我知道这个问题已被问过几次,但我无法找到解决问题的方法。
我没有比观察更多的变量,我的矩阵中没有NAN
个值。这是我的功能:
function [ind, idx_ran] = fselect(features_f, class_f, dir)
idx = linspace(1,size(features_f, 2), size(features_f, 2));
idx_ran = idx(:,randperm(size(features_f, 2)));
features_t_ran = features_f(:,idx_ran); % randomize colums
len = length(class_f);
r = randi(len, [1, round(len*0.15)]);
x = features_t_ran;
y = class_f;
xtrain = x;
ytrain = y;
xtrain(r,:) = [];
ytrain(r,:) = [];
xtest = x(r,:);
ytest = y(r,:);
f = @(xtrain, ytrain, xtest, ytest)(sum(~strcmp(ytest, classify(xtest, xtrain, ytrain))));
fs = sequentialfs(f, x, y, 'direction', dir);
ind = find(fs < 1);
end
这是我的测试和训练数据。
>> whos xtest
Name Size Bytes Class Attributes
xtest 524x42 176064 double
>> whos xtrain
Name Size Bytes Class Attributes
xtrain 3008x42 1010688 double
>> whos ytest
Name Size Bytes Class Attributes
ytest 524x1 32488 cell
>> whos ytrain
Name Size Bytes Class Attributes
ytrain 3008x1 186496 cell
>>
这是错误,
Error using crossval>evalFun (line 465)
The function
'@(xtrain,ytrain,xtest,ytest)(sum(~strcmp(ytest,classify(xtest,xtrain,ytrain))))' generated
the following error:
The pooled covariance matrix of TRAINING must be positive definite.
Error in crossval>getFuncVal (line 482)
funResult = evalFun(funorStr,arg(:));
Error in crossval (line 324)
funResult = getFuncVal(1, nData, cvp, data, funorStr, []);
Error in sequentialfs>callfun (line 485)
funResult = crossval(fun,x,other_data{:},...
Error in sequentialfs (line 353)
crit(k) = callfun(fun,x,other_data,cv,mcreps,ParOptions);
Error in fselect (line 26)
fs = sequentialfs(f, x, y, 'direction', dir);
Error in workflow_forward (line 31)
[ind, idx_ran] = fselect(features_f, class_f, 'forward');
这是昨天的工作。 :/
答案 0 :(得分:2)
如果检查函数classify
,您会发现当程序检查从训练矩阵的QR分解中获得的矩阵R的条件数时会产生错误。换句话说,它对您提供的训练矩阵不满意。它发现这个矩阵是病态的,因此任何解决方案都是不稳定的(该函数执行等效的矩阵求逆,这将导致相当于病态训练矩阵的非常小的除数)。
似乎通过缩小训练集的大小,稳定性降低了。我的建议是尽可能使用更大的训练集。
修改强>
您可能想知道如何获得比变量更多的观察结果并且仍然存在病态问题。答案是不同的观察结果可以是彼此的线性组合。