Trouble implementing part of selection sort

Question

I have implemented the selection sort in python, but couldn't understand this tiny part of inner for loop.

The for loop for j, I have the range from outer index i to max length-1, this makes a lot of sense to me, but the result wouldn't count the last number in the array, and I can't think of the reason.

However, If I change the range from i to max length, it would work. (which in my knowledge should be exceeding the array since alist[len(alist)] would count 1 digit pass the max number).

#Go over the loop, find the smallest num
def swap(arr, num_1, num_2):
    temp = arr[num_1]
    arr[num_1] = arr[num_2]
    arr[num_2] = temp

def selectionSort(alist):
    for i in range(0, len(alist)-1):
        min = i
        # for j in range(i+1, len(alist)):
        # Why len(alist)-1 doesn't work?
        for j in range(i, len(alist)-1):
            if alist[j] < alist[min]:
                min = j
        if min != i :
            swap(alist,i,min)
    return alist


# Test
print "-------------Test--- ----------"
A = [2,1,9,3,4,100,99,30]
print selectionSort(A)

Tom Zych · Accepted Answer · 2017-02-03 22:57:06Z

2

Read about ranges in Python again; you don't have a clear concept of them. range(0, 3), for example, is roughly equivalent to [0, 1, 2]. It stops just short of the second value. The same is true of slices.

answered Feb 3, 2017 at 22:57

Tom Zych

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5 Comments

Yu Zhou Over a year ago

I read over it, and it makes sense now! However if this is the case, say I want to traverse through this array A in the test case, to interate through them all in the outer loop, can I just have "for i in range(0, len(alist))" ? since len(alist) is 8, and it will only loop to A[7]

Tom Zych Over a year ago

Well, that would work, because of the way your program is set up. But if I were writing a selection sort, I'd stop the outer loop at the next-to-last item, and start j at i+1. Note, also, that starting at zero is the default, so you can just say for i in range(len(alist)).

Yu Zhou Over a year ago

Would you mind elaborate on your thought? I was looking at someone's implementation on selection sort, it implements exactly as how you described it. If say, you stop the outer loop @ next-to-last item, wouldn't you then ignore the last item in the array?also to start j at i+1, does that mean so you want to exclude the case where j'th item compares to itself?

Tom Zych Over a year ago

There's no point in comparing a number to itself; it's always equal. As for the last item, we keep swapping lower-valued items upward, so the last item will always be the highest-valued item; if it were not, it would have been swapped upward at some point.

Yu Zhou Over a year ago

It makes sense now!! Thank you

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