Insertion sort implementation in Python
TL;DR

Be as specific as you can:


When naming functions,
And when setting up loops.


Details:



Here's what I've noticed in your code, line-by-line:

def run(input)


Avoid common names for your functions. While it's not always possible to prevent having the same name as functions from other modules (there are methods to deal with that), naming a function run is asking for trouble. For example, a commonly used command in Python test-suites is also run.

Your function name should match as closely as possible the specific goal of the function. e.g. maybe call it insertion_sort.



for i in range(1,len(input)):


Try to use the most specific iterators possible.

The general construction for i in range(... gets used a lot where it doesn't have to be. Any time you see it, there is usually a better way iterate, whether using a list comprehension, using Itertools, or other more specific methods. In your case, replacing for i in range( with for i in input results helps a lot (as in my example below).

The next two lines j = 1 and saved = input[i] are only there to convert i from the non-specific `for I in range (...' to the values you actually want. They can be elimited by changing how the loop is set up.



    while(j &gt; 0 and input[j-1] &gt; saved):


The same goes for your while loop. You will never need a i+=1 (or j-=1) to increment a loop in Python. It's always best to think of what information the inside of the loop needs, and set up an iterator, list comprehension, or for loop that immediately provides that information.

Here is a modified version of your code, to show what I mean about being specific with loops:

def insertion_sort(input):
    for i,n in enumerate(input):
        ins_pos = 0               # default location to insert n
        for j,m in reversed([r for r in enumerate(input[:i]])):
            if m &gt; n:
                ins_pos = j + 1   # insert n one position after first larger number found
                break
        input.insert(ins_pos, input.pop(i))            
    return input


Once inside the loop, you only need:


i and j, the indices of the values to pop and insert.
n and m, the values to compare.


Once you know what the inside of the loop needs, set up the loop statements (e.g. In my example, for i,n in enumerate(input), and for m in reversed(input[i])) so they return only what is needed. This eliminates the extra lines in your code. Also, after this modification, the purpose of each loop is more clear.



    input.insert(j,input.pop(i))


PEP8 specifies spaces between arguments in function calls.



And a note on the last line:

Try not to modify a list that is the source of your iterator.

In general, it's a bad idea to modify the same list that is being iterated over. Your example only works because the pop and insert functions get applied to the parts of the input list that have already been read through by the iterator. If you need to do it in future, be aware of why this works, and how other examples of the same practice might go wrong. I've assumed that part of your spec is to modify the list in place, so I've left it how it is.

Performance:

Nice work with eliminating the extra slice steps. Re-writing the list with each step in the inner while loop is indeed more expensive than just comparing values and moving on. However, both pop, and insert within a list are both \$\mathcal{O} (n)\$, not \$\mathcal{O} (1)\$, sad but true. There may be a small speed-up of a constant factor, doing away with the slices, but both your versions of insertion sort are ultimately limited by the time complexity of the algorithm itself, which is \$\mathcal{O} (n^2)\$, where n is the length of the list.

This line:

    saved = input[i]


Won't change anything as, in Python, this sets up saved as a pointer to the value in input[i], rather than making a copy and storing it in saved. You could explicitly force a copy, but this would probably be self-defeating, as reading an index from an array is no more expensive than accessing any other int.