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Variations in different Sorting techniques in Python

Last Updated : 17 Nov, 2020
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These are all different types for sorting techniques that behave very differently. Let's study which technique works how and which one to use. Let 'a' be a numpy array
  • a.sort() (i) Sorts the array in-place & returns None (ii) Return type is None (iii) Occupies less space. No copy created as it directly sorts the original array (iv) Faster than sorted(a) Python3 
    # Python code to sort an array in-place
# using a.sort
import numpy as np

# Numpy array created
a = np.array([9, 3, 1, 7, 4, 3, 6])

# unsorted array print
print('Original array:\n', a)

# Return type is None
print('Return type:', a.sort())

# Sorted array output
print('Original array sorted->', a)

    OUTPUT: For a.sort()
Original array:
 [9 3 1 7 4 3 6]
Return type: None
Original array sorted-> [1 3 3 4 6 7 9]
  • sorted(a) (i) Creates a new list from the old & returns the new one, sorted (ii) Return type is a list (iii) Occupies more space as copy of original array is created and then sorting is done (iv) Slower than a.sort() Python3 
    # Python code to create a sorted copy using
# sorted()
import numpy as np

# Numpy array created
a = np.array([9, 3, 1, 7, 4, 3, 6])

# unsorted array print
print('Original array:\n', a)
b = sorted(a)

# sorted list returned to b, b type is
# <class 'list'> 
print('New array sorted->', b)

# original array no change
print('Original array->', a)

    OUTPUT:a.sorted()
Original array:
 [9 3 1 7 4 3 6]
New array sorted-> [1, 3, 3, 4, 6, 7, 9]
Original array-> [9 3 1 7 4 3 6]
  • np.argsort(a) (i) Returns the indices that would sort an array (ii) Return type is numpy array (iii) Occupies space as a new array of sorted indices is returned Python3 
    # Python code to demonstrate working of np.argsort
import numpy as np

# Numpy array created
a = np.array([9, 3, 1, 7, 4, 3, 6])

# unsorted array print
print('Original array:\n', a)

# Sort array indices
b = np.argsort(a)
print('Sorted indices of original array->', b)

# To get sorted array using sorted indices
# c is temp array created of same len as of b
c = np.zeros(len(b), dtype = int)
for i in range(0, len(b)):
    c[i]= a[b[i]]
print('Sorted array->', c)

    OUTPUT:np.argsort(a)
Original array:
 [9 3 1 7 4 3 6]
Sorted indices of original array-> [2 1 5 4 6 3 0]
Sorted array-> [1 3 3 4 6 7 9]
  • np.lexsort((b, a)) (i) Perform an indirect sort using a sequence of keys (ii) Sort by a, then by b (iii) Return type ndarray of ints Array of indices that sort the keys along the specified axis (iv) Occupies space as a new array of sorted indices pair wise is returned. Python3 
    # Python code to demonstrate working of 
# np.lexsort()
import numpy as np

# Numpy array created
a = np.array([9, 3, 1, 3, 4, 3, 6]) # First column
b = np.array([4, 6, 9, 2, 1, 8, 7]) # Second column
print('column a, column b')
for (i, j) in zip(a, b):
    print(i, ' ', j)

ind = np.lexsort((b, a)) # Sort by a then by b
print('Sorted indices->', ind)

    OUTPUT:np.lexsort((b, a))
column a, column b
9   4
3   6
1   9
3   2
4   1
3   8
6   7
Sorted indices-> [2 3 1 5 4 6 0]

S

Shaurya Uppal
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