Python: sort list of numbers without built-in sorting, min, max function

Question

If I have a list that changes in length each time, and I want to sort it from lowest to highest, how would I do it?

If I have: [-5, -23, 5, 0, 23, -6, 23, 67]

I want: [-23, -6, -5, 0, 5, 23, 23, 67]

I'll start with this:

 data_list = [-5, -23, 5, 0, 23, -6, 23, 67] new_list = [] minimum = data_list[0] # arbitrary number in list for x in data_list: if x < minimum: minimum = value new_list.append(i) 

BUT this only goes through once, and I get:

 new_list = [-23] 

This is where I am stuck.

How can I continue the cycle until len(new_list) = len(data_list) (i.e. all numbers will not be in the new list) with everything sorted without using the built-in functions max, min, sort? I am not sure whether to create a new list.

Answer 1

I think you are trying to do something like this:

 data_list = [-5, -23, 5, 0, 23, -6, 23, 67] new_list = [] while data_list: minimum = data_list[0] # arbitrary number in list for x in data_list: if x < minimum: minimum = x new_list.append(minimum) data_list.remove(minimum) print new_list 

Answer 2

Here is what I am trying to do. (Insertion sort is not the best way to sort, but it does the job)

 def sort(list): for index in range(1,len(list)): value = list[index] i = index-1 while i>=0: if value < list[i]: list[i+1] = list[i] list[i] = value i -= 1 else: break 

Answer 3

 l = [64, 25, 12, 22, 11, 1,2,44,3,122, 23, 34] for i in range(len(l)): for j in range(i + 1, len(l)): if l[i] > l[j]: l[i], l[j] = l[j], l[i] print l 

Output:

 [1, 2, 3, 11, 12, 22, 23, 25, 34, 44, 64, 122] 

Answer 4

This strictly follows your requirements not to use sort() , min() , max() , but also uses Python best practices without inventing the wheel.

 data_list = [-5, -23, 5, 0, 23, -6, 23, 67] import heapq heapq.heapify(data_list) new_list = [] while data_list: new_list.append(heapq.heappop(data_list))) 

I suggest looking in the Python library for heapq.py to find out how it works. Heapsort is a pretty fun sorting algorithm as it allows you to sort an infinite stream, i.e. You can quickly get the smallest item, but also effectively add new items to the sorted data.

Answer 5

 def bubble_sort(seq): """Inefficiently sort the mutable sequence (list) in place. seq MUST BE A MUTABLE SEQUENCE. As with list.sort() and random.shuffle this does NOT return """ changed = True while changed: changed = False for i in xrange(len(seq) - 1): if seq[i] > seq[i+1]: seq[i], seq[i+1] = seq[i+1], seq[i] changed = True return None if __name__ == "__main__": """Sample usage and simple test suite""" from random import shuffle testset = range(100) testcase = testset[:] # make a copy shuffle(testcase) assert testcase != testset # we've shuffled it bubble_sort(testcase) assert testcase == testset # we've unshuffled it back into a copy 

From: http://rosettacode.org/wiki/Bubble_Sort#Python

Answer 6

try sorting the list, char have ascii code, the same can be used to sort char list.

 aw=[1,2,2,1,1,3,5,342,345,56,2,35,436,6,576,54,76,47,658,8758,87,878] for i in range(aw.__len__()): for j in range(aw.__len__()): if aw[i] < aw[j] :aw[i],aw[j]=aw[j],aw[i] 

Answer 7

 data = [3, 1, 5, 2, 4] n = len(data) for i in range(n): for j in range(1,n): if data[j-1] > data[j]: (data[j-1], data[j]) = (data[j], data[j-1]) print(data) 

Answer 8

Here is a not very efficient sorting algorithm :)

 >>> data_list = [-5, -23, 5, 0, 23, -6, 23, 67] >>> from itertools import permutations >>> for p in permutations(data_list): ... if all(i<=j for i,j in zip(p,p[1:])): ... print p ... break ... (-23, -6, -5, 0, 5, 23, 23, 67) 

Answer 9

My method is

 s = [-5, -23, 5, 0, 23, -6, 23, 67] nl = [] for i in range(len(s)): a = min(s) nl.append(a) s.remove(a) print nl 

Answer 10

Decision

 mylist = [1, 6, 7, 8, 1, 10, 15, 9] print(mylist) n = len(mylist) for i in range(n): for j in range(1, ni): if mylist[j-1] > mylist[j]: (mylist[j-1], mylist[j]) = (mylist[j], mylist[j-1]) print(mylist) 

Answer 11

Here is a more readable example of the Insert Sort algorithm.

 a = [3, 1, 5, 2, 4] for i in a[1:]: j = a.index(i) while j > 0 and a[j-1] > a[j]: a[j], a[j-1] = a[j-1], a[j] j = j - 1 

Answer 12

 def my_sort(lst): a = [] for i in range(len(lst)): a.append(min(lst)) lst.remove(min(lst)) return a def my_revers_sort(lst):#in revers!!!!! a = [] for i in range(len(lst)): a.append(max(lst)) lst.remove(max(lst)) return a 

Answer 13

You can do this easily with the min () function

 'def asc(a): b=[] l=len(a) for i in range(l): x=min(a) b.append(x) a.remove(x) return b print asc([2,5,8,7,44,54,23])' 

Answer 14

 # Your current setup data_list = [-5, -23, 5, 0, 23, -6, 23, 67] new_list = [] # Sort function for i in data_list: new_list = [ x for x in new_list if i > x ] + [i] + [ x for x in new_list if i <= x ] 

Answer 15

 n = int(input("Input list lenght: ")) lista = [] for i in range (1,n+1): print ("A[",i,"]=") ele = int(input()) lista.append(ele) print("The list is: ",lista) invers = True while invers == True: invers = False for i in range (n-1): if lista[i]>lista[i+1]: c=lista[i+1] lista[i+1]=lista[i] lista[i]=c invers = True print("The sorted list is: ",lista) 

Answer 16

How can I continue the loop until len (new_list) = len (data_list)

 while len(new_list) != len(data_list): # ... 

Maybe so?

And no, there is no need to create a new list; most sorting algorithms work by modifying the list in place.

What you're probably trying to do is Sort sort with a separate list. See the Wikipedia article for more information on this sorting algorithm, and you will see how it works with a single list and how efficient it is (spoiler: it isnt).

Answer 17

 def getIndexOfMaximum(list1): index = 0 emptyList = [] value = list1[0] c = 0 while (c == 0): for cell in list1: index += 1 if (cell >= value): value = cell hold = index -1 if (len(list1) == index): emptyList += [value] del list1[hold] index = 0 value = 0 if (len(list1) == 1): newList = emptyList + list1 del list1[index] c = 1 return newList print(getIndexOfMaximum([2,5,8,7,44,54,23])) #TRY THIS!!!