Median code explanation

My professor wrote this median function and I don't understand it very well. Can someone please explain the part about i = len(list)/2 and median = avg() and the else statement?

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开发者_运维百科

To add from an example I saw, for 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 odd list length:

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

Since x has a list length of odd, wouldn't the [len(x)/2] be a floating number index? I'm not getting this all the way? Any explanation better than mine is much appreciated.

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Why this is is 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 data types, namely list.

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

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

#3 and #5 are calculated the same way regardless of the if branch taken. Regardless, i is 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]

Also, the avg_list() function (which is not used nor could be used in median()) could 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.

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We're missing some code here, but we can puzzle it out.

The comments here are instructive. When we check:

    if len(list)%2 == 0:

Then we're checking to see if the list is of even length. If a list has an even number of members, then there is no true "middle" element, and so:

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

We assume that the avg() function is going to return the average of the two middle elements. Since you didn't include a definition of an avg function, it's possible that this is really supposed to be an avg_list function taking the middle two elements of the list.

Now, if the list is of 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 this looks kinda wrong to me too, but my guess is that the intention is that this should read:

median = list[i]

That would be us returning the middle element of the list. Since the list has been sorted, that middle element is the true median of the list.

Hope this helps!

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I'm sure it's trying to say, "If the list is of odd size, just take the central element; otherwise take the mean of the central two elements" - but I can't see that that's what the code is actually doing at all.

In particular:

• It's calling an avg() function (not avg_list, note) but without any arguments
• It's ignoring the value of i after computing it in the same way in both branches

Are you sure that's the complete code which is meant to work?

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You can also decide to always return the average of the middle sub-array of the ordered list: For instance return the average of [4,5] out of [1,2,3,4,5,6,7,8], and that of [5] out of [1,2,3,4,5,6,7,8,9].

A python implementation would be:

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