The maximum stress that a bar will withstand before failing is called • Rapture Strength • Yield Strength • Tensile Strength • B
1 answer:
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Answer:
import numpy as np
import time
def matrixMul(m1,m2):
if m1.shape[1] == m2.shape[0]:
t1 = time.time()
r1 = np.zeros((m1.shape[0],m2.shape[1]))
for i in range(m1.shape[0]):
for j in range(m2.shape[1]):
r1[i,j] = (m1[i]*m2.transpose()[j]).sum()
t2 = time.time()
print("Native implementation: ",r1)
print("Time: ",t2-t1)
t1 = time.time()
r2 = m1.dot(m2)
t2 = time.time()
print("\nEfficient implementation: ",r2)
print("Time: ",t2-t1)
else:
print("Wrong dimensions!")
Explanation:
We define a function (matrixMul) that receive two arrays representing the two matrices to be multiplied, then we verify is the dimensions are appropriated for matrix multiplication if so we proceed with the native implementation consisting of two for-loops and prints the result of the operation and the execution time, then we proceed with the efficient implementation using .dot method then we return the result with the operation time. As you can see from the image the execution time is appreciable just for large matrices, in such a case the execution time of the efficient implementation can be 1000 times faster than the native implementation.
