What does x(1) mean in matlab? - matlab
I'm really new to matlab and I need some help, I know java and python. Here's a code I want to understand
x(1) = 0
y(1) = 0
i = 1
x(i+1)=x(i)+vx*t+.5*a*t^2;
i=i+1;
end
I want to know what is happening here, "vx", "t" and "a" are variables tho
x(1) = 0 and y(1) = 0 is very similar to initializing a list / array in Python where x and y are the list variables and the first position is 1. MATLAB starts indexing at 1, and not 0 like in Java and Python. The similar syntax in Java or Python is: x[0] = 0; y[0] = 0. MATLAB uses round braces to index into an array / vector / list.
i = 1
x(i+1)=x(i)+vx*t+.5*a*t^2;
i=i+1;
This is pretty simple. i is some sort of loop variable or index variable... which you failed to show us that this code is probably part of a loop. The code just sets the next value of x or the second element in the array or list to x(i) + vx*t + 0.5*a*t^2. The * operator is multiplication and the ^ is the exponentiation operator. In Python, this is equivalent to saying x[i] + vx*t + 0.5*a*(t**2). Now the origin of this equation actually comes from calculating the displacement of a body mass using Newtonian physics - the Kinematic Equations actually. As such vx is the velocity of the body mass and a is the acceleration. t would be the time point you are considering. Specifically, the displacement can be calculated as follows:
Source: The Physics Classroom - Kinematic Equations
Look at the top left equation as that is what the statement is doing in the code. This only calculates the displacement at one point in time. Therefore, what you are doing is adding the displacement that this body mass encounters at some fixed point t a certain amount of times. x captures the total displacement overall from the beginning to end. The next statement after is just incrementing the loop counter. The thing about MATLAB is that you can dynamically extend the length of a list / array / vector whereas Python or Java will give you an out-of-bounds error. However, it's recommended you should pre-allocate the right amount of memory to use before using it for efficiency. See this informative post for more details: Efficient Array Preallocation in MATLAB.
Related
How to make vectors be of same length in my Matlab script?
I am trying to plot some data. The script I wrote below has worked fine before, but now I have no idea why it's not working. Here is the code: x = [335,41,14,18,15,9,7,9,20607,5,5,143,3,5,72,134,2,28,172,3,72,173,280,186,20924,1,1,22,3,3,1,2,13,1,3,2,11,66,12983,176,123,192,64,258,182,123,299,58,198,7,113,342,72,8376,122,20,19,2,3,28,8,36,8,56,43,2,48,127,395,4664,186,46,236,219,258,69,203,189,169,72,100,78,109,46,112,3929,272,40,4,31,2,97,36,5,35,56,2,237,1672,256,224,28,163,341,151,263,157,397,94,380,173,75,87,272,1194,133,6,112,1,6,2,26,25,64,8,40,57,106,525,150,248,125,269,264,256,357,153,64,152,283,1,2,2,454,154,39,1,1,64,151,242,1,18,99,1,36,607,55,54,110,225,108,37,1,144,162,137,107,21,360,362,18,51,25,43,1,3,6,1,27,7,45,326,32,103,50,124,155,39,180,143,33,116,46,7,151,120,19,4,2,4,110,2,7,4,9,4,27,216,323,148,1,1,2,1,47,113,150,1,2,144,16,4827,1,1,1,14]; size = length(x); disp(size); z = 0; for i = 1:size z = z + 1; y(i) = z; end scatter(x,y); This code should ensure that y is of same length as x as we are only filling in y as long as x is (since we are using a for loop from 1 through to size, where size is basically the number of indices in x), but I keep getting this error. I checked with disp and it turns out that my x and y vectors have different lengths, x is 227 and y is 256. Can anyone help me out with this trivial issue?
This is most likely because y was created to be a different size before you ran that piece of code you showed us. Somewhere in your script before you call this piece of code, y was created to be a 256 element vector and now you are reusing this variable in this part of the code by populating elements in the y vector. The variable x has 227 elements and the loop you wrote will change y's first 227 elements as you have looped for as many times as there are elements in x. However, the remaining 29 elements are still there from before. The reusing of the variable y is probably why your script is failing as now the sizes between both variables are not the same. As such, explicitly recreate y before calling scatter. Actually, that for loop is not needed at all. The purpose of the loop is to create an increasing array from 1 up to as many elements as you have in x. Just do this instead: y = 1:size; I also do not like that you are creating a variable called size. It is overshadowing the function size, which finds the number of elements in each dimension that the input array contains. With the recommendations I've stated above, replace your entire code with this: x = [335,41,14,18,15,9,7,9,20607,5,5,143,3,5,72,134,2,28,172,3,72,173,280,186,20924,1,1,22,3,3,1,2,13,1,3,2,11,66,12983,176,123,192,64,258,182,123,299,58,198,7,113,342,72,8376,122,20,19,2,3,28,8,36,8,56,43,2,48,127,395,4664,186,46,236,219,258,69,203,189,169,72,100,78,109,46,112,3929,272,40,4,31,2,97,36,5,35,56,2,237,1672,256,224,28,163,341,151,263,157,397,94,380,173,75,87,272,1194,133,6,112,1,6,2,26,25,64,8,40,57,106,525,150,248,125,269,264,256,357,153,64,152,283,1,2,2,454,154,39,1,1,64,151,242,1,18,99,1,36,607,55,54,110,225,108,37,1,144,162,137,107,21,360,362,18,51,25,43,1,3,6,1,27,7,45,326,32,103,50,124,155,39,180,143,33,116,46,7,151,120,19,4,2,4,110,2,7,4,9,4,27,216,323,148,1,1,2,1,47,113,150,1,2,144,16,4827,1,1,1,14]; numX = numel(x); y = 1 : numX; scatter(x,y); The vector y is now explicitly created instead of reusing the variable that was created with a previous size in the past. It also uses the colon operator to explicitly create this sequence instead of using a for loop. That for loop is just not needed. numel determines the total number of elements for an input matrix. I don't like using length as a personal preference because it finds the number of elements in the largest dimension. This may work fine for vectors, but it has really made some hard to spot bugs in code that I've written in the past.
Calculating vector lengths within my loop
The code is meant to draw a line with each click of the mouse made within the figure. Here is my code; b=1 while b>0; axis([-10, 10, -10, 10]) b=b+1 [x(b),y(b)]=ginput(1) plot(x,y,x,y) end However I cant get my head around how I can add the vectors seeing as some are + and some are -, I need to turn the negatives into positives. I need to add code that will give me an overall combined length after all the mouse clicks. Maybe I am just thinking about it completely wrong. I have tried; length=(sqrt(x.^2)+(y.^2)) I was hoping this would give me the correct vector length accept unless I click an exact straight line.
So assuming that x and y are vectors of sequential points, if you want to get the total distance then you need to take the sum of the distance between each point i.e. Σi(sqrt((xi-xi-1)2+(yi-yi-1)2)) in Matlab we can calculate all the x (and y) differences in one step using the diff function so in the end we get length = sum(sqrt(diff(x).^2 + diff(y).^2))
Your problem is twofold: there's a typo, I think instead of length=(sqrt(x.^2)+(y.^2)) you mean length=sqrt((x.^2)+(y.^2)) The second problem is, that this does not actually calculate the length of your path, instead, you probably want something like clear all b=1; radius = 3; while b>0; axis([-10, 10, -10, 10]) b=b+1; [x(b),y(b)]=ginput(1); plot(x,y,x,y); length(b)=sqrt((x(b)-x(b-1))^2+(y(b)-y(b-1))^2); if sqrt(x(b)^2 + y(b)^2) < radius; break; end end sum(length) which calculates the length for each new piece you add and sums them all up. As soon as you click within "radius" distance of 0 the while loop breaks. Also, generally it's good practice to preallocate variables, no big deal here, cause your arrays are small, just saying. Note: Dan's solution gives you a vectorized way of calculating the total length in one step, so in case you don't need the individual path lengths this is the more concise way to go.
Indexing issue in MATLAB (image processing)
I'm dealing with a predictive block-matching motion estimation algorithm. This means, the values of motion vectors are found using the previously found values and I am stuck with a really trivial thing. I'm dealing with images divided into blocks, so I should have a motion vector for each block. I created a 2xN matrix motion_vectors, where N is the number of all blocks (blocks_in_the_first_row*blocks_in_the_first_column). The first row is the x coordinate and second row the y coordinate of the motion vector. I have 2 predictors to help to estimate the motion vector of the current block. If my current block is at position (i,j) then the positions of the predictors are (i, j-1) (the block "on top)" and (i-1, j) (the block on the left). My problem is, that I can't figure out a way how to adress the predictor blocks (in for loops) in motion_vectors since the dimensions are different (one is a 2xN matrix, the other blocks_in_row x blocks_in_column). I also wouldn't like to change the dimensions of motion_vectors, since then I would need a two-"layer" array. One for the x coordinates and one for y, but that doesn't fit to the further design. I hope I made myself understandable, if not, please let me know. Thanks for any clues!
If you're accessing an element of motion_vectors, you're getting data about a corresponding block. That means that there's a system of translating between an index 1 through N, where N is blocks_in_row*blocks_in_column, and a specific block. If index 1 is the top-left block, index 2 is the block to its right, and you increment as reading a book (left-to-right and wrap to the next row), then you would translate as follows: row_of_block = floor((index_in_motion_vector-1)/number_of_columns) + 1 col_of_block = mod((index_in_motion_vector-1), number_of_columns) + 1 (This is called row-major ordering.) If instead index 1 is the top-left block and index 2 is the block below it, and you wrap to the top of the next column when done with one, then the conversion would be row_of_block = mod((index_in_motion_vector-1), number_of_rows) + 1 col_of_block = floor((index_in_motion_vector-1)/number_of_rows) + 1 (This is called column-major ordering, and is what MATLAB uses by default.) So, if you're iterating 1 to N, you can just use those conversions. If you'd like to iterate 1 through number_of_rows, and 1 through number_of_columns you would do the opposite. If you're using the book-like indexing of blocks (row-major ordering), the conversion to index of the motion vector would be col_in_motion_vector = (row_of_block-1)*number_of_columns + column_of_block If you're using the second, top-to-bottom-and-wrap method of indexing blocks (column-major ordering), the conversion would instead be col_in_motion_vector = (column_of_block-1)*number_of_rows + row_of_block
How to generate random matlab vector with these constraints
I'm having trouble creating a random vector V in Matlab subject to the following set of constraints: (given parameters N,D, L, and theta) The vector V must be N units long The elements must have an average of theta No 2 successive elements may differ by more than +/-10 D == sum(L*cosd(V-theta)) I'm having the most problems with the last one. Any ideas? Edit Solutions in other languages or equation form are equally acceptable. Matlab is just a convenient prototyping tool for me, but the final algorithm will be in java. Edit From the comments and initial answers I want to add some clarifications and initial thoughts. I am not seeking a 'truly random' solution from any standard distribution. I want a pseudo randomly generated sequence of values that satisfy the constraints given a parameter set. The system I'm trying to approximate is a chain of N links of link length L where the end of the chain is D away from the other end in the direction of theta. My initial insight here is that theta can be removed from consideration until the end, since (2) in essence adds theta to every element of a 0 mean vector V (shifting the mean to theta) and (4) simply removes that mean again. So, if you can find a solution for theta=0, the problem is solved for all theta. As requested, here is a reasonable range of parameters (not hard constraints, but typical values): 5<N<200 3<D<150 L==1 0 < theta < 360
I would start by creating a "valid" vector. That should be possible - say calculate it for every entry to have the same value. Once you got that vector I would apply some transformations to "shuffle" it. "Rejection sampling" is the keyword - if the shuffle would violate one of your rules you just don't do it. As transformations I come up with: switch two entries modify the value of one entry and modify a second one to keep the 4th condition (Theoretically you could just shuffle two till the condition is fulfilled - but the chance that happens is quite low) But maybe you can find some more. Do this reasonable often and you get a "valid" random vector. Theoretically you should be able to get all valid vectors - practically you could try to construct several "start" vectors so it won't take that long.
Here's a way of doing it. It is clear that not all combinations of theta, N, L and D are valid. It is also clear that you're trying to simulate random objects that are quite complex. You will probably have a hard time showing anything useful with respect to these vectors.
The series you're trying to simulate seems similar to the Wiener process. So I started with that, you can start with anything that is random yet reasonable. I then use that as a starting point for an optimization that tries to satisfy 2,3 and 4. The closer your initial value to a valid vector (satisfying all your conditions) the better the convergence.
function series = generate_series(D, L, N,theta)
s(1) = theta;
for i=2:N,
s(i) = s(i-1) + randn(1,1);
end
f = #(x)objective(x,D,L,N,theta)
q = optimset('Display','iter','TolFun',1e-10,'MaxFunEvals',Inf,'MaxIter',Inf)
[sf,val] = fminunc(f,s,q);
val
series = sf;
function value= objective(s,D,L,N,theta)
a = abs(mean(s)-theta);
b = abs(D-sum(L*cos(s-theta)));
c = 0;
for i=2:N,
u =abs(s(i)-s(i-1)) ;
if u>10,
c = c + u;
end
end
value = a^2 + b^2+ c^2;
It seems like you're trying to simulate something very complex/strange (a path of a given curvature?), see questions by other commenters. Still you will have to use your domain knowledge to connect D and L with a reasonable mu and sigma for the Wiener to act as initialization.
So based on your new requirements, it seems like what you're actually looking for is an ordered list of random angles, with a maximum change in angle of 10 degrees (which I first convert to radians), such that the distance and direction from start to end and link length and number of links are specified? Simulate an initial guess. It will not hold with the D and theta constraints (i.e. specified D and specified theta) angles = zeros(N, 1) for link = 2:N angles (link) = theta(link - 1) + (rand() - 0.5)*(10*pi/180) end Use genetic algorithm (or another optimization) to adjust the angles based on the following cost function: dx = sum(L*cos(angle)); dy = sum(L*sin(angle)); D = sqrt(dx^2 + dy^2); theta = atan2(dy/dx); the cost is now just the difference between the vector given by my D and theta above and the vector given by the specified D and theta (i.e. the inputs). You will still have to enforce the max change of 10 degrees rule, perhaps that should just make the cost function enormous if it is violated? Perhaps there is a cleaner way to specify sequence constraints in optimization algorithms (I don't know how). I feel like if you can find the right optimization with the right parameters this should be able to simulate your problem.
You don't give us a lot of detail to work with, so I'll assume the following:
random numbers are to be drawn from [-127+theta +127-theta]
all random numbers will be drawn from a uniform distribution
all random numbers will be of type int8
Then, for the first 3 requirements, you can use this:
N = 1e4;
theta = 40;
diffVal = 10;
g = #() randi([intmin('int8')+theta intmax('int8')-theta], 'int8') + theta;
V = [g(); zeros(N-1,1, 'int8')];
for ii = 2:N
V(ii) = g();
while abs(V(ii)-V(ii-1)) >= diffVal
V(ii) = g();
end
end
inline the anonymous function for more speed.
Now, the last requirement,
D == sum(L*cos(V-theta))
is a bit of a strange one...cos(V-theta) is a specific way to re-scale the data to the [-1 +1] interval, which the multiplication with L will then scale to [-L +L]. On first sight, you'd expect the sum to average out to 0.
However, the expected value of cos(x) when x is a random variable from a uniform distribution in [0 2*pi] is 2/pi (see here for example). Ignoring for the moment the fact that our limits are different from [0 2*pi], the expected value of sum(L*cos(V-theta)) would simply reduce to the constant value of 2*N*L/pi.
How you can force this to equal some other constant D is beyond me...can you perhaps elaborate on that a bit more?
Matlab problem with writing equations
i am having problem with writing equations. r = 25, k= 2, R = 50:25:600, DR = 0.5:0.5:4.0 h= r*[1-cos(asin((sqrt(2*R*DR+DR^2))+r*sin(acos(r-k)/r)/r))]-k but as a resault i get this: h = 1.9118e+001 +1.7545e+002i. I just start with Matlab. Thanks
What I get from what you've written is actually ??? Error using ==> mtimes Inner matrix dimensions must agree. which is correct because you're trying to multiply two row vectors by one another. Could you please show us the actual code you used? Anyway, supposing that's dealt with somehow, it looks to me as if you're feeding something to asin that's much bigger than 1. That'll give you complex results. Is the thing you're passing to asin perhaps meant to be divided by R^2 or DR^2 or something of the kind? You have a similar issue a bit later with the argument to acos. I also suspect that some of your * and ^ and / operators should actually be elementwise ones .*, .^, ./.
If you're trying to do as you said: so in first equation i used R= 50, DR = 0.5, r= 25, k=2 and i need to get h. In second equation i used R=75, DR=1.0, r=25, k=2...for a last equation i used R=600,DR=4.0,r=25,k=2. DR and R need to be the same length... so if R goes between 50 and 600 in increments of 25, DR should go from 0.5 to 12.5 in increments of 0.5, or 0.5 to 4.0 in increments of 0.1522... once you figure that out, be sure the add a period before every matrix multiplication operation (e.g. * or ^)
EDIT: formula adjusted slightly (bracketing) to reflect success in comment. When you say you want a table, I guess it is to be an R by DR table (since you have to vectors of different length). To do that you need to use R as a column vector (R' below) and multiply with * (not .*). When R doesn't appear in a term multiply by ones(size(R)) (or use repmat) to get DR into the right shape. To square DR by element, you need DR.^2. There seems to be a misplaced bracket for the acos, surely you divide by r before taking the acos. There must be a division by something like r in the asin (not r^2 because you've taken the sqrt). Finally, the last division by r is redundant as written, since you multiply by r at the same level just before. Anyway, if I do the following: h= r*(1-cos(asin((sqrt(2*R'*DR+ones(size(R))'*DR.^2)/r)+sin(acos((r-k)/r)))))-k I get an R by DR table. Results for small R,DR are real; higher R,DR are complex due to the argument of the first asin being >1. The first entry in the table is 4.56, as you require.