I've been trying to finish Andrew Ng's Machine Learning course, I am at the part about logistic regression now. I am trying to discover the parameters and also calculate the cost without using the MATLAB function fminunc. However, I am not converging to the correct results as posted by other students who have finished the assignment using fminunc. Specifically, my problems are:
the parameters theta are incorrect
my cost seems to be blowing up
I get many NaNs in my cost vector (I just create a vector of the costs to keep track)
I attempted to discover the parameters via Gradient Descent as how I understood the content. However, my implementation still seems to be giving me incorrect results.
dataset = load('dataStuds.txt');
x = dataset(:,1:end-1);
y = dataset(:,end);
m = length(x);
% Padding the the 1's (intercept term, the call it?)
x = [ones(length(x),1), x];
thetas = zeros(size(x,2),1);
% Setting the learning rate to 0.1
alpha = 0.1;
for i = 1:100000
% theta transpose x (tho why in MATLAB it needs to be done the other way
% round? :)
ttrx = x * thetas;
% the hypothesis function h_x = g(z) = sigmoid(-z)
h_x = 1 ./ (1 + exp(-ttrx));
error = h_x - y;
% the gradient (aka the derivative of J(\theta) aka the derivative
% term)
for j = 1:length(thetas)
gradient = 1/m * (h_x - y)' * x(:,j);
% Updating the parameters theta
thetas(j) = thetas(j) - alpha * gradient;
end
% Calculating the cost, just to keep track...
cost(i) = 1/m * ( -y' * log(h_x) - (1-y)' * log(1-h_x) );
end
% Displaying the final theta's that I obtained
thetas
The parameters theta that I get are:
thetas =
-482.8509
3.7457
2.6976
The results below is from one example that I downloaded, but the author used fminunc for this one.
Cost at theta found by fminunc: 0.203506
theta:
-24.932760
0.204406
0.199616
The data:
34.6236596245170 78.0246928153624 0
30.2867107682261 43.8949975240010 0
35.8474087699387 72.9021980270836 0
60.1825993862098 86.3085520954683 1
79.0327360507101 75.3443764369103 1
45.0832774766834 56.3163717815305 0
61.1066645368477 96.5114258848962 1
75.0247455673889 46.5540135411654 1
76.0987867022626 87.4205697192680 1
84.4328199612004 43.5333933107211 1
95.8615550709357 38.2252780579509 0
75.0136583895825 30.6032632342801 0
82.3070533739948 76.4819633023560 1
69.3645887597094 97.7186919618861 1
39.5383391436722 76.0368108511588 0
53.9710521485623 89.2073501375021 1
69.0701440628303 52.7404697301677 1
67.9468554771162 46.6785741067313 0
70.6615095549944 92.9271378936483 1
76.9787837274750 47.5759636497553 1
67.3720275457088 42.8384383202918 0
89.6767757507208 65.7993659274524 1
50.5347882898830 48.8558115276421 0
34.2120609778679 44.2095285986629 0
77.9240914545704 68.9723599933059 1
62.2710136700463 69.9544579544759 1
80.1901807509566 44.8216289321835 1
93.1143887974420 38.8006703371321 0
61.8302060231260 50.2561078924462 0
38.7858037967942 64.9956809553958 0
61.3792894474250 72.8078873131710 1
85.4045193941165 57.0519839762712 1
52.1079797319398 63.1276237688172 0
52.0454047683183 69.4328601204522 1
40.2368937354511 71.1677480218488 0
54.6351055542482 52.2138858806112 0
33.9155001090689 98.8694357422061 0
64.1769888749449 80.9080605867082 1
74.7892529594154 41.5734152282443 0
34.1836400264419 75.2377203360134 0
83.9023936624916 56.3080462160533 1
51.5477202690618 46.8562902634998 0
94.4433677691785 65.5689216055905 1
82.3687537571392 40.6182551597062 0
51.0477517712887 45.8227014577600 0
62.2226757612019 52.0609919483668 0
77.1930349260136 70.4582000018096 1
97.7715992800023 86.7278223300282 1
62.0730637966765 96.7688241241398 1
91.5649744980744 88.6962925454660 1
79.9448179406693 74.1631193504376 1
99.2725269292572 60.9990309984499 1
90.5467141139985 43.3906018065003 1
34.5245138532001 60.3963424583717 0
50.2864961189907 49.8045388132306 0
49.5866772163203 59.8089509945327 0
97.6456339600777 68.8615727242060 1
32.5772001680931 95.5985476138788 0
74.2486913672160 69.8245712265719 1
71.7964620586338 78.4535622451505 1
75.3956114656803 85.7599366733162 1
35.2861128152619 47.0205139472342 0
56.2538174971162 39.2614725105802 0
30.0588224466980 49.5929738672369 0
44.6682617248089 66.4500861455891 0
66.5608944724295 41.0920980793697 0
40.4575509837516 97.5351854890994 1
49.0725632190884 51.8832118207397 0
80.2795740146700 92.1160608134408 1
66.7467185694404 60.9913940274099 1
32.7228330406032 43.3071730643006 0
64.0393204150601 78.0316880201823 1
72.3464942257992 96.2275929676140 1
60.4578857391896 73.0949980975804 1
58.8409562172680 75.8584483127904 1
99.8278577969213 72.3692519338389 1
47.2642691084817 88.4758649955978 1
50.4581598028599 75.8098595298246 1
60.4555562927153 42.5084094357222 0
82.2266615778557 42.7198785371646 0
88.9138964166533 69.8037888983547 1
94.8345067243020 45.6943068025075 1
67.3192574691753 66.5893531774792 1
57.2387063156986 59.5142819801296 1
80.3667560017127 90.9601478974695 1
68.4685217859111 85.5943071045201 1
42.0754545384731 78.8447860014804 0
75.4777020053391 90.4245389975396 1
78.6354243489802 96.6474271688564 1
52.3480039879411 60.7695052560259 0
94.0943311251679 77.1591050907389 1
90.4485509709636 87.5087917648470 1
55.4821611406959 35.5707034722887 0
74.4926924184304 84.8451368493014 1
89.8458067072098 45.3582836109166 1
83.4891627449824 48.3802857972818 1
42.2617008099817 87.1038509402546 1
99.3150088051039 68.7754094720662 1
55.3400175600370 64.9319380069486 1
74.7758930009277 89.5298128951328 1
I ran your code and it does work fine. However, the tricky thing about gradient descent is ensuring that your costs don't diverge to infinity. If you look at your costs array, you will see that the costs definitely diverge and this is why you are not getting the correct results.
The best way to eliminate this in your case is to reduce the learning rate. Through experimentation, I have found that a learning rate of alpha = 0.003 is the best for your problem. I've also increased the number of iterations to 200000. Changing these two things gives me the following parameters and associated cost:
>> format long g;
>> thetas
thetas =
-17.6287417780435
0.146062780453677
0.140513170941357
>> cost(end)
ans =
0.214821863463963
This is more or less in line with the magnitudes of the parameters you see when you are using fminunc. However, they get slightly different parameters as well as different costs because of the actual minimization method itself. fminunc uses a variant of L-BFGS which finds the solution in a much faster way.
What is most important is the actual accuracy itself. Remember that to classify whether an example belongs to label 0 or 1, you take the weighted sum of the parameters and examples, run it through the sigmoid function and threshold at 0.5. We find what the average amount of times each expected label and predicted label match.
Using the parameters we found with gradient descent gives us the following accuracy:
>> ttrx = x * thetas;
>> h_x = 1 ./ (1 + exp(-ttrx)) >= 0.5;
>> mean(h_x == y)
ans =
0.89
This means that we've achieved an 89% classification accuracy. Using the labels provided by fminunc also gives:
>> thetas2 = [-24.932760; 0.204406; 0.199616];
>> ttrx = x * thetas2;
>> h_x = 1 ./ (1 + exp(-ttrx)) >= 0.5;
>> mean(h_x == y)
ans =
0.89
So we can see that the accuracy is the same so I wouldn't worry too much about the magnitude of the parameters but it's more in line with what we see when we compare the costs between the two implementations.
As a final note to you, I would suggest looking at this post of mine for some tips on how to make logistic regression work over long-term. I would definitely recommend normalizing your features prior to finding the parameters to make the algorithm run faster. It also addresses why you were finding the wrong parameters (namely the cost blowing up): Cost function in logistic regression gives NaN as a result.
normalizing the data using mean and standard deviation as follows enables you to use large learning rate and get a similar answer
clear; clc
data = load('ex2data1.txt');
m = length(data);
alpha = 0.1;
theta = [0; 0; 0];
y = data(:,3);
% Normalizing the data
xm1 = mean(data(:,1)); xm2 = mean(data(:,2));
xs1 = std(data(:,1)); xs2 = std(data(:,2));
x1 = (data(:,1)-xm1)./xs1; x2 = (data(:,2)-xm2)./xs2;
X = [ones(m, 1) x1 x2];
for i=1:10000
h = 1./(1+exp(-(X*theta)));
theta = theta - (alpha/m)* (X'*(h-y));
J(i) = (1/m)*(-y'*log(h)-(1-y)'*log(1-h));
end
theta
J(end)
figure
plot(J)
function A=obj(b,Y, YL, tempcyc, cyc, Yzero, age, agesq, educ, ageav, agesqav, educav, qv, year1, year2, year3, year4, year5, year6, year7, workregion1, workregion2, workregion3, workregion4, workregion5, workregion6, workregion7, workregion8, workregion9, workregion10, workregion11, workregion12, workregion13, workregion14, workregion15, workregion16, qw)
A=0;
for i=1:715
S=0;
for m=1:12
P=1;
for t=1:8
P=P*(normcdf((2*Y(i,t)-1)*(b(1)*YL(i,t)+b(2)*tempcyc(i,t)+b(3)*cyc(i,t)+b(4)*Yzero(i,t)+b(5)*age(i,t)+b(6)*agesq(i,t)+b(7)*educ(i,t)+b(8)*ageav(i,t)+b(9)*agesqav(i,t)+b(10)*educav(i,t)+b(11)*1+b(12)*sqrt(2)*qv(m,1)+b(13)*year1(i,t)+b(14)*year2(i,t)+b(15)*year3(i,t)+b(16)*year4(i,t)+b(17)*year5(i,t)+b(18)*year6(i,t)+b(19)*year7(i,t)+b(20)*workregion1(i,t)+b(21)*workregion2(i,t)+b(22)*workregion3(i,t)+b(23)*workregion4(i,t)+b(24)*workregion5(i,t)+b(25)*workregion6(i,t)+b(26)*workregion7(i,t)+b(27)*workregion8(i,t)+b(28)*workregion9(i,t)+b(29)*workregion10(i,t)+b(30)*workregion11(i,t)+b(31)*workregion12(i,t)+b(32)*workregion13(i,t)+b(33)*workregion14(i,t)+b(34)*workregion15(i,t)+b(35)*workregion16(i,t))));
end
S=S+qw(m,1)*P;
end
A=A+log(S/sqrt(pi));
A=-A;
end
This is my objective function I am minimizing (actually maximizing; I am minimizing -A) and the parameters I am estimating is b=[b(1)............b(35)]. Y, YL, tempcyc,.........., qw are matrices that I am imported as data. The objective function consists of a product(across t=1:8) nested within a sum (across m=1:12) that is in turn nested within a sum (across n=1:715). And below is my code using fminsearch for my unconstrained minimization.
% %% Minimization
start=[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0];
iter=50000;
options=optimset('Display','iter','MaxIter',iter,'MaxFunEvals',100000);
[b, fval, exitflag, output]=fminsearch(#(b)obj(b,Y, YL, tempcyc, cyc, Yzero, age, agesq, educ, ageav, agesqav, educav, qv, year1, year2, year3, year4, year5, year6, year7, workregion1, workregion2, workregion3, workregion4, workregion5, workregion6, workregion7, workregion8, workregion9, workregion10, workregion11, workregion12, workregion13, workregion14, workregion15, workregion16, qw), start, options);%
%% Results
fprintf('[b] : % 1.4e % 1.4e %1.4e % 1.4e % 1.4e % 1.4e \n',...b(1),b(2),b(3),b(4), b(5),b(6),b(7),b(8), b(9),b(10),b(11),b(12), b(13),b(14),b(15),b(16), b(17),b(18),b(19),b(20), b(21),b(22),b(23),b(24), b(25),b(26),b(27),b(28), b(29),b(30),b(31),b(32), b(33),b(34), b(35));
end
The problem is that the optmization results do not show up even after 12+ hours (It is still reflecting, expanding, contracting inside, etc.) Could someone give me an idea on how to make the process faster?
Thank you.
This is not a complete answer to your question; this is a suggestion on how to better write your code, which is probably half the answer.
I would write your calling script something like this:
%% Minimization
b0 = zeros(1,35);
iter = 5e4;
options=optimset(...
'Display' , 'iter',...
'MaxIter' , iter,...
'MaxFunEvals', 1e5);
data = {...
YL, tempcyc, cyc, Yzero, age, agesq, educ, ageav, agesqav, educav, qv, ...
year1, year2, year3, year4, year5, year6, year7, ...
workregion1, workregion2, workregion3, workregion4, workregion5, ...
workregion6, workregion7, workregion8, workregion9, workregion10, ...
workregion11, workregion12, workregion13, workregion14, workregion15, ...
workregion16
};
[b, fval, exitflag, output] = fminsearch(#(b)obj(b, Y,data,qw), b0, options);
%% Print results
fprintf('[b]:');
fprintf('%1.4e', b);
fprintf('\n');
(actually, I would also wrap the data in cells instead of having 16 workregions and 7 years etc. But you may not be the creator of the data, so this may not always be possible)
You could re-write your objective function something like this:
function A = obj(b, Y, others, qw)
A = 0;
sPi = -0.5*log(pi);
bB = num2cell(b);
for ii = 1:715
P = prod( normcdf((2*Y(ii,1:8)-1) .* cellfun(#(x,y) x(ii,1:8).*y, others, bB)) );
S = sum( P*qw(1:12,1) );
A = A + log(S) + sPi;
end
A = -A;
end
Probably also the outer loop can be simplified into vectorized form. But I cannot test the correctness of all this (or whether it runs at all), because I don't have access to your data. But I trust you get the idea -- you can now at least read your code, and probably spot the error yourself.
I need to convert a Decimal Number to a Binary Vector
For example, Something like this:
length=de2bi(length_field,16);
Unfortunately, because of licensing, I cannot use this command. Is there any quick short technique for converting binary to a vector.
Here is what I am looking for,
If
Data=12;
Bin_Vec=Binary_To_Vector(Data,6) should return me
Bin_Vec=[0 0 1 1 0 0]
Thanks
You mention not being able to use the function de2bi, which is likely because it is a function in the Communications System Toolbox and you don't have a license for it. Luckily, there are two other functions that you can use that are part of the core MATLAB toolbox: BITGET and DEC2BIN. I generally lean towards using BITGET since DEC2BIN can be significantly slower when converting many values at once. Here's how you would use BITGET:
>> Data = 12; %# A decimal number
>> Bin_Vec = bitget(Data,1:6) %# Get the values for bits 1 through 6
Bin_Vec =
0 0 1 1 0 0
A single call to Matlab's built-in function dec2bin can achieve this:
binVec = dec2bin(data, nBits)-'0'
Here's a solution that is reasonably fast:
function out = binary2vector(data,nBits)
powOf2 = 2.^[0:nBits-1];
%# do a tiny bit of error-checking
if data > sum(powOf2)
error('not enough bits to represent the data')
end
out = false(1,nBits);
ct = nBits;
while data>0
if data >= powOf2(ct)
data = data-powOf2(ct);
out(ct) = true;
end
ct = ct - 1;
end
To use:
out = binary2vector(12,6)
out =
0 0 1 1 0 0
out = binary2vector(22,6)
out =
0 1 1 0 1 0
Are you using this for IEEE 802.11 SIGNAL field? I noticed "length_field" and "16".
Anyway here's how I do it.
function [Ibase2]= Convert10to2(Ibase10,n)
% Convert the integral part by successive divisions by 2
Ibase2=[];
if (Ibase10~=0)
while (Ibase10>0)
q=fix(Ibase10/2);
r=Ibase10-2*q;
Ibase2=[r Ibase2];
Ibase10=q;
end
else
Ibase2=0;
end
o = length(Ibase2);
% append redundant zeros
Ibase2 = [zeros(1,n-o) Ibase2];