I have used a similar method used in numpy lecture, but using "time.perf_counter()" instead of "datetime.now()", why? because I did not find any difference and I usually use that.
The code is shown as follows:
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import numpy as np import time A = np.array([[1,2,3], [4,5,6], [7,8,9]]) #A = np.ones((3,3)) * 3 B = np.ones((3,3)) * 2 X = np.zeros((3,3)) #Matrix dot solution sum1 = 0 sum2 = 0 med1 = 0 med2 = 0 #Manual method for loop in range(0,500): t1 = time.perf_counter() for k in range(0,2+1): for i in range(0,2+1): e = 0 for j in range(0,2+1): e = e + (A[k,j] * B[j,i]) X[k,i] = e t2 = time.perf_counter() sum1 = sum1 + (t2 - t1) med1 = sum1/50 for loop2 in range(0,500): t1 = time.perf_counter() A.dot(B) t2 = time.perf_counter()