Code Med

Question

My professor wrote this median function, and I do not understand it very well. Can someone explain the part i = len(list)/2 and median = avg() and the else ?

 def avg_list(numbers): sum = 0 for num in numbers: sum += num avg = float(sum)/len(numbers) print avg def median(list): list.sort() if len(list)%2 == 0: #have to take avg of middle two i = len(list)/2 median = avg() else: #find the middle (remembering that lists start at 0) i = len(list)/2 median = list return median 

To add from the example I saw for an even list length:

 def median(s): i = len(s) if not i%2: return (s[(i/2)-1]+s[i/2])/2.0 return s[i/2] 

This works very well, but I don’t understand the last return s[i/2] ?

For the length of the odd list:

 x = [1,2,5,2,3,763,234,23,1,234,21,3,2134,23,54] median = sorted(x)[len(x)/2] 

Since x is an odd-numbered list, is [len(x)/2] a floating-point index? I do not get it completely? Any explanation better than mine is greatly appreciated.

Answer 1

We are missing the code here, but we can figure it out.

The comments here are instructive. When we check:

  if len(list)%2 == 0: 

Then we check if the list has an even length. If the list has an even number of members, then there is no true "middle" element, and therefore:

  #have to take avg of middle two i = len(list)/2 median = avg() 

Assume that the function avg () returns the average of two middle elements. Since you did not specify a definition for the avg function, it is possible that it really should be an avg_list function that uses two items from a list.

Now, if the list has an odd length, there is a middle element, and so:

  else: #find the middle (remembering that lists start at 0) i = len(list)/2 median = list 

Now it looks wrong for me too, but I assume that this intention is that it should read:

 median = list[i] 

This means that we are returning the middle element of the list. Since the list is sorted, this middle element is the true median of the list.

Hope this helps!

Answer 2

Why is this very wrong, line by line:

 def median(list): # 1 list.sort() # 2 if len(list)%2 == 0: #have to take avg of middle two i = len(list)/2 # 3 median = avg() # 4 else: #find the middle (remembering that lists start at 0) i = len(list)/2 # 5 median = list # 6 return median 

#1 : It’s a bad idea to give your variables the same name as the data types, namely list .

#2 : list.sort() will change the list that is passed. One would expect a getter like median() not do this.

#4 It calls the avg() function with no arguments, which is completely pointless, even if such a function is defined.

#3 and #5 are evaluated the same regardless of the if branch. Despite this, i never used.

#6 It sets median to the original list , which makes zero sense.

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Here's how I would rewrite this (while maintaining clarity):

 def median(alist): srtd = sorted(alist) # returns a sorted copy mid = len(alist)/2 # remember that integer division truncates if len(alist) % 2 == 0: # take the avg of middle two return (srtd[mid-1] + srtd[mid]) / 2.0 else: return srtd[mid] 

In addition, the avg_list() function (which is not used and cannot be used in median() ) can be rewritten as:

 def avg_list(numbers): return float(sum(numbers))/len(numbers) 

sum() is a function that returns the sum of all elements in an iterable version.

Answer 3

I'm sure he is trying to say: "If the list is odd in size, just take the center element, otherwise take the middle of the center two elements" - but I don’t see that this is what the code actually is.

In particular:

• It calls the avg() function (not avg_list , note), but without any arguments
• It ignores the value of i after calculating it in the same way in both branches

Are you sure the complete code should work?

Answer 4

You can also always return the average value of the average auxiliary array of an ordered list: For example, return the average value [4,5] from [1,2,3,4,5,6,7,8], and the average from [5] from [ 1,2,3,4,5, 6,7,8,9].

The python implementation will be:

 def median(a): ordered = sorted(a) length = len(a) return float((ordered[length/2] + ordered[-(length+1)/2]))/2