>>> import numpy as np>>> x = np.array([1, 2, 3])>>> xarray([1, 2, 3])>>> y = np.arange(10) # seperti fungsi Python list(range(10)), tetapi menghasilkan array>>> yarray([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>>> carray([ 1. , 1.33333333, 1.66666667, 4. ])>>> a**2array([ 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 = barray([-4.83050847, 2.13559322, 1.18644068])>>> c = rand(3, 3) * 20 # buat matriks nilai acak 3x3 dalam [0,1] dengan skala 20>>> carray([[ 3.98732789, 2.47702609, 4.71167924], [ 9.24410671, 5.5240412 , 10.6468792 ], [ 10.38136661, 8.44968437, 15.17639591]])>>> np.dot(a, c) # matrix multiplicationarray([[ 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 1.10array([[ 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]