implement selection sort in python? [duplicate]

Question

This may seem like a simple question but when I attempted to implement selection sort in Python, I do not get a sorted list. Is there something wrong with my implementation?

def selectionSort (B, annotate=True):
    for i in range(len(A)):
    for j in range(1,len(A)):
        if(A[i] > A [j]):
            A[i], A[j] = A[j], A[i]



A = [5, 4, 3, 2, 1]
A_sorted = selectionSort (A)
print ("Sorted " + str(A) + " = " + str(A_sorted))

A = [10, 7, 8, 40, 2, 5]
A_sorted = selectionSort (A)
print ("Sorted " + str(A) + " = " + str(A_sorted))

Here's what I get:

>>> (executing lines 1 to 74 of "selection_sort_103_v2.py")

Sorted [1, 5, 4, 3, 2] = None

Sorted [2, 40, 10, 8, 7, 5] = None

you don't return anything from your function. Besides you don't use parameter B: all the work is done on global A — Jean-François Fabre
– Jean-François Fabre ♦, Commented Oct 10, 2016 at 20:39

Priyom · Accepted Answer · 2016-10-10 21:01:51Z

0

Could try this. Your index for j is fixed from 1 too which needs to vary according to i.

def selectionSort (A, annotate=True):
    for i in range(len(A)):
      for j in range(i+1,len(A)):
          if(A[i] > A [j]):
              A[i], A[j] = A[j], A[i]

    return A

Checking for values:

A = [5, 4, 3, 2, 1]
print selectionSort(A)
>>[1, 2, 3, 4, 5]

Also, if you want to print both the old and new arrays might want to save the old array by referencing it.

A = [5, 4, 3, 2, 1]
Arr=A[:]
A_sorted=selectionSort(A)
print ("Sorted " + str(Arr) + " = " + str(A_sorted))

answered Oct 10, 2016 at 21:01

Priyom

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