Question

我很困惑，为什么当我尝试最小化的函数明确返回并接受向量时，我收到错误ValueError: setting an array element with a sequence.。

这是我的实现，它是http://czep.net/stat/mlelr.pdf的多项逻辑回归的改编

from __future__ import division
import os
import re
import numpy as np
import pandas as pd
from collections import defaultdict, Counter, OrderedDict
cont_data = pd.read_csv('data.csv', sep=' ')
cont_data['age'] = [np.mean(map(int, x.split('_'))) for x in cont_data['age']]

d = cont_data.drop('age', axis=1)

# vector n contains the number of members in each group
n = d.sum(axis=1)

K = 1  # we are using 0 based indexing
J = 2

# matrix y is the input data with the baseline class dropped
y = d.drop('none', axis=1)

# design matrix
X = np.array([[age ** k for k in xrange(K + 2)] for age in cont_data.age])


# calculate pi following equations 24 and 25


def calculate_pi_i_j(i, j, b):
    try:
        return np.exp(np.sum([X[i][k] * b[k][j] for k in xrange(K)]))
    except Exception, e:
        assert False, (e, i, j, k, b)


def calc_prob(i, j, b):
    num = calculate_pi_i_j(i, j, b)
    denom = 1 + np.sum([calculate_pi_i_j(i, j, b) for j in xrange(J - 1)])
    return num / denom


def calculate_pi(b):
    # calculate pi for the first 2 columns
    pi = np.array([[calc_prob(i, j, b) for j in xrange(J - 1)] for i in xrange(len(n))])
    # calculate pi for the last column where J=J
    pi_j = np.array([1 / (1 + np.sum([calculate_pi_i_j(i, j, b) for j in xrange(J - 1)])) for i in xrange(len(n))])
    return np.hstack([pi, pi_j.reshape(len(n), 1)])


# equation to optimize
def eqn_32(b):
    # b comes in as a vector
    b = b.reshape(3, 2)
    pi = calculate_pi(b)
    r = []  # will hold the result of the gradient calculation for each member of beta
    try:
        for k in xrange(b.shape[0]):
            for j in xrange(b.shape[1]):
                r.append(np.sum([(y.iloc[i][j] * X[i][k]) * (n.iloc[i] * pi[i][j] * X[i][k]) for i in xrange(len(n))]))
    except Exception, e:
        assert False, (e, k, j, b)
    print r
    return np.array(r)


from scipy.optimize import *
# starting values for coefficients
b = np.zeros(6)
#minimize(eqn_32, x0=b, method='Nelder-Mead')
fmin_bfgs(eqn_32, x0=b)

data.csv包含的位置：

age ster other none
15_19 3 61 232
20_24 80 137 400
25_29 216 131 301
30_34 268 76 203
35_39 197 50 188
40_44 150 24 164
45_49 91 10 183

如您所见，b最初是一个6的数组，返回值r是一个6成员列表，然后转换为数组。当我使用debug print语句运行代码时可以看到这一点：

[249525.5, 130873.5, 270069249.5, 100471089.0, 342109256925.5, 94938609711.0]
[249525.50185910985, 130873.49902491644, 270069251.5121727, 100471088.25143206, 342109259474.41266, 94938609003.652222]
---------------------------------------------------------------------------
ValueError                                Traceback (most recent call last)
<ipython-input-8-edd793c643e2> in <module>()
     67 b = np.zeros(6)
     68 #minimize(eqn_32, x0=b, method='Nelder-Mead')
---> 69 fmin_bfgs(eqn_32, x0=b)

/Users/ifiddes/anaconda/lib/python2.7/site-packages/scipy/optimize/optimize.pyc in fmin_bfgs(f, x0, fprime, args, gtol, norm, epsilon, maxiter, full_output, disp, retall, callback)
    791             'return_all': retall}
    792 
--> 793     res = _minimize_bfgs(f, x0, args, fprime, callback=callback, **opts)
    794 
    795     if full_output:

/Users/ifiddes/anaconda/lib/python2.7/site-packages/scipy/optimize/optimize.pyc in _minimize_bfgs(fun, x0, args, jac, callback, gtol, norm, eps, maxiter, disp, return_all, **unknown_options)
    845     else:
    846         grad_calls, myfprime = wrap_function(fprime, args)
--> 847     gfk = myfprime(x0)
    848     k = 0
    849     N = len(x0)

/Users/ifiddes/anaconda/lib/python2.7/site-packages/scipy/optimize/optimize.pyc in function_wrapper(*wrapper_args)
    287     def function_wrapper(*wrapper_args):
    288         ncalls[0] += 1
--> 289         return function(*(wrapper_args + args))
    290 
    291     return ncalls, function_wrapper

/Users/ifiddes/anaconda/lib/python2.7/site-packages/scipy/optimize/optimize.pyc in approx_fprime(xk, f, epsilon, *args)
    620 
    621     """
--> 622     return _approx_fprime_helper(xk, f, epsilon, args=args)
    623 
    624 

/Users/ifiddes/anaconda/lib/python2.7/site-packages/scipy/optimize/optimize.pyc in _approx_fprime_helper(xk, f, epsilon, args, f0)
    560         ei[k] = 1.0
    561         d = epsilon * ei
--> 562         grad[k] = (f(*((xk + d,) + args)) - f0) / d[k]
    563         ei[k] = 0.0
    564     return grad

ValueError: setting an array element with a sequence.

为什么在2次迭代后失败？

Answer 1

我认为eqn_32必须返回一个标量。

尽管显式返回1d数组，但使用scipy优化进行2次迭代后的ValueError

1 个答案: