我正在尝试按照这里的课程http://cs231n.github.io/optimization-1/,在 以数字方式计算渐变 的部分中,他们提供了一个应该的代码片段计算机给出函数和数组的渐变。我尝试使用我自己的函数和numpy数组作为输入来运行它,我得到以下错误:
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我理解错误是因为它无法为一个序列指定grad [ix],我也尝试使用列数组并得到相同的错误。
以下是代码:
ValueError Traceback (most recent call last)
<ipython-input-18-31c1f1d6169c> in <module>()
2 return a
3
----> 4 eval_numerical_gradient(f,np.array([1,2,3,4,5]))
<ipython-input-12-d6bea4220895> in eval_numerical_gradient(f, x)
28 print(x[ix])
29 # compute the partial derivative
---> 30 grad[ix] = (fxh - fx) / h # the slope
31 it.iternext() # step to next dimension
32
ValueError: setting an array element with a sequence.
我的问题是 :我输入的numpy数组(行和列)错了吗?有人可以解释为什么会这样吗?
示例输入:
def eval_numerical_gradient(f, x):
"""
a naive implementation of numerical gradient of f at x
- f should be a function that takes a single argument
- x is the point (numpy array) to evaluate the gradient at
"""
fx = f(x) # evaluate function value at original point
print(x)
print(fx)
grad = np.zeros(x.shape)
h = 0.00001
# iterate over all indexes in x
it = np.nditer(x, flags=['multi_index'], op_flags=['readwrite'])
while not it.finished:
print(it)
# evaluate function at x+h
ix = it.multi_index
print(ix)
old_value = x[ix]
print(old_value)
x[ix] = old_value + h # increment by h
print(x)
fxh = f(x) # evalute f(x + h)
print(fxh)
x[ix] = old_value # restore to previous value (very important!)
print(x[ix])
# compute the partial derivative
grad[ix] = (fxh - fx) / h # the slope
it.iternext() # step to next dimension
return grad
和
def f(a):
return a
eval_numerical_gradient(f,np.array([[1],[2],[3]]))
答案 0 :(得分:2)
我建议对eval_numerical_gradient(f, x)
进行以下修复:
fxh = f(x)
替换为fxh = f(x[ix])
grad[ix] = (fxh - fx) / h
替换为grad[ix] = (fxh - fx[ix]) / h
并使用浮点数条目输入输入矩阵,例如,
eval_numerical_gradient(f,np.array([[1],[2],[3]], dtype=np.float))