Question

这是使用玩具数据集的逻辑回归的实现。来自@dermen的一些反馈帮助我修复了使用scipy.optimize.minimize的基本问题，但即使在修复该问题之后，优化也无法收敛，即使只使用测试数据集的前五行也是如此。这是代码的独立版本：

import numpy as np
from scipy.optimize import minimize

# `data` is a subset of a toy dataset. The full dataset is ~100 rows, linearly seperable and located at 
# https://github.com/liavkoren/ng-redux/blob/master/ex2/ex2data1.txt
data = np.array([
    [ 34.62365962, 78.02469282],
    [ 30.28671077, 43.89499752],
    [ 35.84740877, 72.90219803],
    [ 60.18259939, 86.3085521 ],
    [ 79.03273605, 75.34437644],
    [ 45.08327748, 56.31637178],
    [ 61.10666454, 96.51142588],
    [ 75.02474557, 46.55401354],
    [ 76.0987867,  87.42056972],
    [ 84.43281996, 43.53339331],
])
# Ground truth
y = np.array([0., 0., 0., 1., 1., 0., 1., 1., 1., 1.])


def sigmoid(z):
    return 1/(1 + np.power(np.e, -z))

h = lambda theta, x: sigmoid(x.dot(theta))


def cost(theta, X, y):
    m = X.shape[0]
    j = y.dot(np.log(h(theta, X)))  + (1 - y).dot(np.log(1 - h(theta, X)))
    return (-j/m)


def grad(theta, X, y):
    m = X.shape[0]
    return ((h(theta, X) - y).dot(X))/m

# Add a column of ones:
m, features = np.shape(X_initial)
features += 1
X = np.concatenate([np.ones((m, 1)), X_initial], axis=1)
initial_theta = np.zeros((features))


def check_functions(grad_func, cost_func):
    '''
    Asserts that the cost and gradient functions return known corret values for a given theta, X, y.  
    Test case from https://www.coursera.org/learn/machine-learning/discussions/weeks/3/threads/tA3ESpq0EeW70BJZtLVfGQ
    The expected cost is 4.6832.
    The expected gradient = [0.31722, 0.87232, 1.64812, 2.23787]
    '''
    test_X = np.array([[1, 8, 1, 6], [1, 3, 5, 7], [1, 4, 9, 2]])  # X
    test_y = np.array([[1, 0, 1]])  # y
    test_theta = np.array([-2, -1, 1, 2])

    grad_diff = grad_func(test_theta, test_X, test_y) - np.array([0.31722, 0.87232, 1.64812, 2.23787])
    assert grad_diff.dot(grad_diff.T) < 0.0001
    assert abs(cost_func(test_theta, test_X, test_y, debug=False) - 4.6832) < 0.0001

check_functions(grad, cost)

# `cutoff` slices out a subset of rows.
cutoff = 2
print minimize(fun=cost, x0=initial_theta, args=(X[0:cutoff, :], y[0:cutoff]), jac=grad)

此代码失败：

   fun: nan
 hess_inv: array([[1, 0, 0],
       [0, 1, 0],
       [0, 0, 1]])
      jac: array([ 0.,  0.,  0.])
  message: 'Desired error not necessarily achieved due to precision loss.'
     nfev: 32
      nit: 1
     njev: 32
   status: 2
  success: False
        x: array([ -0.5       , -16.2275926 , -30.47992258])

/Users/liavkoren/Envs/data-sci/lib/python2.7/site-packages/ipykernel/__main__.py:25: RuntimeWarning: overflow encountered in power
/Users/liavkoren/Envs/data-sci/lib/python2.7/site-packages/ipykernel/__main__.py:38: RuntimeWarning: divide by zero encountered in log
/Users/liavkoren/Envs/data-sci/lib/python2.7/site-packages/ipykernel/__main__.py:42: RuntimeWarning: divide by zero encountered in log

Answer 1

np.power函数内对sigma的调用发生溢出。我在cost函数中添加了调试消息，并看到了以下内容：

theta: [ 0.  0.  0.]
--
X: [[  1.          34.62365962  78.02469282]
 [  1.          30.28671077  43.89499752]]
--
y=1: [ 0.5  0.5]  y=0: [ 0.5  0.5]
log probabilities: 
y=1: [-0.69314718 -0.69314718]
y=0: [-0.69314718 -0.69314718]
=======

theta: [ -0.5        -16.2275926  -30.47992258]
--
X: [[  1.          34.62365962  78.02469282]
 [  1.          30.28671077  43.89499752]]
--
y=1: [ 0.  0.]  y=0: [ 1.  1.]
log probabilities: 
y=1: [-inf -inf]
y=0: [ 0.  0.]

第二次迭代溢出!!

我很快就确认这似乎是通过将数据集除以1/10来解决问题并且它收敛了。我想我将不得不考虑功能扩展/规范化或其他一些避免溢出的策略。

SciPy最小化渐变

1 个答案: