# 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]

# 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]

# 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]

# 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]