# How to use matrix_to_string method in autotest

Best Python code snippet using autotest_python

Karten_Seth_HW3_2.py

Source:Karten_Seth_HW3_2.py

`...7def cosd(x):8 return cos(x*pi/180)9def sind(x):10 return sin(x*pi/180)11def matrix_to_string(m):12 s = ""13 for i in range(len(m)-1):14 for j in range(m.shape[1]):15 if (j != m.shape[1]-1):16 s += str(m[i,j]) + "\t"17 else:18 s += str(m[i,j])19 if i != len(m) - 2:20 s += "\n"21 return s22def main():23 try:24 theta1 = float(input("Enter first angle value: "))25 theta2 = float(input("Enter second angle value: "))26 theta3 = float(input("Enter third angle value: "))27 except:28 print("Bad user input")29 exit(0)30 L1 = 1031 L2 = 1032 L3 = 333 T1 = np.matrix([[cosd(theta1), -sind(theta1), 0, L1*cosd(theta1)],34 [sind(theta1), cosd(theta1), 0, L1*sind(theta1)],35 [0, 0, 1, 0],36 [0, 0, 0, 1]])37 T2 = np.matrix([[cosd(theta2), -sind(theta2), 0, L2*cosd(theta2)],38 [sind(theta2), cosd(theta2), 0, L2*sind(theta2)],39 [0, 0, 1, 0],40 [0, 0, 0, 1]])41 T3 = np.matrix([[cosd(theta3), -sind(theta3), 0, L3*cosd(theta3)],42 [sind(theta3), cosd(theta3), 0, L3*sind(theta3)],43 [0, 0, 1, 0],44 [0, 0, 0, 1]])45 print("0T1:\n", matrix_to_string(T1))46 print("1T2:\n", matrix_to_string(T2))47 print("2T3:\n", matrix_to_string(T3))48 T03 = T1 * T2 * T349 print("0T3:\n", matrix_to_string(T03))50 print("The location of the wrist with respect to the reference coordinate frame is...\nx = %f\ty = %f" % (T03[0,3], T03[1,3]))51 return52if __name__ == "__main__":...`

day11.py

Source:day11.py

`1import sys2import copy3import itertools4lines = open("%s_input.txt" % sys.argv[0].split('.')[0], "r")5def matrix_to_string(m):6 return '|'.join(list(map(lambda x: ''.join(x), m)))7def adjacent_seats(m, i, j):8 return sum(map(lambda x: (x[max(0, j - 1):min(j + 1, len(x) - 1) + 1]), m[max(0, i - 1):min(i + 1, len(m) - 1) + 1]), list())9def first_seen(m, i, j):10 a = []11 if m[i][j] == '.':12 return a 13 for k, l in itertools.product(range(-1,2), range(-1,2)):14 n = 115 while 0 <= i + n*k < len(m) and 0 <= j + n*l < len(m[0]) and m[i + n*k][j + n*l] == ".":16 n += 117 if (0 <= i + n*k < len(m) and 0 <= j + n*l < len(m[0])):18 a.append(m[i + n*k][j + n*l]) 19 return a20def apply_round(m, func, tolerance):21 n = [[''] * len(m[0]) for _ in range(len(m))]22 23 for i in range(len(m)):24 for j in range(len(m[i])):25 a = func(m, i, j)26 if m[i][j] == "L" and "#" not in a:27 n[i][j] = "#"28 elif m[i][j] == "#" and a.count('#') > tolerance:29 n[i][j] = "L"30 else:31 n[i][j] = m[i][j]32 return n33# init34m1 = []35for line in lines:36 m1.append(list(line.rstrip()))37m2 = copy.deepcopy(m1) # Copy pour ex238# ex 139m_str = ''40while matrix_to_string(m1) != m_str:41 m_str = matrix_to_string(m1)42 m1 = apply_round(m1, adjacent_seats, 4)43print("ex1: %d" % sum(m1, list()).count('#'))44# ex 245m_str = ''46while matrix_to_string(m2) != m_str:47 m_str = matrix_to_string(m2)48 m2 = apply_round(m2, first_seen, 5)...`

Hillsypher.py

Source:Hillsypher.py

`...29 l += 130 ans.append(sum_ % 27)31 return ans32# print(matrix_mul([[2],[0],[17]]))33def matrix_to_string(m):34 b=list('abcdefghijklmnopqrstuvwxyz ')35 st=''36 for i in m:37 st+=b[i]38 return st39#print(matrix_to_string([1,26,24]))40def endcoder(s):41 ans=''42 l1=string_to_matrix(s)43 for i in l1:44 ans+=matrix_to_string(matrix_mul(i)).upper()45 return ans...`

## Automation Testing Tutorials

Learn to execute automation testing from scratch with LambdaTest Learning Hub. Right from setting up the prerequisites to run your first automation test, to following best practices and diving deeper into advanced test scenarios. LambdaTest Learning Hubs compile a list of step-by-step guides to help you be proficient with different test automation frameworks i.e. Selenium, Cypress, TestNG etc.