Question

在Octave成功解决问题后，我试图将Octave代码转换为Python。

这是我的代码，它没有使用fmin_cg。我看到人们在Python中使用各种各样的最小化器，但我会坚持使用相同名称的那个。

代码：

# Not yet working
#
#
# Reimplementation of Octave code
#
# 
# X = [ones(m, 1) X]
#
# [theta] = trainLinearReg([ones(m, 1) X], y, lambda);
#
# initial_theta = zeros(size(X, 2), 1);
# costFunction = @(t) linearRegCostFunction(X, y, t, lambda);
# options = optimset('MaxIter', 200, 'GradObj', 'on');
# % Minimize using fmincg
# theta = fmincg(costFunction, initial_theta, options);


def lr_cost(X, y, theta, lambda_):
    m = X.shape[0]
    cost = np.sum(np.power(((X * theta.T) - y), 2))/2/m + np.sum(lambda_/2/m * np.power(theta[:, 0:-1], 2))
    return cost

def lr_grad(X, y, theta, lambda_):
    m = X.shape[0]
    grad = 1/m * X.T * (X*theta.T - y) + lambda_/m * np.vstack((0,theta[:, 0:-1]))
    return grad.flatten()

def train_lr(X, y, lambda_):
    from scipy.optimize import fmin_cg

    initial_theta = np.zeros((X.shape[1], 1)) 
    #initial_theta = np.zeros((1, 2))
    print(f'initial_theta {initial_theta.shape}')
    print(f'X {X.shape}\n{X}')
    print(f'y {y.shape}\n{y}')

    theta = fmin_cg(lr_cost, x0=initial_theta, args=(X, y, lambda_), maxiter=200)
    #theta = fmin_cg(lr_cost, x0=initial_theta, fprime=lr_grad , args=(X, y, lambda_), maxiter=200)
    print(f'theta {theta}')
    # return theta

lambda_ = 0
train_lr(np.insert(X, 0 , 1, axis=1), y, lambda_)


---------------------------------------------------------------------------
ValueError                                Traceback (most recent call last)
<ipython-input-171-6fc8b0780a49> in <module>()
     39 
     40 lambda_ = 0
---> 41 train_lr(np.insert(X, 0 , 1, axis=1), y, lambda_)

<ipython-input-171-6fc8b0780a49> in train_lr(X, y, lambda_)
     35     print(f'y {y.shape}\n{y}')
     36 
---> 37     theta = fmin_cg(lr_cost, x0=initial_theta, args=(X, y, lambda_), maxiter=200)
     38     print(f'theta {theta}')
     39 

/Users/apple/anaconda/lib/python3.6/site-packages/scipy/optimize/optimize.py in fmin_cg(f, x0, fprime, args, gtol, norm, epsilon, maxiter, full_output, disp, retall, callback)
   1175             'return_all': retall}
   1176 
-> 1177     res = _minimize_cg(f, x0, args, fprime, callback=callback, **opts)
   1178 
   1179     if full_output:

/Users/apple/anaconda/lib/python3.6/site-packages/scipy/optimize/optimize.py in _minimize_cg(fun, x0, args, jac, callback, gtol, norm, eps, maxiter, disp, return_all, **unknown_options)
   1226     else:
   1227         grad_calls, myfprime = wrap_function(fprime, args)
-> 1228     gfk = myfprime(x0)
   1229     k = 0
   1230     xk = x0

/Users/apple/anaconda/lib/python3.6/site-packages/scipy/optimize/optimize.py in function_wrapper(*wrapper_args)
    290     def function_wrapper(*wrapper_args):
    291         ncalls[0] += 1
--> 292         return function(*(wrapper_args + args))
    293 
    294     return ncalls, function_wrapper

/Users/apple/anaconda/lib/python3.6/site-packages/scipy/optimize/optimize.py in approx_fprime(xk, f, epsilon, *args)
    686 
    687     """
--> 688     return _approx_fprime_helper(xk, f, epsilon, args=args)
    689 
    690 

/Users/apple/anaconda/lib/python3.6/site-packages/scipy/optimize/optimize.py in _approx_fprime_helper(xk, f, epsilon, args, f0)
    620     """
    621     if f0 is None:
--> 622         f0 = f(*((xk,) + args))
    623     grad = numpy.zeros((len(xk),), float)
    624     ei = numpy.zeros((len(xk),), float)

/Users/apple/anaconda/lib/python3.6/site-packages/scipy/optimize/optimize.py in function_wrapper(*wrapper_args)
    290     def function_wrapper(*wrapper_args):
    291         ncalls[0] += 1
--> 292         return function(*(wrapper_args + args))
    293 
    294     return ncalls, function_wrapper

<ipython-input-171-6fc8b0780a49> in lr_cost(X, y, theta, lambda_)
     18 def lr_cost(X, y, theta, lambda_):
     19     m = X.shape[0]
---> 20     cost = np.sum(np.power(((X * theta.T) - y), 2))/2/m + np.sum(lambda_/2/m * np.power(theta[:, 0:-1], 2))
     21     return cost
     22 

ValueError: operands could not be broadcast together with shapes (2,) (1,12)

lr_cost和lr_grad使用：

J = lr_cost(np.insert(X, 0 , 1, axis=1), y, theta, 1)
grad = lr_grad(np.insert(X, 0 , 1, axis=1), y, theta, 1)
J, grad

输出：

(303.99319222026429, matrix([[ -15.30301567,  598.25074417]]))

有没有更好的方法来调试它？

Answer 1

---> 20     cost = np.sum(np.power(((X * theta.T) - y), 2))/2/m + np.sum(lambda_/2/m * np.power(theta[:, 0:-1], 2))

ValueError: operands could not be broadcast together with shapes (2,) (1,12)

这意味着对于其中一个操作*，/或`+＆＃39;，一个参数是1d，包含2个元素（shape（2，）），另一个参数是是2d（1,12）。

在numpy数组中，数组可以是0或1d或更高。在Octave / MATLAB中，2d是最小的。 numpy可以广播。 Octave复制了那个。但是numpy可以在开始时自动添加维度，例如把a（2，）变成（1,2）。但是对于这对，这没有任何帮助。 A（2,1）可以用（1,12）广播以形成（2,12）结果。

所以你/我们需要打印一些形状，或者从之前的代码中推断其他形式的各种元素的形状。

如果我计算了（），sum都没有指定轴，所以它们对所有值求和，产生标量。所以+不是问题所在。

debug numpy octave conversion，可能的矩阵问题

1 个答案: