NumPy

>>> import numpy as np >>> x = np.array([1, 2, 3]) >>> x array([1, 2, 3]) >>> y = np.arange(10) # like Python's range, but returns an 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) # create an array with four equally spaced points starting with 0 and ending with 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) # solve the equation ax = b array([-4.83050847, 2.13559322, 1.18644068]) >>> c = rand(3, 3) * 20 # create a 3x3 random matrix of values within [0,1] scaled by 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]]) 

>>> import numpy as np >>> import cv2 >>> r = np.reshape(np.arange(256*256)%256,(256,256)) # 256x256 pixel array with a horizontal gradient from 0 to 255 for the red color channel >>> g = np.zeros_like(r) # array of same size and type as r but filled with 0s for the green color channel >>> b = r.T # transposed r will give a vertical gradient for the blue color channel >>> cv2.imwrite('gradients.png', np.dstack([b,g,r])) # OpenCV images are interpreted as BGR, the depth-stacked array will be written to an 8bit RGB PNG-file called '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: ', 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: ', points[minIdx] Nearest point to q: [3 4 4]