我经常fminunc用于逻辑回归问题。我在网上读到Andrew Ng用相同的论点fmincg代替, 。fminunc结果是不同的，通常fmincg更准确，但不会太多。（我将 fmincg 函数 fminunc 的结果与相同的数据进行比较）
所以，我的问题是：这两个功能有什么区别？每个函数都实现了什么算法？（现在，我只是使用这些函数而不知道它们是如何工作的）。
谢谢 ：）
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8 回答
54
您将不得不查看代码的内部，fmincg因为它不是 Octave 的一部分。经过一番搜索，我发现它是 Coursera 机器学习类提供的一个函数文件，作为作业的一部分。阅读有关此问题的评论和答案，以讨论算法。
于 2012-08-24T20:47:11.400 回答
24
与此处暗示 fmincg 和 fminunc 之间的主要区别是准确性或速度的其他答案相反，对于某些应用程序而言，最重要的区别可能是内存效率。在 Coursera 吴恩达的机器学习课程的编程练习 4（即神经网络训练）中，ex4.m 中关于 fmincg 的评论是
%% =================== 第 8 部分：训练 NN ===================
% 你现在已经实现了训练神经
网络所需的所有代码。为了训练你的神经网络，我们现在将使用“fmincg”，它
是一个类似于“fminunc”的函数。 回想一下，只要我们为它们提供梯度计算 ，这些
高级优化器就能够有效地训练我们的成本函数。
像原始海报一样，我也很好奇 ex4.m 的结果使用 fminunc 而不是 fmincg 可能会有什么不同。所以我尝试替换 fmincg 调用
options = optimset('MaxIter', 50);
[nn_params, cost] = fmincg(costFunction, initial_nn_params, options);
通过以下对 fminunc 的调用
options = optimset('GradObj', 'on', 'MaxIter', 50);
[nn_params, cost, exit_flag] = fminunc(costFunction, initial_nn_params, options);
但从在 Windows 上运行的 32 位 Octave 构建中收到以下错误消息：
错误：内存已用完或请求的大小对于 Octave 的索引类型范围来说太大——试图返回提示
在 Windows 上运行的 32 位版本的 MATLAB 提供了更详细的错误消息：
使用查找内存不足时出错。键入 HELP MEMORY 作为您的选项。
Spones 错误（第 14 行）
[i,j] = find(S);
颜色错误（第 26 行）
J = spones(J)；
sfminbx 中的错误（第 155 行）
组 = 颜色（Hstr，p）；
fminunc 中的错误（第 408 行）
[x,FVAL,~,EXITFLAG,OUTPUT,GRAD,HESSIAN] = sfminbx(funfcn,x,l,u, ...
ex4 中的错误（第 205 行）
[nn_params, cost, exit_flag] = fminunc（成本函数，initial_nn_params，选项）；
我笔记本电脑上的 MATLAB memory 命令报告：
最大可能数组：2046 MB（2.146e+09 字节）*
可用于所有数组的内存：3402 MB（3.568e+09 字节）**
MATLAB 使用的内存：373 MB（3.910e+08 字节）
物理内存 (RAM) : 3561 MB (3.734e+09 bytes)
* 受限于可用的连续虚拟地址空间。
** 受限于可用的虚拟地址空间。
之前我在想 Ng 教授选择使用 fmincg 来训练 ex4.m 神经网络（有 400 个输入特征，401 包括偏置输入）来提高训练速度。然而，现在我相信他使用 fmincg 的原因是为了提高内存效率，足以允许在 Octave/MATLAB 的 32 位构建上执行训练。关于获得在 Windows 操作系统上运行的 64 位 Octave 构建的必要工作的简短讨论在这里。
于 2014-08-23T20:20:39.020 回答
14
根据 Andrew Ng 本人的说法，fmincg它不是用来获得更准确的结果（请记住，您的成本函数在任何一种情况下都是相同的，并且您的假设不会更简单或更复杂），而是因为它在对特别复杂的情况下进行梯度下降时更有效假设。他自己似乎使用fminunc了假设很少有特征但fmincg有数百个特征的地方。
于 2013-11-23T22:36:29.427 回答
9
为什么 fmincg 有效？
这是源代码的副本，其中包含解释所使用的各种算法的注释。这真是太棒了，就像孩子的大脑在学习区分狗和椅子时所做的一样。
这是 fmincg.m 的 Octave 源代码。
function [X, fX, i] = fmincg(f, X, options, P1, P2, P3, P4, P5)
% Minimize a continuous differentialble multivariate function. Starting point
% is given by "X" (D by 1), and the function named in the string "f", must
% return a function value and a vector of partial derivatives. The Polack-
% Ribiere flavour of conjugate gradients is used to compute search directions,
% and a line search using quadratic and cubic polynomial approximations and the
% Wolfe-Powell stopping criteria is used together with the slope ratio method
% for guessing initial step sizes. Additionally a bunch of checks are made to
% make sure that exploration is taking place and that extrapolation will not
% be unboundedly large. The "length" gives the length of the run: if it is
% positive, it gives the maximum number of line searches, if negative its
% absolute gives the maximum allowed number of function evaluations. You can
% (optionally) give "length" a second component, which will indicate the
% reduction in function value to be expected in the first line-search (defaults
% to 1.0). The function returns when either its length is up, or if no further
% progress can be made (ie, we are at a minimum, or so close that due to
% numerical problems, we cannot get any closer). If the function terminates
% within a few iterations, it could be an indication that the function value
% and derivatives are not consistent (ie, there may be a bug in the
% implementation of your "f" function). The function returns the found
% solution "X", a vector of function values "fX" indicating the progress made
% and "i" the number of iterations (line searches or function evaluations,
% depending on the sign of "length") used.
%
% Usage: [X, fX, i] = fmincg(f, X, options, P1, P2, P3, P4, P5)
%
% See also: checkgrad
%
% Copyright (C) 2001 and 2002 by Carl Edward Rasmussen. Date 2002-02-13
%
%
% (C) Copyright 1999, 2000 & 2001, Carl Edward Rasmussen
%
% Permission is granted for anyone to copy, use, or modify these
% programs and accompanying documents for purposes of research or
% education, provided this copyright notice is retained, and note is
% made of any changes that have been made.
%
% These programs and documents are distributed without any warranty,
% express or implied. As the programs were written for research
% purposes only, they have not been tested to the degree that would be
% advisable in any important application. All use of these programs is
% entirely at the user's own risk.
%
% [ml-class] Changes Made:
% 1) Function name and argument specifications
% 2) Output display
%
% Read options
if exist('options', 'var') && ~isempty(options) && isfield(options, 'MaxIter')
length = options.MaxIter;
else
length = 100;
end
RHO = 0.01; % a bunch of constants for line searches
SIG = 0.5; % RHO and SIG are the constants in the Wolfe-Powell conditions
INT = 0.1; % don't reevaluate within 0.1 of the limit of the current bracket
EXT = 3.0; % extrapolate maximum 3 times the current bracket
MAX = 20; % max 20 function evaluations per line search
RATIO = 100; % maximum allowed slope ratio
argstr = ['feval(f, X']; % compose string used to call function
for i = 1:(nargin - 3)
argstr = [argstr, ',P', int2str(i)];
end
argstr = [argstr, ')'];
if max(size(length)) == 2, red=length(2); length=length(1); else red=1; end
S=['Iteration '];
i = 0; % zero the run length counter
ls_failed = 0; % no previous line search has failed
fX = [];
[f1 df1] = eval(argstr); % get function value and gradient
i = i + (length<0); % count epochs?!
s = -df1; % search direction is steepest
d1 = -s'*s; % this is the slope
z1 = red/(1-d1); % initial step is red/(|s|+1)
while i < abs(length) % while not finished
i = i + (length>0); % count iterations?!
X0 = X; f0 = f1; df0 = df1; % make a copy of current values
X = X + z1*s; % begin line search
[f2 df2] = eval(argstr);
i = i + (length<0); % count epochs?!
d2 = df2'*s;
f3 = f1; d3 = d1; z3 = -z1; % initialize point 3 equal to point 1
if length>0, M = MAX; else M = min(MAX, -length-i); end
success = 0; limit = -1; % initialize quanteties
while 1
while ((f2 > f1+z1*RHO*d1) | (d2 > -SIG*d1)) & (M > 0)
limit = z1; % tighten the bracket
if f2 > f1
z2 = z3 - (0.5*d3*z3*z3)/(d3*z3+f2-f3); % quadratic fit
else
A = 6*(f2-f3)/z3+3*(d2+d3); % cubic fit
B = 3*(f3-f2)-z3*(d3+2*d2);
z2 = (sqrt(B*B-A*d2*z3*z3)-B)/A; % numerical error possible - ok!
end
if isnan(z2) | isinf(z2)
z2 = z3/2; % if we had a numerical problem then bisect
end
z2 = max(min(z2, INT*z3),(1-INT)*z3); % don't accept too close to limits
z1 = z1 + z2; % update the step
X = X + z2*s;
[f2 df2] = eval(argstr);
M = M - 1; i = i + (length<0); % count epochs?!
d2 = df2'*s;
z3 = z3-z2; % z3 is now relative to the location of z2
end
if f2 > f1+z1*RHO*d1 | d2 > -SIG*d1
break; % this is a failure
elseif d2 > SIG*d1
success = 1; break; % success
elseif M == 0
break; % failure
end
A = 6*(f2-f3)/z3+3*(d2+d3); % make cubic extrapolation
B = 3*(f3-f2)-z3*(d3+2*d2);
z2 = -d2*z3*z3/(B+sqrt(B*B-A*d2*z3*z3)); % num. error possible - ok!
if ~isreal(z2) | isnan(z2) | isinf(z2) | z2 < 0 % num prob or wrong sign?
if limit < -0.5 % if we have no upper limit
z2 = z1 * (EXT-1); % the extrapolate the maximum amount
else
z2 = (limit-z1)/2; % otherwise bisect
end
elseif (limit > -0.5) & (z2+z1 > limit) % extraplation beyond max?
z2 = (limit-z1)/2; % bisect
elseif (limit < -0.5) & (z2+z1 > z1*EXT) % extrapolation beyond limit
z2 = z1*(EXT-1.0); % set to extrapolation limit
elseif z2 < -z3*INT
z2 = -z3*INT;
elseif (limit > -0.5) & (z2 < (limit-z1)*(1.0-INT)) % too close to limit?
z2 = (limit-z1)*(1.0-INT);
end
f3 = f2; d3 = d2; z3 = -z2; % set point 3 equal to point 2
z1 = z1 + z2; X = X + z2*s; % update current estimates
[f2 df2] = eval(argstr);
M = M - 1; i = i + (length<0); % count epochs?!
d2 = df2'*s;
end % end of line search
if success % if line search succeeded
f1 = f2; fX = [fX' f1]';
fprintf('%s %4i | Cost: %4.6e\r', S, i, f1);
s = (df2'*df2-df1'*df2)/(df1'*df1)*s - df2; % Polack-Ribiere direction
tmp = df1; df1 = df2; df2 = tmp; % swap derivatives
d2 = df1'*s;
if d2 > 0 % new slope must be negative
s = -df1; % otherwise use steepest direction
d2 = -s'*s;
end
z1 = z1 * min(RATIO, d1/(d2-realmin)); % slope ratio but max RATIO
d1 = d2;
ls_failed = 0; % this line search did not fail
else
X = X0; f1 = f0; df1 = df0; % restore point from before failed line search
if ls_failed | i > abs(length) % line search failed twice in a row
break; % or we ran out of time, so we give up
end
tmp = df1; df1 = df2; df2 = tmp; % swap derivatives
s = -df1; % try steepest
d1 = -s'*s;
z1 = 1/(1-d1);
ls_failed = 1; % this line search failed
end
if exist('OCTAVE_VERSION')
fflush(stdout);
end
end
fprintf('\n');
于 2012-08-28T14:04:14.440 回答
1
fmincg 比 fminunc 更准确。他们两个所花费的时间几乎相同。在神经网络或一般情况下，没有更多的权重 fminunc 可以给出内存不足错误。所以 fmincg 内存效率更高。
使用 fminunc，准确度为 93.10，所用时间为 11.3794 秒。
Testing lrCostFunction() with regularization
Cost: 2.534819
Expected cost: 2.534819
Gradients:
0.146561
-0.548558
0.724722
1.398003
Expected gradients:
0.146561
-0.548558
0.724722
1.398003
Program paused. Press enter to continue.
Training One-vs-All Logistic Regression...
id = 1512324857357
Elapsed time is 11.3794 seconds.
Program paused. Press enter to continue.
Training Set Accuracy: 93.100000
使用 fmincg，准确度为 95.12，所用时间为 11.7978 秒。
Testing lrCostFunction() with regularization
Cost: 2.534819
Expected cost: 2.534819
Gradients:
0.146561
-0.548558
0.724722
1.398003
Expected gradients:
0.146561
-0.548558
0.724722
1.398003
Program paused. Press enter to continue.
Training One-vs-All Logistic Regression...
id = 1512325280047
Elapsed time is 11.7978 seconds.
Training Set Accuracy: 95.120000
于 2017-12-03T18:36:13.273 回答
0
使用 fmincg 优化成本函数。fmincg 的工作方式与 fminunc 类似，但在处理大量参数时效率更高。在任何一种情况下，您的成本函数都是相同的，并且您的假设不会更简单或更复杂），但是因为它在对特别复杂的假设进行梯度下降时更有效。
它用于有效地利用内存。
于 2018-11-10T12:16:06.403 回答
0
fmincg 是 Coursera 上 course 开发的内部函数，不像 fminunc 是内置的 Octave 函数。由于它们都用于逻辑回归，因此它们仅在一个方面有所不同。当要考虑的参数数量相当大时（如果与训练集的大小相比），fmincg 比 fminunc 工作得更快并且处理更准确。而且，当传递给它的参数没有限制（不受约束）时，首选 fminunc。
于 2019-08-10T10:34:29.573 回答