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];
Related
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)
output is a 3d matrix with size(output) == [height width N] and input is a 2d matrix with size(input) == [height width] . I need to implement the following code in one line.
for k = 1:size(output,3)
f = output(:,:,k);
i_zero = (f==0);
f(is_zero) = input(is_zero);
output(:,:,k) = f;
end
bsxfun approach -
output = bsxfun(#times,output==0,input) + output
Alternative approach -
output = (output==0).*input(:,:,ones(1,N))+ output
I hope the "I need to implement" is not a homework.
Here goes a solution that should solve your problem although not in one line.
new_input=repmat(input,1,1,size(output,3));
output(output==0)=new_input(output==0);
All the answers solve the problem when there is a exact comparison to 0 (as OP required) but for the sake of generalization if you intend to change for another comparison be aware that not all methods work in the same way.
Example below:
CODE:
%Simulation
output=rand(10,10,3);
input=rand(10,10);
% output=randi(9,10,10,3);
% input=randi(9,10,10);
%OP code
output2=[]
for k = 1:size(output,3)
f = output(:,:,k);
i_zero = (f<0.5);
f(i_zero) = input(i_zero);
output2(:,:,k) = f;
end
%repmat code
output3=output;
new_input=repmat(input,1,1,size(output,3));
output3(output<0.5)=new_input(output<0.5);
any(output2(:)-output3(:))
%bsxfun code
output4 = bsxfun(#times,output<0.5,input) + output;
any(output2(:)-output4(:))
%other variation code
output5 = (output<0.5).*input(:,:,ones(1,size(output,3)))+ output;
any(output2(:)-output5(:))
% bultin code
output6=output;
output6(output<0.5)=builtin('_paren',repmat(input,[1,1,size(output,3)]),output<0.5);
any(output2(:)-output6(:))
'-----'
any(abs(output2(:)-output3(:))>eps)
any(abs(output2(:)-output4(:))>eps)
any(abs(output2(:)-output5(:))>eps)
any(abs(output2(:)-output6(:))>eps)
'-----'
sum(abs(output2(:)-output3(:)))
sum(abs(output2(:)-output4(:)))
sum(abs(output2(:)-output5(:)))
sum(abs(output2(:)-output6(:)))
OUTPUT:
ans =
0
ans =
1
ans =
1
ans =
0
-----
ans =
0
ans =
1
ans =
1
ans =
0
-----
ans =
0
ans =
150.5088
ans =
150.5088
ans =
0
If you insist on a single line solution, you can use the (:) operator along with the mod command:
output(output(:)==0) = input(mod(find(output(:)==0)-1,height*width)+1)
where the -1 and +1 are in order to avoid index 0
Here is in a one liner... but uses the undocumented builtin('_paren',... to subscript reference to the output of a function
output(output==0)=builtin('_paren',repmat(input,[1,1,N]),output==0)
without the the undocumented builtin this method gets messy if you want it in one line...
output=subsasgn(output,struct('type','()','subs',{{output==0}}),...
subsref(repmat(input,[1,1,N]),struct('type','()','subs',{{output==0}})))
...sadly I forgot using masks and adding two matrices together was an option...
You should have the 'cat' function:
http://www.mathworks.nl/help/matlab/ref/cat.html
cat(3,matrix1,matrix2,...) to concatenate along the 3rd dimension.
I'm implementing some sort of flood fill algorithm in Matlab that, given a starting pixel in a binary image, will output a binary image containing only the pixels that can be directly connected to it.
Basically, let
foo =
1 1 1
1 0 0
1 0 1
calling flood_fill(foo,1,1) would yield
1 1 1
1 0 0
1 0 0
Now, I'm pretty new to Matlab. I initially implemented flood_fill in a recursive style, but the pass-by-value behaviour of Matlab made working with large images very inefficient. To fix this, I reimplemented flood_fill like this
function [outImage] = flood_fill(inImage,start_x,start_y)
width = size(inImage,2);
height = size(inImage,1);
outImage = zeros(height,width);
points = [];
points = [points; [start_x start_y]];
while size(points,1)>0
point = points(1,:);
points = points(2:end,:);
y=point(2);
x=point(1);
outImage(y,x)=1;
if (y>1 && (outImage(y-1,x)==0) && (inImage(y-1,x)==1))
points = [points; [x y-1]];
end
if (y<height && (outImage(y+1,x)==0) && (inImage(y+1,x)==1))
points = [points; [x y+1]];
end
if (x>1 && (outImage(y,x-1)==0) && (inImage(y,x-1)==1))
points = [points; [x-1 y]];
end
if (x<width && (outImage(y,x+1)==0) && (inImage(y,x+1)==1))
points = [points; [x+1 y]];
end
end
end
Now, this works on small matrices/images but takes forever on large images as well. I suspect the reason why is the large amount of array resizes going on. Normally (in C++ for example), I'd use an unordered linked list and use it as a stack (remove from and insert at the head) to avoid costly array resizes.
Is there an equivalent to such a data structure in Matlab? If not, what's a Matlab idiom I could use? Perhaps a built-in function?
The function that you search for called bwselect:
foo=[1 1 1; 1 0 0; 1 0 1]
b=bwselect(foo,1, 1)
Note that you can define also fourth input n (like that: bwselect(foo,1,1,n)), that can have a value of 4 to specify 4-connected region, or 8 to specify 8-connected region.
Adiel answered your second question "Perhaps a built-in function?". As for the first part:
I'm not familiar with linked lists in MATLAB. However, you can speed up your function significantly by initializing the size of the points-matrix and don't change the size after that. Pre-initialization should always be done in MATLAB. If the function won't work with matrices of fixed size, I would always recommend you to try to rewrite the function.
For your specific case:
function [outImage] = flood_fill(inImage,start_x,start_y)
width = size(inImage,2);
height = size(inImage,1);
outImage = zeros(height,width);
points = zeros(nnz(inImage),2); % I take it this is the maximum size
points(1,:) = [start_x start_y];
k = 1; % Increment row number in points
while size(points,1)>0
k = k + 1;
y=points(k, 2);
x=points(k, 1);
I understand it you have programming skills in general, so I believe you should be able to adapt the remaining code to the new format. (I don't have time to go through it and rewrite it). I'm quite sure it will run much faster!
i actually want to generate crc in matlab for Modbus protocol and i have used following code in matlab. I have also given message array as message=uint16([hex2dec('01') hex2dec('02') hex2dec('00') hex2dec('C4') hex2dec('00') hex2dec('16')]); and done bitand with 0xffff at the end, but it is unable to give correct crc..
My code is as below and the expected crc is B839 as per the Modbus crc calculator but it is giving B8DD(47325 decimal). Please help me if there is anything to change in the code. Thank you.
function crc_val = crc3 (~)
crc = uint16(hex2dec('1D0F')); % Non-augmented initial value equivalent to augmented initial value 0xFFFF
polynomial = hex2dec('1021'); % Polynomial
message=uint16([hex2dec('01') hex2dec('02') hex2dec('00') hex2dec('C4') hex2dec('00') hex2dec('16') hex2dec('00') hex2dec('00')]);
for i = 1:(length(message)-2) % Not taking the last 2 bytes because they are the CRC.
crc = bitxor(crc, bitsll(message(i), 8));
for j = 1:8
if (bitand(crc, hex2dec('8000')) > 0);
crc = bitxor(bitsll(crc, 1), polynomial);
else
crc = bitsll(crc, 1);
end
end
end
crc_val = bitand(crc, hex2dec('ffff'));
end
Did you try this. It is available as BSD license. You would not face any possible licensing issues. The following explains how CRC actually works. The following also helps understand the concept.
%usage: crc16(input vector).
//
function [resto] = crc16(h)
% g(X) = X^16+X^15+X^2+1
gx = [1 1 0 0 0 0 0 0 0 0 0 0 0 1 0 1];
% P(X) is given input vector
px = h;
%Calculate P(x)x^r
pxr=[px zeros(1,length(gx)-1)];
% deconvolve (r), entre pxr y gx
[c r]=deconv(pxr,gx);
r=mod(abs(r),2);
% returncrc-16
resto=r(length(px)+1:end);
end
I'm simulating equations of motion for a (somewhat odd) system with mass-springs and double pendulum, for which I have a mass matrix and function f(x), and call ode45 to solve
M*x' = f(x,t);
I have 5 state variables, q= [ QDot, phi, phiDot, r, rDot]'; (removed Q because nothing depends on it, QDot is current.)
Now, to calculate some forces, I would like to also save the calculated values of rDotDot, which ode45 calculates for each integration step, however ode45 doesn't give this back. I've searched around a bit, but the only two solutions I've found are to
a) turn this into a 3rd order problem and add phiDotDot and rDotDot to the state vector. I would like to avoid this as much as possible, as it's already non-linear and this really makes matters a lot worse and blows up computation time.
b) augment the state to directly calculate the function, as described here. However, in the example he says to make add a line of zeros in the mass matrix. It makes sense, since otherwise it will integrate the derivative and not just evaluate it at the one point, but on the other hand it makes the mass matrix singular. Doesn't seem to work for me...
This seems like such a basic thing (to want the derivative values of the state vector), is there something quite obvious that I'm not thinking of? (or something not so obvious would be ok too....)
Oh, and global variables are not so great because ode45 calls the f() function several time while refining it's step, so the sizes of the global variable and the returned state vector q don't match at all.
In case someone needs it, here's the code:
%(Initialization of parameters are above this line)
options = odeset('Mass',#massMatrix);
[T,q] = ode45(#f, tspan,q0,options);
function dqdt = f(t,q,p)
% q = [qDot phi phiDot r rDot]';
dqdt = zeros(size(q));
dqdt(1) = -R/L*q(1) - kb/L*q(3) +vs/L;
dqdt(2) = q(3);
dqdt(3) = kt*q(1) + mp*sin(q(2))*lp*g;
dqdt(4) = q(5);
dqdt(5) = mp*lp*cos(q(2))*q(3)^2 - ks*q(4) - (mb+mp)*g;
end
function M = massMatrix(~,q)
M = [
1 0 0 0 0;
0 1 0 0 0;
0 0 mp*lp^2 0 -mp*lp*sin(q(2));
0 0 0 1 0;
0 0 mp*lp*sin(q(2)) 0 (mb+mp)
];
end
The easiest solution is to just re-run your function on each of the values returned by ode45.
The hard solution is to try to log your DotDots to some other place (a pre-allocated matrix or even an external file). The problem is that you might end up with unwanted data points if ode45 secretly does evaluations in weird places.
Since you are using nested functions, you can use their scoping rules to mimic the behavior of global variables.
It's easiest to (ab)use an output function for this purpose:
function solveODE
% ....
%(Initialization of parameters are above this line)
% initialize "global" variable
rDotDot = [];
% Specify output function
options = odeset(...
'Mass', #massMatrix,...
'OutputFcn', #outputFcn);
% solve ODE
[T,q] = ode45(#f, tspan,q0,options);
% show the rDotDots
rDotDot
% derivative
function dqdt = f(~,q)
% q = [qDot phi phiDot r rDot]';
dqdt = [...
-R/L*q(1) - kb/L*q(3) + vs/L
q(3)
kt*q(1) + mp*sin(q(2))*lp*g
q(5)
mp*lp*cos(q(2))*q(3)^2 - ks*q(4) - (mb+mp)*g
];
end % q-dot function
% mass matrix
function M = massMatrix(~,q)
M = [
1 0 0 0 0;
0 1 0 0 0;
0 0 mp*lp^2 0 -mp*lp*sin(q(2));
0 0 0 1 0;
0 0 mp*lp*sin(q(2)) 0 (mb+mp)
];
end % mass matrix function
% the output function collects values for rDotDot at the initial step
% and each sucessful step
function status = outputFcn(t,q,flag)
status = 0;
% at initialization, and after each succesful step
if isempty(flag) || strcmp(flag, 'init')
deriv = f(t,q);
rDotDot(end+1) = deriv(end);
end
end % output function
end
The output function only computes the derivatives at the initial and all successful steps, so it's basically doing the same as what Adrian Ratnapala suggested; re-run the derivative at each of the outputs of ode45; I think that would even be more elegant (+1 for Adrian).
The output function approach has the advantage that you can plot the rDotDot values while the integration is being run (don't forget a drawnow!), which can be very useful for long-running integrations.