我的python选择排序有什么问题？

Question

I can't figure out why the python program produces the following output:

c:\Python Programs>selection_sort.py

[7, 4, 2, 9, 6]

[2, 4, 7, 9, 6]
[2, 6, 7, 9, 4]
[2, 6, 4, 9, 7]
Traceback (most recent call last):
  File "J:\Python Programs\Python Practice\selection_sort.py", line 11, in <modu
le>
    num_list[i], num_list[min_num] = num_list[min_num], num_list[i]
IndexError: list index out of range

c:\Python Programs>

I think I understand the list index out of range part, but I'm not sure about why the 6 becomes the second element when i = 1. Didn't the machine read my if statement?

Here is the code below:

num_list = [7,4,2,9,6]
len_num_list = len(num_list)
print num_list
print""#print empty string to separate the original list from the following iterations 
for i in range(0,len_num_list):
    min_num = min(num_list[i:]) #finds minimum number in list to the right of i
    if min_num>num_list[i]:
        min_num = num_list[i]
    num_list[i], num_list[min_num] = num_list[min_num], num_list[i]
    print num_list  

Answer 1

First, let's note that in your snippet:

min_num = min(num_list[i:]) #finds minimum number in list to the right of i
if min_num>num_list[i]:
    min_num = num_list[i]

the if will never, ever match -- since min_num is the minimum of a sub-list that starts with num_list[i] , it can't possibly, ever, under any circumstance, be greater than the latter.

So, lose the last two of these three statements -- they're about as useful as checking if 2+2 != 4: :-).

Next, let's note that you don't really want min_num to be a value (which is what your call to min gives you) -- you want it to be an index into the list, in order to perform the swap:

num_list[i], num_list[min_num] = num_list[min_num], num_list[i]

But trying to turn a value into an index via the index method is quite an iffy path: if the input list can have any duplicates, index will always locate the first one of them, and that might quite possibly tangle you up. I personally would choose not to go there.

Rather consider the more direct path of finding the minimum index using the corresponding value via the key= feature of min ! Ie:

for i in range(0,len_num_list):
    min_ind = min(range(i, len_num_list),
                  key=lambda j: num_list[j])
    num_list[i], num_list[min_ind] = num_list[min_ind], num_list[i]
    print num_list

If you're not familiar with the key= feature of many Python built-ins ( min , max , sorted , ...), it's really a good thing to learn.

It sorts (or gives the min, or max, or) a certain sequence, with the comparisons done after passing each item of the sequence through the "key extraction function" you pass as key= . Here, you want "the index of the minimum" and you get that by picking the min index with a key= of the corresponding look-up of each index into the list.

I personally dislike lambda and might use key=numlist.__getitem__ , but that's not very readable either -- most readable is always to use def (and I'd do the same for that swap functionality), eg..:

def item_in_list(index): return num_list[index]
def swap(i, j): num_list[i], num_list[j] = num_list[j], num_list[i]
for i in range(0,len_num_list):
    min_ind = min(range(i, len_num_list), key=item_in_list)
    swap(i, min_ind)
    print num_list

which I find to be the most readable and elegant approach to this task.

Answer 2

The problem is that min(num_list[i:]) returns a number from the list, not an index into that list. You can use the index method to get the index corresponding to min(num_list[i:]) . Thus, try:

num_list = [7,4,2,9,6]
len_num_list = len(num_list)
print num_list
print""#print empty string to separate the original list from the following iterations_
for i in range(0,len_num_list):
    min_num = min(num_list[i:]) #finds minimum number in list to the right of i
    j = num_list.index(min_num)
    if min_num>num_list[i]:
        min_num = num_list[i]
    num_list[i], num_list[j] = num_list[j], num_list[i]
    print num_list

This produces the output:

[7, 4, 2, 9, 6]

[2, 4, 7, 9, 6]
[2, 4, 7, 9, 6]
[2, 4, 6, 9, 7]
[2, 4, 6, 7, 9]
[2, 4, 6, 7, 9]