Quicksort with Python
I am totally new to python and I am trying to implement quicksort in it. Could someone please help me complete my code?
I do not know how to concatenate the three arrays and printing them.
def sort(array=[12,4,5,6,7,3,1,15]):
less = []
equal = []
greater = []
if len(array) > 1:
pivot = array[0]
for x in array:
if x < pivot:
less.append(x)
if x == pivot:
equal.append(x)
if x > pivot:
greater.append(x)
sort(less)
sort(pivot)
sort(greater)
Solution 1:
def sort(array=[12,4,5,6,7,3,1,15]):
"""Sort the array by using quicksort."""
less = []
equal = []
greater = []
if len(array) > 1:
pivot = array[0]
for x in array:
if x < pivot:
less.append(x)
elif x == pivot:
equal.append(x)
elif x > pivot:
greater.append(x)
# Don't forget to return something!
return sort(less)+equal+sort(greater) # Just use the + operator to join lists
# Note that you want equal ^^^^^ not pivot
else: # You need to handle the part at the end of the recursion - when you only have one element in your array, just return the array.
return array
Solution 2:
Quick sort without additional memory (in place)
Usage:
array = [97, 200, 100, 101, 211, 107]
quicksort(array)
print(array)
# array -> [97, 100, 101, 107, 200, 211]
def partition(array, begin, end):
pivot = begin
for i in range(begin+1, end+1):
if array[i] <= array[begin]:
pivot += 1
array[i], array[pivot] = array[pivot], array[i]
array[pivot], array[begin] = array[begin], array[pivot]
return pivot
def quicksort(array, begin=0, end=None):
if end is None:
end = len(array) - 1
def _quicksort(array, begin, end):
if begin >= end:
return
pivot = partition(array, begin, end)
_quicksort(array, begin, pivot-1)
_quicksort(array, pivot+1, end)
return _quicksort(array, begin, end)
Solution 3:
There is another concise and beautiful version
def qsort(arr):
if len(arr) <= 1:
return arr
else:
return qsort([x for x in arr[1:] if x < arr[0]])
+ [arr[0]]
+ qsort([x for x in arr[1:] if x >= arr[0]])
Let me explain the above codes for details
-
pick the first element of array
arr[0]
as pivot[arr[0]]
-
qsort
those elements of array which are less than pivot withList Comprehension
qsort([x for x in arr[1:] if x < arr[0]])
-
qsort
those elements of array which are larger than pivot withList Comprehension
qsort([x for x in arr[1:] if x >= arr[0]])
Solution 4:
This answer is an in-place QuickSort for Python 2.x
. My answer is an interpretation of the in-place solution from Rosetta Code which works for Python 3
too:
import random
def qsort(xs, fst, lst):
'''
Sort the range xs[fst, lst] in-place with vanilla QuickSort
:param xs: the list of numbers to sort
:param fst: the first index from xs to begin sorting from,
must be in the range [0, len(xs))
:param lst: the last index from xs to stop sorting at
must be in the range [fst, len(xs))
:return: nothing, the side effect is that xs[fst, lst] is sorted
'''
if fst >= lst:
return
i, j = fst, lst
pivot = xs[random.randint(fst, lst)]
while i <= j:
while xs[i] < pivot:
i += 1
while xs[j] > pivot:
j -= 1
if i <= j:
xs[i], xs[j] = xs[j], xs[i]
i, j = i + 1, j - 1
qsort(xs, fst, j)
qsort(xs, i, lst)
And if you are willing to forgo the in-place property, below is yet another version which better illustrates the basic ideas behind quicksort. Apart from readability, its other advantage is that it is stable (equal elements appear in the sorted list in the same order that they used to have in the unsorted list). This stability property does not hold with the less memory-hungry in-place implementation presented above.
def qsort(xs):
if not xs: return xs # empty sequence case
pivot = xs[random.choice(range(0, len(xs)))]
head = qsort([x for x in xs if x < pivot])
tail = qsort([x for x in xs if x > pivot])
return head + [x for x in xs if x == pivot] + tail