好的......因爲沒有人給一個答案,我結束了我自己寫它......這裏的Python代碼:
import numpy as np
# Max-Min Composition given by Zadeh
def maxMin(x, y):
z = []
for x1 in x:
for y1 in y.T:
z.append(max(np.minimum(x1, y1)))
return np.array(z).reshape((x.shape[0], y.shape[1]))
# Max-Product Composition given by Rosenfeld
def maxProduct(x, y):
z = []
for x1 in x:
for y1 in y.T:
z.append(max(np.multiply(x1, y1)))
return np.array(z).reshape((x.shape[0], y.shape[1]))
# 3 arrays for the example
r1 = np.array([[1, 0, .7], [.3, .2, 0], [0, .5, 1]])
r2 = np.array([[.6, .6, 0], [0, .6, .1], [0, .1, 0]])
r3 = np.array([[1, 0, .7], [0, 1, 0], [.7, 0, 1]])
print "R1oR2 => Max-Min :\n" + str(maxMin(r1, r2)) + "\n"
print "R1oR2 => Max-Product :\n" + str(maxProduct(r1, r2)) + "\n\n"
print "R1oR3 => Max-Min :\n" + str(maxMin(r1, r3)) + "\n"
print "R1oR3 => Max-Product :\n" + str(maxProduct(r1, r3)) + "\n\n"
print "R1oR2oR3 => Max-Min :\n" + str(maxMin(r1, maxMin(r2, r3))) + "\n"
print "R1oR2oR3 => Max-Product :\n" + str(maxProduct(r1, maxProduct(r2, r3))) + "\n\n"
,這裏是它給出了答案:
R1oR2 => Max-Min :
[[ 0.6 0.6 0. ]
[ 0.3 0.3 0.1]
[ 0. 0.5 0.1]]
R1oR2 => Max-Product :
[[ 0.6 0.6 0. ]
[ 0.18 0.18 0.02]
[ 0. 0.3 0.05]]
R1oR3 => Max-Min :
[[ 1. 0. 0.7]
[ 0.3 0.2 0.3]
[ 0.7 0.5 1. ]]
R1oR3 => Max-Product :
[[ 1. 0. 0.7 ]
[ 0.3 0.2 0.21]
[ 0.7 0.5 1. ]]
R1oR2oR3 => Max-Min :
[[ 0.6 0.6 0.6]
[ 0.3 0.3 0.3]
[ 0.1 0.5 0.1]]
R1oR2oR3 => Max-Product :
[[ 0.6 0.6 0.42 ]
[ 0.18 0.18 0.126]
[ 0.035 0.3 0.05 ]]