Skip to content

Instantly share code, notes, and snippets.

@bee-san
Last active August 15, 2021 10:28
Show Gist options
  • Star 21 You must be signed in to star a gist
  • Fork 4 You must be signed in to fork a gist
  • Save bee-san/f6ccc000ab6527769999fd0a9ebf59de to your computer and use it in GitHub Desktop.
Save bee-san/f6ccc000ab6527769999fd0a9ebf59de to your computer and use it in GitHub Desktop.
An Python implementation of Timsort
# based off of this code https://gist.github.com/nandajavarma/a3a6b62f34e74ec4c31674934327bbd3
# Brandon Skerritt
# https://skerritt.tech
def binary_search(the_array, item, start, end):
if start == end:
if the_array[start] > item:
return start
else:
return start + 1
if start > end:
return start
mid = round((start + end)/ 2)
if the_array[mid] < item:
return binary_search(the_array, item, mid + 1, end)
elif the_array[mid] > item:
return binary_search(the_array, item, start, mid - 1)
else:
return mid
"""
Insertion sort that timsort uses if the array size is small or if
the size of the "run" is small
"""
def insertion_sort(the_array):
l = len(the_array)
for index in range(1, l):
value = the_array[index]
pos = binary_search(the_array, value, 0, index - 1)
the_array = the_array[:pos] + [value] + the_array[pos:index] + the_array[index+1:]
return the_array
def merge(left, right):
"""Takes two sorted lists and returns a single sorted list by comparing the
elements one at a time.
[1, 2, 3, 4, 5, 6]
"""
if not left:
return right
if not right:
return left
if left[0] < right[0]:
return [left[0]] + merge(left[1:], right)
return [right[0]] + merge(left, right[1:])
def timsort(the_array):
runs, sorted_runs = [], []
length = len(the_array)
new_run = [the_array[0]]
# for every i in the range of 1 to length of array
for i in range(1, length):
# if i is at the end of the list
if i == length - 1:
new_run.append(the_array[i])
runs.append(new_run)
break
# if the i'th element of the array is less than the one before it
if the_array[i] < the_array[i-1]:
# if new_run is set to None (NULL)
if not new_run:
runs.append([the_array[i]])
new_run.append(the_array[i])
else:
runs.append(new_run)
new_run = []
# else if its equal to or more than
else:
new_run.append(the_array[i])
# for every item in runs, append it using insertion sort
for item in runs:
sorted_runs.append(insertion_sort(item))
# for every run in sorted_runs, merge them
sorted_array = []
for run in sorted_runs:
sorted_array = merge(sorted_array, run)
print(sorted_array)
timsort([2, 3, 1, 5, 6, 7])
@ajnauleau
Copy link

ajnauleau commented May 10, 2019

I was having some issues when sorting where integers were being duplicated, based off of the above code you wrote.

Looking things over, I noticed that the issue was potentially coming from insertion_sort and therefore went ahead and rewrote it. Here's the code snippet for anyone interested or having similar issues:

def insertion_sort(the_array):
    l = len(the_array)
    for index in range(1, l):
        value = the_array[index]
        while index>0 and the_array[index-1]>value:
            the_array[index] = the_array[index-1]
            index = index-1
            the_array[index] = value
    return the_array

@ConorHogan
Copy link

Hi. I have made all the changes listed above, but when I run this on an array of random number of length 5000+ I get a recursion error.
It identifies line 47 as the problem.
Anybody got any ideas how to fix it?

It looks like the runs are not being allowed to grow to a large enough size so the merge element has way too much work to do.
For example, I had 2461 runs for an array of 5000 elements.

@rxdravi
Copy link

rxdravi commented May 16, 2019

@ConorHogan, use this in your timsort method:: sys.setrecursionlimit(max(sys.getrecursionlimit(), len(the_array)))

Sign up for free to join this conversation on GitHub. Already have an account? Sign in to comment