C.T. Matsumoto wrote:
Dave Angel wrote:
(You forgot to do a Reply-All, so your message went to just me, rather
than to me and the list )
C.T. Matsumoto wrote:
Dave Angel wrote:
C.T. Matsumoto wrote:
Hello,
This is follow up on a question I had about algorithms. In the
thread it was suggested I make my own sorting algorithm.
Here are my results.
#!/usr/bin/python
def sort_(list_):
for item1 in list_:
pos1 = list_.index(item1)
pos2 = pos1 + 1
try:
item2 = list_[pos2]
except IndexError:
pass
if item1 >= item2:
try:
list_.pop(pos2)
list_.insert(pos1, item2)
return True
except IndexError:
pass
def mysorter(list_):
while sort_(list_) is True:
sort_(list_)
I found this to be a great exercise. In doing the exercise, I got
pretty stuck. I consulted another programmer (my dad) who described
how to go about sorting. As it turned out the description he
described was the Bubble sort algorithm. Since coding the solution
I know the Bubble sort is inefficient because of repeated
iterations over the entire list. This shed light on the quick sort
algorithm which I'd like to have a go at.
Something I haven't tried is sticking in really large lists. I was
told that with really large list you break down the input list into
smaller lists. Sort each list, then go back and use the same
swapping procedure for each of the different lists. My question is,
at what point to you start breaking things up? Is that based on
list elements or is it based on memory(?) resources python is using?
One thing I'm not pleased about is the while loop and I'd like to
replace it with a for loop.
Thanks,
T
There are lots of references on the web about Quicksort, including a
video at:
http://www.youtube.com/watch?v=y_G9BkAm6B8
which I think illustrates it pretty well. It would be a great
learning exercise to implement Python code directly from that
description, without using the sample C++ code available.
(Incidentally, there are lots of variants of Quicksort, so I'm not
going to quibble about whether this is the "right" one to be called
that.)
I don't know what your earlier thread was, since you don't mention
the subject line, but there are a number of possible reasons you
might not have wanted to use the built-in sort. The best one is for
educational purposes. I've done my own sort for various reasons in
the past, even though I had a library function, since the library
function had some limits. One time I recall, the situation was that
the library sort was limited to 64k of total data, and I had to work
with much larger arrays (this was in 16bit C++, in "large" model).
I solved the size problem by using the C++ sort library on 16k
subsets (because a pointer was 2*2 bytes). Then I merged the
results of the sorts. At the time, and in the circumstances
involved, there were seldom more than a dozen or so sublists to
merge, so this approach worked well enough.
Generally, it's better for both your development time and the
efficiency and reliabilty of the end code, to base a new sort
mechanism on the existing one. In my case above, I was replacing
what amounted to an insertion sort, and achieved a 50* improvement
for a real customer. It was fast enough that other factors
completely dominated his running time.
But for learning purposes? Great plan. So now I'll respond to your
other questions, and comment on your present algorithm.
It would be useful to understand about algorithmic complexity, the
so called Order Function. In a bubble sort, if you double the size
of the array, you quadruple the number of comparisons and swaps.
It's order N-squared or O(n*n). So what works well for an array of
size 10 might take a very long time for an array of size 10000 (like
a million times as long). You can do much better by sorting smaller
lists, and then combining them together. Such an algorithm can be
O(n*log(n)).
You ask at what point you consider sublists? In a language like C,
the answer is when the list is size 3 or more. For anything larger
than 2, you divide into sublists, and work on them.
Now, if I may comment on your code. You're modifying a list while
you're iterating through it in a for loop. In the most general
case, that's undefined. I think it's safe in this case, but I would
avoid it anyway, by just using xrange(len(list_)-1) to iterate
through it. You use the index function to find something you would
already know -- the index function is slow. And the first
try/except isn't needed if you use a -1 in the xrange argument, as I
do above.
You use pop() and push() to exchange two adjacent items in the
list. Both operations copy the remainder of the list, so they're
rather slow. Since you're exchanging two items in the list, you can
simply do that:
list[pos1], list[pos2] = list[pos2], list[pos1]
That also eliminates the need for the second try/except.
You mention being bothered by the while loop. You could replace it
with a simple for loop with xrange(len(list_)), since you know that
N passes will always be enough. But if the list is partially
sorted, your present scheme will end sooner. And if it's fully
sorted, it'll only take one pass over the data.
There are many refinements you could do. For example, you don't
have to stop the inner loop after the first swap. You could finish
the buffer, swapping any other pairs that are out of order. You'd
then be saving a flag indicating if you did any swaps. You could
keep a index pointing to the last pair you swapped on the previous
pass, and use that for a limit next time. Then you just terminate
the outer loop when that limit value is 1. You could even keep two
limit values, and bubble back and forth between them, as they
gradually close into the median of the list. You quit when they
collide in the middle.
The resultant function should be much faster for medium-sized lists,
but it still will slow down quadratically as the list size
increases. You still need to divide and conquer, and quicksort is
just one way of doing that.
DaveA
Thanks a lot Dave,
Sorry the original thread is called 'Python and algorithms'.
Yes, I think it's best to use what python provides and build on top
of that. I got to asking my original question based on trying to
learn more about algorithms in general, through python. Of late many
people have been asking me how well I can 'build' algorithms, and
this prompted me to start the thread. This is for learning purposes
(which the original thread will give you and indication where I'm
coming from).
The refactored code looks like this. I have tackled a couple items.
First the sub-listing (which I'll wait till I can get the full sort
working), then the last couple of paragraphs about refinements.
Starting with the first refinement, I'm not sure how *not* to stop
the inner loop?
def s2(list_):
for pos1 in xrange(len(list_)-1):
item1 = list_[pos1]
pos2 = pos1 + 1
item2 = list_[pos2]
if item1 >= item2:
list_[pos1], list_[pos2] = list_[pos2], list_[pos1]
return True
def mysorter(list_):
# This is the outer loop?
while s2(list_) is True:
# Calling s2 kicks off the inner loop?
s2(list_)
if __name__ == '__main__':
from random import shuffle
foo = range(10)
shuffle(foo)
mysorter(foo)
Thanks again.
As before, I'm not actually trying this code, so there may be typos.
But assuming your code here works, the next refinement would be:
In s2() function, add a flag variable, initially False. Then instead
of the return True, just say flag=True
Then at the end of the function, return flag
About the while loop. No need to say 'is True' just use while
s2(list_): And no need to call s2() a second time.
while s2(list_):
pass
Okay up to here I follow. This all makes sense.
def s2(list_):
flag = False
for pos1 in xrange(len(list_)-1):
item1 = list_[pos1]
pos2 = pos1 + 1
item2 = list_[pos2]
if item1 >= item2:
list_[pos1], list_[pos2] = list_[pos2], list_[pos1]
flag = True
return flag
def mysorter(list_):
while s2(list_):
pass
Before you can refine the upper limit, you need a way to preserve it
between calls. Simplest way to do that is to combine the two
functions, as a nested loop. Then, instead of flag, you can have a
value "limit" which indicates what index was last swapped. And the
inner loop uses that as an upper limit on its xrange.
Where I start to get confused is refining the 'upper limit'. What is the
upper limit defining? I'm guessing it is the last position processed.
T
DaveA
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Take out the = in the following line
if item1 >= item2:
and it will sort like items together, which I think is what you
originally wanted.
Also change:
> return flag
to:
> return flag
So that False gets returned if you don't make a swap.
This worked for me. Thank you for the interesting thread!
--
Jeff
Jeff Johnson
j...@dcsoftware.com
Phoenix Python User Group - sunpigg...@googlegroups.com
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