NumPy

>>> import numpy as np
>>> x = np.array([1, 2, 3])
>>> x
array([1, 2, 3])
>>> y = np.arange(10)  # seperti fungsi Python list(range(10)), tetapi menghasilkan array
>>> y
array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9])

>>> a = np.array([1, 2, 3, 6])
>>> b = np.linspace(0, 2, 4)  # buat array dengan empat titik dengan jarak yang sama dimulai dengan 0 dan diakhiri dengan 2.
>>> c = a - b
>>> c
array([ 1.        ,  1.33333333,  1.66666667,  4.        ])
>>> a**2
array([ 1,  4,  9, 36])

>>> a = np.linspace(-np.pi, np.pi, 100) 
>>> b = np.sin(a)
>>> c = np.cos(a)

>>> from numpy.random import rand
>>> from numpy.linalg import solve, inv
>>> a = np.array([[1, 2, 3], [3, 4, 6.7], [5, 9.0, 5]])
>>> a.transpose()
array([[ 1. ,  3. ,  5. ],
       [ 2. ,  4. ,  9. ],
       [ 3. ,  6.7,  5. ]])
>>> inv(a)
array([[-2.27683616,  0.96045198,  0.07909605],
       [ 1.04519774, -0.56497175,  0.1299435 ],
       [ 0.39548023,  0.05649718, -0.11299435]])
>>> b =  np.array([3, 2, 1])
>>> solve(a, b)  # menyelesaikan persamaan ax = b
array([-4.83050847,  2.13559322,  1.18644068])
>>> c = rand(3, 3) * 20  # buat matriks nilai acak 3x3 dalam [0,1] dengan skala 20
>>> c
array([[  3.98732789,   2.47702609,   4.71167924],
       [  9.24410671,   5.5240412 ,  10.6468792 ],
       [ 10.38136661,   8.44968437,  15.17639591]])
>>> np.dot(a, c)  # matrix multiplication
array([[  53.61964114,   38.8741616 ,   71.53462537],
       [ 118.4935668 ,   86.14012835,  158.40440712],
       [ 155.04043289,  104.3499231 ,  195.26228855]])
>>> a @ c # Starting with Python 3.5 and NumPy 1.10
array([[  53.61964114,   38.8741616 ,   71.53462537],
       [ 118.4935668 ,   86.14012835,  158.40440712],
       [ 155.04043289,  104.3499231 ,  195.26228855]])

>>> M = np.zeros(shape=(2, 3, 5, 7, 11))
>>> T = np.transpose(M, (4, 2, 1, 3, 0))
>>> T.shape
(11, 5, 3, 7, 2)

>>> import numpy as np
>>> import cv2
>>> r = np.reshape(np.arange(256*256)%256,(256,256))  # array 256x256 piksel dengan gradien horizontalfrom 0 to 255 for the red color channel
>>> g = np.zeros_like(r)  # array dengan ukuran dan tipe yang sama dengan r tetapi diisi dengan 0 untuk kanal warna hijau
>>> b = r.T # transposed r will give a vertical gradient for the blue color channel
>>> cv2.imwrite('gradients.png', np.dstack([b,g,r]))  # Gambar OpenCV ditafsirkan sebagai BGR, array tumpuk kedalaman akan ditulis ke file PNG RGB 8bit yang dinamai 'gradients.png'
True

>>> # # # Pure iterative Python # # #
>>> points = [[9,2,8],[4,7,2],[3,4,4],[5,6,9],[5,0,7],[8,2,7],[0,3,2],[7,3,0],[6,1,1],[2,9,6]]
>>> qPoint = [4,5,3]
>>> minIdx = -1
>>> minDist = -1
>>> for idx, point in enumerate(points):  # iterate over all points
...     dist = sum([(dp-dq)**2 for dp,dq in zip(point,qPoint)])**0.5  # compute the euclidean distance for each point to q
...     if dist < minDist or minDist < 0:  # if necessary, update minimum distance and index of the corresponding point
...         minDist = dist
...         minIdx = idx

>>> print('Nearest point to q: {0}'.format(points[minIdx]))
Nearest point to q: [3, 4, 4]

>>> # # # Equivalent NumPy vectorization # # #
>>> import numpy as np
>>> points = np.array([[9,2,8],[4,7,2],[3,4,4],[5,6,9],[5,0,7],[8,2,7],[0,3,2],[7,3,0],[6,1,1],[2,9,6]])
>>> qPoint = np.array([4,5,3])
>>> minIdx = np.argmin(np.linalg.norm(points-qPoint,axis=1))  # compute all euclidean distances at once and return the index of the smallest one
>>> print('Nearest point to q: {0}'.format(points[minIdx]))
Nearest point to q: [3 4 4]