Reputation: 63
I want to define a recursive function can sort any list of ints:
def sort_l(l):
if l==[]:
return []
else:
if len(l)==1:
return [l[-1]]
elif l[0]<l[1]:
return [l[0]]+sort_l(l[1:])
else:
return sort_l(l[1:])+[l[0]]
Calling this function on a list [3, 1, 2,4,7,5,6,9,8] should give me:
[1,2,3,4,5,6,7,8,9]
But I get:
print(sort_l([3, 1, 2,4,7,5,6,9,8]))--> [1, 2, 4, 5, 6, 8, 9, 7, 3]
Please help me to fix the problem, actual code would be appreciated. Thanks!
Upvotes: -2
Views: 26351
Reputation: 1
Just index access, condition, and recursions:
def ordena(vector):
if len(vector)==1:
return vector
if(vector[0]>vector[1]):
vector[1], vector[0] = vector[0], vector[1]
vector = vector[0:1] + ordena(vector[1:])
return ordena(vector[:-1]) + vector[-1:]
v = [3,2,5,1,10,8,4]
print (v)
print(ordena(v))
Upvotes: 0
Reputation: 11
This is a recursive solution. For an explanation, refer to this video:
arr = [3,1,2,4,7,5,6,9,8]
def insert_fn(arr, temp): # Hypothesis
if len(arr) == 0 or arr[-1] <= temp: # Base - condition
arr.append(temp)
return arr
# Induction
val = arr[-1]
arr.pop()
insert_fn(arr, temp) # Call function on a smaller input.
arr.append(val) # Induction step
return arr
def sort_fn(arr): # Hypothesis
if len(arr) == 1: # Base - condition
return arr
# Induction
val = arr[-1]
arr.pop()
sort_fn(arr) # Call function on a smaller input.
insert_fn(arr, val) # Induction step
return arr
print(sort_fn(arr))
Upvotes: 0
Reputation: 119
This is a complementary answer since both quicksort
and complexity are already covered in previous answers. Although, I believe an easy-to-get sort
function that covers Python's identity is missing*.
def sort(xs: list) -> list:
if not xs:
return xs
else:
xs.remove(num := min(xs))
return [num] + sort(xs)
[*] Python is a (slow) interpreted language, but it has become famous because of its readability and easiness to learn. It doesn't really "reward" its developers for using immutable objects nor it is a language that should be used for computation intensive applications
Upvotes: 0
Reputation: 9
def maximum(lis):
if len(lis) == 1:
return lis[0]
return maximum(lis[1:]) if lis[0] < lis[1] else maximum(lis[:1] + lis[2:])
def sorter(lis):
if len(lis) == 1:
return lis
x = maximum(lis)
lis.remove(x)
return sorter(lis) + [x]
with functional programming:
sor = lambda lis: lis if len(lis) == 1 else [lis.pop(lis.index(reduce(lambda x, y: x if x > y else y, lis)))] + sor(lis)
Upvotes: 1
Reputation: 381
Here i am explaining recursive approach to sort a list. we can follow "Induction Base-condition Hypothesis" recursion approach. so basically we consider our hypothesis here sort_l(nums) function which sorts for given list and Base condition will be found when we have singly number list available which is already sorted. Now in induction step, we insert the temp element (last element of list) in the correct position of given list.
example-
sort_l([1,5,0,2])
will make below recursively callsort_l([1]) <-- 5
(here you need to insert 5 in correct position)sort_l([1,5]) <-- 0
(here you need to insert 0 in correct position)sort_l([0,1,5]) <-- 2
(here you need to insert 2 in correct position)sort_l([0,1,5,2])
Finally it will be in sorted list.====== Below is working code=======
def insert_element(nums, temp):
if len(nums) == 1:
if nums[0] > temp:
nums.insert(0, temp)
elif nums[0] < temp:
nums.append(temp)
else:
for i in range(len(nums)):
if nums[i] > temp:
nums.insert(i, temp)
break
if nums[-1] < temp:
nums.append(temp)
def sort_l(nums): ## hypothesis
if len(nums)==1: ## base condition
return nums
temp = nums[-1]
nums.pop()
sort_l(nums)
insert_element(nums, temp) ## induction
return nums
Upvotes: 0
Reputation: 1
#sort an int list using recursion
global array
array=[5,3,8,4,2,6,1]
def sort1(array:[])->[]:
if len(array)==1:
return
temp=array[-1]
array.pop()
sort1(array)
sort2(array,temp)
def sort2(array:[],temp):
if len(array)==0 or temp>=array[-1]:
array.append(temp)
return
a=array[-1]
array.pop()
sort2(array,temp)
array.append(a)
sort1(array)
print(array)
Upvotes: 0
Reputation: 1
def sort(array, index = 0, bigNumber = 0):
if len(array) == index:
return array
elif bigNumber > array[index]:
array[index - 1] = array[index]
array[index] = bigNumber
bigNumber = array[0]
index = 0
else:
bigNumber = array[index]
return sort(array, (index + 1), bigNumber)
Upvotes: 0
Reputation: 133
def quicksort(lst):
"Quicksort over a list-like sequence"
if len(lst) == 0:
return lst
pivot = lst[0]
pivots = [x for x in lst if x == pivot]
small = quicksort([x for x in lst if x < pivot])
large = quicksort([x for x in lst if x > pivot])
return small + pivots + large
Above is a more readable recursive implementation of Quick Sort Algorithm. Above piece of code is from book Functional programing in python by O'REILLY.
Above function will produce.
list=[9,8,7,6,5,4]
quicksort(list)
>>[4,5,6,7,8,9]
Upvotes: 0
Reputation: 1408
The quick sort is recursive and easy to implement in Python:
def quick_sort(l):
if len(l) <= 1:
return l
else:
return quick_sort([e for e in l[1:] if e <= l[0]]) + [l[0]] +\
quick_sort([e for e in l[1:] if e > l[0]])
will give:
>>> quick_sort([3, 1, 2, 4, 7, 5, 6, 9, 8])
[1, 2, 3, 4, 5, 6, 7, 8, 9]
Upvotes: 7
Reputation: 181
For this you would want to use merge sort. Essentially in a merge sort you recursively split the list in half until you have single elements and than build it back up in the correct order. merge sort on has a complexity of O(n log(n))
and is an extremely stable sorting method.
Here are some good in depth explanations and visuals for merge sorting:
Upvotes: 1