Question

Okay, so I have a bit of a problem, Im new to python sorry.

I am trying to sort a list by score, which is a number, but if there is a draw I need to sort them by the length of their name, the shorter the name the higher the rank.

So I have this list here

11 Jenny
8 Adam
10 Mark
8 Ada

and when I use this code here on it, it just comes back with

11 Jenny
10 Mark
10 Mark
10 Mark

def sort(names, counts):
    newArr = []
    newNames = names
    newCount = counts
    for x in range(0, len(names)):
        newArr.append(findBiggest(newNames, newCount))
        for z in range(0, len(names)):
            name = newArr[len(newArr) - 1].split(" ")[1]
            print name 
            if names[z] == name:
                tempArr1 = newNames
                tempArr2 = newCount
                newNames = []
                newCount = []
                for y in range(0, len(tempArr1)):
                    if y != z:
                        newNames.append(tempArr1[y])
                        newCount.append(tempArr2[y])
    return newArr
def findBiggest(names, counts):
    biggest = 0;
    for x in range(0, len(counts)):
        if int(counts[x]) > biggest:
            biggest = int(counts[x])
    biggestCountArr = [[], []]
    for x in range(0, len(counts)):
        if int(counts[x]) == biggest:
            biggestCountArr[0].append(counts[x])
            biggestCountArr[1].append(names[x])
    if len(biggestCountArr[0]) == 1:
        return str(biggestCountArr[0][0]) + " " + biggestCountArr[1][0]
    else:
        return smallestLength(biggestCountArr)
def smallestLength(twoDArr):
    names = twoDArr[1]
    shortestLen = 0
    for x in range(0, len(names)):
        if len(names[x]) > shortestLen:
            shortestlen = len(names[x])
    for x in range(0, len(names)):
        if len(names[x]) == shortestLen:
            return str(twoDArr[0][x]) + " " + twoDArr[1][x]

Just so you know

11 Jenny
8 Adam
10 Mark
8 Ada

should come out as

11 Jenny
10 Mark
8 Ada
8 Adam

Answer 1

lst=[(11, "Jenny"),(8, "Adam"),(10, "Mark"),(8, "Ada")]
lst.sort(key=lambda x: (-x[0],len(x[1])) )
print (lst) # [(11, 'Jenny'), (10, 'Mark'), (8, 'Ada'), (8, 'Adam')]

The list method sort and the builtin function sorted accept a keyword argument key which is given a callable. Basically, for every element in the sequence, that element is passed to the key function and the return value of that function is actually what python uses to determine ordering when sorting. So, in the above, I use lambda to construct a function which returns a tuple from the input elements. The tuple is ordered first_element, lenth_of_second_element.

When tuples (or lists for that matter) are compared, it’s much like comparing a string. You look at the first element, if they’re the same, you continue on to look at the second element, then the third and so on until one element is greater than the other. e.g.

(1,2,3,4) > (1,2,3,3) #True

This ends up being handy for sorting in very interesting ways.

I suppose to round this out, I should mention that the algorithm that python uses to sort is stable. This means that if you sort by keyA and then you sort by keyB, two elements which compare equal based on keyB will maintain the order they had after the sort using keyA. In other words, a sort doesn’t change the order of equal valued elements. So, the above could also be accomplished like this:

lst.sort(key=lambda x:len(x[1]))  #sort by length of names
lst.sort(key=lambda x:x[0], reversed=True) #sort by score (highest first instead of regular lowest first)

And I suppose no answer is complete without a link to something which explains it more elegantly. (Specifically, see the section on “Key Functions”)