

解决方案我见过：

def median(array)
  sorted = array.sort
  len = sorted.length
  return (sorted[(len - 1) / 2] + sorted[len / 2]) / 2.0
end

推荐答案

数组是零索引的。因此，如果长度为4，则需要取平均指数 1 和 2 。您当前的尝试将平均指数 3 和 2 的长度为4.所以你只需要更改一个小东西（加上减去）：

Arrays are zero-indexed. So if the length was 4, you need to be taking average of indices 1 and 2. Your current attempt would take average of indices 3 and 2 for a length of 4. So you just need to change one small thing (plus into minus):

(array[(array.length/2) - 1] + array[array.length/2]) / 2.to_f

对于偶数编号的Fixnum n ，这总是如此：（n - 1）/ 2 ==（n / 2） - 1 ，这意味着你已经找到了与你找到的方法类似的方法。这并不太令人惊讶，有效地计算中位数的方法有限。

For an even numbered Fixnum n, this is always true: ( n - 1 ) / 2 == ( n / 2 ) - 1, which means you have figured out a similar approach to the one you found. This is not too surprising, there are a limited number of ways to calculate medians efficiently.

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