我试图找到一个具有多个变量(500+(的Python函数的全局最大值。为此，我尝试使用JAX grad((来计算这个MyFunction的梯度函数。
但我显然做错了什么——因为每次我试图得到MyFunction的导数时，我只得到一个零列表，这毫无意义。(注意：MyFunction按预期工作(
知道吗？
from jax import grad
import jax.numpy as jnp
import numpy as np
import json
# Example data - in real case I have 150,000+ rows
data = jnp.array([[ 1. , 1.06, 9.77, 5. , 3. , 2. , 6. , 12. , 4. ,
10. , 1. , 7. , 1. , 12. , 10. , 12. , 4. , 10. ,
8. , 7. , 11. , 5. , 9. , 3. , 6. , 12. , 6. ,
5. , 3. , 5. , 9. , 8. , 9. , 10. , 11. , 12. ,
1. , 2. , 3. , 4. , 5. , 6. , 7. , 0. , 0. ,
0. , 0. , 0. , 0. , 0. , 0. , 0. , 0. , 0. ,
0. , 0. , 0. ],
[ 1. , 1.33, 3.33, 5. , 3. , 2. , 6. , 12. , 4. ,
10. , 1. , 7. , 1. , 12. , 10. , 12. , 4. , 10. ,
8. , 7. , 11. , 5. , 9. , 3. , 6. , 12. , 6. ,
5. , 3. , 5. , 9. , 8. , 9. , 10. , 11. , 12. ,
1. , 2. , 3. , 4. , 5. , 6. , 7. , 0. , 0. ,
0. , 0. , 0. , 0. , 0. , 0. , 0. , 0. , 0. ,
0. , 0. , 0. ],
[ 2. , 1.65, 2.07, 5. , 3. , 2. , 6. , 12. , 4. ,
10. , 1. , 7. , 1. , 12. , 10. , 12. , 4. , 8. ,
6. , 5. , 9. , 3. , 7. , 1. , 4. , 10. , 4. ,
3. , 1. , 3. , 7. , 10. , 11. , 12. , 1. , 2. ,
3. , 4. , 5. , 6. , 7. , 8. , 9. , 0. , 0. ,
0. , 0. , 0. , 0. , 0. , 0. , 0. , 0. , 0. ,
0. , 0. , 0. ]])
def MyFunction(coefs, data):
balance = float(len(data)*-1000)
for row in data:
result = row[0]
fOdds = row[1]
dOdds = row[2]
h1P = 0.0
h2P = 0.0
h3P = 0.0
h4P = 0.0
h5P = 0.0
h6P = 0.0
h7P = 0.0
h8P = 0.0
h9P = 0.0
h10P = 0.0
h11P = 0.0
h12P = 0.0
for p in range (0, 14):
s = int(row[3+p]-1)
h = int(row[17+p])
r = int(row[43+p])
bCoef = coefs[p]
sCoef = coefs[14 + (p * 12) + s]
hCoef = coefs[182 + (p * 12) + h]
if r == 1:
rCoef = coefs[350 + p]
else:
rCoef = 1.0
pStrength = bCoef * sCoef * hCoef * rCoef
if h == 0:
h1P += pStrength
if h == 1:
h2P += pStrength
if h == 2:
h3P += pStrength
if h == 3:
h4P += pStrength
if h == 4:
h5P += pStrength
if h == 5:
h6P += pStrength
if h == 6:
h7P += pStrength
if h == 7:
h8P += pStrength
if h == 8:
h9P += pStrength
if h == 9:
h10P += pStrength
if h == 10:
h11P += pStrength
if h == 11:
h12P += pStrength
for h in range (0, 12):
hSign = int(row[31+h]-1)
if h == 0:
h1P *= coefs [364 + (h*12) + hSign]
if h == 1:
h2P *= coefs [364 + (h*12) + hSign]
if h == 2:
h3P *= coefs [364 + (h*12) + hSign]
if h == 3:
h4P *= coefs [364 + (h*12) + hSign]
if h == 4:
h5P *= coefs [364 + (h*12) + hSign]
if h == 5:
h6P *= coefs [364 + (h*12) + hSign]
if h == 6:
h7P *= coefs [364 + (h*12) + hSign]
if h == 7:
h8P *= coefs [364 + (h*12) + hSign]
if h == 8:
h9P *= coefs [364 + (h*12) + hSign]
if h == 9:
h10P *= coefs [364 + (h*12) + hSign]
if h == 10:
h11P *= coefs [364 + (h*12) + hSign]
if h == 11:
h12P *= coefs [364 + (h*12) + hSign]
fPoints = 0.0
dPoints = 0.0
fPoints += h1P * coefs[508]
fPoints += h2P * coefs[509]
fPoints += h3P * coefs[510]
fPoints += h4P * coefs[511]
fPoints += h5P * coefs[512]
fPoints += h6P * coefs[513]
fPoints += h7P * coefs[514]
fPoints += h8P * coefs[515]
fPoints += h9P * coefs[516]
fPoints += h10P * coefs[517]
fPoints += h11P * coefs[518]
fPoints += h12P * coefs[519]
dPoints += h1P * coefs[520]
dPoints += h2P * coefs[521]
dPoints += h3P * coefs[522]
dPoints += h4P * coefs[523]
dPoints += h5P * coefs[524]
dPoints += h6P * coefs[525]
dPoints += h7P * coefs[526]
dPoints += h8P * coefs[527]
dPoints += h9P * coefs[528]
dPoints += h10P * coefs[529]
dPoints += h11P * coefs[530]
dPoints += h12P * coefs[531]
if result == 1:
if fPoints >= dPoints:
balance += fOdds*1000
elif result == 2:
if dPoints > fPoints:
balance += dOdds*1000
return balance
derivFunction = grad (MyFunction)
coefs = np.random.sample(532)
# here I just get a list of 532 zeros instead of the derivatives...
print (derivFunction(coefs, data))
coefs = np.random.sample(532)
print (derivFunction(coefs, data))
似乎得到了零梯度，因为这是正确的结果：函数在输入值处的局部梯度为零。一种方法是通过扰动系数并观察它不会改变输出值：
print(MyFunction(coefs, data))
# -3000.0
print(MyFunction(coefs + 0.1, data))
# -3000.0
print(MyFunction(coefs - 0.1, data))
# -3000.0