【发布时间】:2017-02-07 19:48:10
【问题描述】:
我正在尝试在 python 中实现梯度下降,我的损失/成本随着每次迭代而不断增加。
我看到一些人发布了关于此的帖子,并在此处看到了答案:gradient descent using python and numpy
我相信我的实现是相似的,但看不出我做错了什么以获得爆炸性的成本价值:
Iteration: 1 | Cost: 697361.660000
Iteration: 2 | Cost: 42325117406694536.000000
Iteration: 3 | Cost: 2582619233752172973298548736.000000
Iteration: 4 | Cost: 157587870187822131053636619678439702528.000000
Iteration: 5 | Cost: 9615794890267613993157742129590663647488278265856.000000
我正在网上找到的数据集(LA Heart Data)上对此进行测试:http://www.umass.edu/statdata/statdata/stat-corr.html
导入代码:
dataset = np.genfromtxt('heart.csv', delimiter=",")
x = dataset[:]
x = np.insert(x,0,1,axis=1) # Add 1's for bias
y = dataset[:,6]
y = np.reshape(y, (y.shape[0],1))
梯度下降:
def gradientDescent(weights, X, Y, iterations = 1000, alpha = 0.01):
theta = weights
m = Y.shape[0]
cost_history = []
for i in xrange(iterations):
residuals, cost = calculateCost(theta, X, Y)
gradient = (float(1)/m) * np.dot(residuals.T, X).T
theta = theta - (alpha * gradient)
# Store the cost for this iteration
cost_history.append(cost)
print "Iteration: %d | Cost: %f" % (i+1, cost)
计算成本:
def calculateCost(weights, X, Y):
m = Y.shape[0]
residuals = h(weights, X) - Y
squared_error = np.dot(residuals.T, residuals)
return residuals, float(1)/(2*m) * squared_error
计算假设:
def h(weights, X):
return np.dot(X, weights)
实际运行它:
gradientDescent(np.ones((x.shape[1],1)), x, y, 5)
【问题讨论】:
-
我最好的选择是琐碎的签名问题,因为它似乎走错了方向。
标签: python numpy machine-learning regression gradient-descent