I am familiar with iterative methods on paper, but MATLAB coding is relatively new to me and I cannot seem to find a way to code this.
In code language...
This is essentially what I have:
A = { [1;1] [2;1] [3;1] ... [33;1]
[1;2] [2;2] [3;2] ... [33;2]
... ... ... ... ....
[1;29] [2;29] [3;29] ... [33;29] }
... a 29x33 cell array of 2x1 column vectors, which I got from:
[X,Y] = meshgrid([1:33],[1:29])
A = squeeze(num2cell(permute(cat(3,X,Y),[3,1,2]),1))
[ Thanks to members of stackOverflow who helped me do this ]
I have a function that calls each of these column vectors and returns a single value. I want to institute a 2-D 5-point stencil method that evaluates a column vector and its 4 neighbors and finds the maximum value attained through the function out of those 5 column vectors.
i.e. if I was starting from the middle, the points evaluated would be :
1.
A{15,17}(1)
A{15,17}(2)
2.
A{14,17}(1)
A{14,17}(2)
3.
A{15,16}(1)
A{15,16}(2)
4.
A{16,17}(1)
A{16,17}(2)
5.
A{15,18}(1)
A{15,18}(2)
Out of these 5 points, the method would choose the one with the largest returned value from the function, move to that point, and rerun the method. This would continue on until a global maximum is reached. It's basically an iterative optimization method (albeit a primitive one). Note: I don't have access to the optimization toolbox.
Thanks a lot guys.
EDIT: sorry I didn't read the iterative part of your Q properly. Maybe someone else wants to use this as a template for a real answer, I'm too busy to do so now.
One solution using for loops (there might be a more elegant one):
overallmax=0;
for v=2:size(A,1)-1
for w=2:size(A,2)-1
% temp is the horizontal part of the "plus" stencil
temp=A((v-1):(v+1),w);
tmpmax=max(cat(1,temp{:}));
temp2=A(v,(w-1):(w+1));
% temp2 is the vertical part of the "plus" stencil
tmpmax2=max(cat(1,temp2{:}));
mxmx=max(tmpmax,tmpmax2);
if mxmx>overallmax
overallmax=mxmx;
end
end
end
But if you're just looking for max value, this is equivalent to:
maxoverall=max(cat(1,A{:}));
Related
I'm trying to change the number of items in array, over which a for loop is running, during the for loop, with the objective that this changes the number of loops. In a very simplified version, the code would look something like this:
var loopArray: [Int] = []
loopArray.append(1)
loopArray.append(2)
loopArray.append(3)
loopArray.append(4)
loopArray.append(5)
for x in 0..<Int(loopArray.count) {
print(x)
if x == 4 {
loopArray.append(6)
}
}
When running this code, 5 numbers are printed, and while the number 6 is added to the Array, the loopArray.count does not seem to update. How can I make the .count dynamic?
This is a very simplified example, in the project I'm working on, appending numbers to the array depends on conditions that may or may not be met.
I have looked for examples online, but have not been able to find any similar cases. Any help or guidance is much appreciated.
sfung3 gives the correct way to do what you want, but I think there needs to be a bit of explanation as to why your solution doesn't work
The line
for x in 0..<Int(loopArray.count)
only evaluates loopArray.count once, the first time it is hit. This is because of the way for works. Conceptually a for loop iterates through the elements of a sequence. The syntax is something like
for x in s
where
s is a sequence, give it type S
x is a let constant (you can also make it a var but that is not relevant to the current discussion) with type S.Element
So the bit after the in is a sequence - any sequence. There's nothing special about the use of ..< here, it's just a convenient way to construct a sequence of consecutive integers. In fact, it constructs a Range (btw, you don't need the cast to Int, Array.count is already an Int).
The range is only constructed when you first hit the loop and it's effectively a constant because Range is a value type.
If you don't want to use Joakim's answer, you could create your own reference type (class) that conforms to Sequence and whose elements are Int and update the upper bound each time through the loop, but that seems like a lot of work to avoid a while loop.
you can use a while loop instead of a for loop.
var i = 0
while i < loopArray.count {
print(i)
if i == 4 {
loopArray.append(6)
}
i += 1
}
which prints
0 1 2 3 4 5
Working in MATLAB R2017a. I'm trying to optimise a piece of code I'm working on. It uses arrays to store field values on a grid.
In order to create a specific function in a field array I originally used the straight forward method of two for loops iterating over all the array elements. But i know for loops are slow so since then I came back and tried my best to remove them. However I could only manage to remove one of the loops; leaving me with this:
for n = 1:1:K
%%% define initial pertubation
t=n*dt;
% create array for source Ez field.
xtemps = (1:Ng)*dX;
for k = 1:Ng
ztemp = k*dX;
Ez0(k,:) = THzamp * (1/(1+exp(-(t-stepuppos)))) * exp(-((xtemps-...
THzstartx).^2)./(bx^2)) .* (t-((ztemp-THzstartz)/vg))*exp(-((t-((ztemp-...
THzstartz)/vg))^2)/(bt^2));
end
The important bit here is the last 5 lines, but I figured the stuff before might be important for context. I've removed the for loop looping over the x coordinates. I want to vectorize the z/k for loop but I can't figure out how to distinguish between the dimensions with the array oporators.
Edit: THzamp, stepuppos, bx, bt, THzstartz, THzstartx are all just scalars, they control the function (Ez0) I'm trying to create. dX and t are also just scalars. Ez0 is a square array of size Ng.
What I want to achieve is to remove the for loop that loops over k, so that that the values of ztemp are defined in a vector (like xtemps already is), rather than individually in the loop. However, I don't know how I'd write the definition of Ez0 in that case.
First time posting here, if I'm doing it wrong let me know. If you need more info just ask.
It isn't clear if n is used in the other headers and as stated in the comments your sizes aren't properly defined so you'll have to ensure the sizes are correct.
However, you can give this vectorize code a try.
n = 1:K
%%% define initial pertubation
t=n*dt;
% create array for source Ez field.
xtemps = (1:Ng)*dX;
for k = 1:Ng
ztemp = k*dX;
Ez0(k,:) = THzamp .* (1./(1+exp(-(t-stepuppos)))) .* exp(-((xtemps-...
THzstartx).^2)./(bx^2)) .* (t-((ztemp-THzstartz)/vg)).*exp(-((t-((ztemp-...
THzstartz)/vg)).^2)/(bt.^2));
end
So now t has the size K you'll need to ensure stepupposand (ztemp-THzstartz)/vg) have the same size K. Also you can take a look at vectors vs array operators here.
I have a typical question in dynamic programming.
My question is given an array = {1,2,3,4,5,6}, I have to find all the arrays whose sum is atmost k. If I consider all the sets, it will become exponential alogorthm. I thought of achiveng this by Dynamic Programming.
Suppose f k =7,
My idea is
Pass 1: {1],{2}....{6}
Pass 2: Pass1 + {1,2},{1,3},{1,4},{1,5}
Pass 3: Pass2 + {1,2,3},
And my algo stops.
Im not able to formulate this with dynamic programming. Any inputs?? How to formulate this algo into program?
A DP solution for the problem should follow the next recursive formula, and build bottom-up:
f(i,0) = {{}} //a set containing only an empty set
f(0,W) = {{}} (W > 0)
f(0,W) = {} (W < 0) //an empty set
f(i,W) = f(i-1,W) [union] extend(f(i-1,w-element[i]),element[i])
Where the function extend(set,e) is:
extend(set,e):
for each s in set: //s is a set itself
s.add(e)
Note that complexity could still be exponential (and not even pseudo-polynomial), since the number of sets generated could be exponential, and is stored in the DP table.
your problem is an instance of the knapsack problem whose related decision problem is known to be NP-complete. this means that most certainly there will be no sub-exponential algorithm (though a mathematical proof is missing ).
ZachLangleys comment shows that the enumeration of all solutions would still be exponential in the worst case even if there was an efficient problem solver since producing the output already requires exponential time.
since the decision problem is NP-complete, counting can not be easier (otherwise you'd count and afterwards test the result whether it equals 0 or not).
Is there any way to "vector" assign an array of struct.
Currently I can
edges(1000000) = struct('weight',1.0); //This really does not assign the value, I checked on 2009A.
for i=1:1000000; edges(i).weight=1.0; end;
But that is slow, I want to do something more like
edges(:).weight=[rand(1000000,1)]; //with or without the square brackets.
Any ideas/suggestions to vectorize this assignment, so that it will be faster.
Thanks in advance.
This is much faster than deal or a loop (at least on my system):
N=10000;
edge(N) = struct('weight',1.0); % initialize the array
values = rand(1,N); % set the values as a vector
W = mat2cell(values, 1,ones(1,N)); % convert values to a cell
[edge(:).weight] = W{:};
Using curly braces on the right gives a comma separated value list of all the values in W (i.e. N outputs) and using square braces on the right assigns those N outputs to the N values in edge(:).weight.
You can try using the Matlab function deal, but I found it requires to tweak the input a little (using this question: In Matlab, for a multiple input function, how to use a single input as multiple inputs?), maybe there is something simpler.
n=100000;
edges(n)=struct('weight',1.0);
m=mat2cell(rand(n,1),ones(n,1),1);
[edges(:).weight]=deal(m{:});
Also I found that this is not nearly as fast as the for loop on my computer (~0.35s for deal versus ~0.05s for the loop) presumably because of the call to mat2cell. The difference in speed is reduced if you use this more than once but it stays in favor of the for loop.
You could simply write:
edges = struct('weight', num2cell(rand(1000000,1)));
Is there something requiring you to particularly use a struct in this way?
Consider replacing your array of structs with simply a separate array for each member of the struct.
weights = rand(1, 1000);
If you have a struct member which is an array, you can make an extra dimension:
matrices = rand(3, 3, 1000);
If you just want to keep things neat, you could put these arrays into a struct:
edges.weights = weights;
edges.matrices = matrices;
But if you need to keep an array of structs, I think you can do
[edges.weight] = rand(1, 1000);
The reason that the structs in your example don't get initialized properly is that the syntax you're using only addresses the very last element in the struct array. For a nonexistent array, the rest of them get implicitly filled in with structs that have the default value [] in all their fields.
To make this behavior clear, try doing a short array with clear edges; edges(1:3) = struct('weight',1.0) and looking at each of edges(1), edges(2), and edges(3). The edges(3) element has 1.0 in its weight like you want; the others have [].
The syntax for efficiently initializing an array of structs is one of these.
% Using repmat and full assignment
edges = repmat(struct('weight', 1.0), [1 1000]);
% Using indexing
% NOTE: Only correct if variable is uninitialized!!!
edges(1:1000) = struct('weight', 1.0); % QUESTIONABLE
Note the 1:1000 instead of just 1000 when indexing in to the uninitialized edges array.
There's a problem with the edges(1:1000) form: if edges is already initialized, this syntax will just update the values of selected elements. If edges has more than 1000 elements, the others will be left unchanged, and your code will be buggy. Or if edges is a different type, you could get an error or weird behavior depending on its existing datatype. To be safe, you need to do clear edges before initializing using the indexing syntax. So it's better to just do full assignment with the repmat form.
BUT: Regardless of how you initialize it, an array-of-structs like this is always going to be inherently slow to work with for larger data sets. You can't do real "vectorized" operations on it because your primitive arrays are all broken up in to separate mxArrays inside each struct element. That includes the field assignment in your question – it is not possible to vectorize that. Instead, you should switch a struct-of-arrays like Brian L's answer suggests.
You can use a reverse struct and then do all operations without any errors
like this
x.E(1)=1;
x.E(2)=3;
x.E(2)=8;
x.E(3)=5;
and then the operation like the following
x.E
ans =
3 8 5
or like this
x.E(1:2)=2
x =
E: [2 2 5]
or maybe this
x.E(1:3)=[2,3,4]*5
x =
E: [10 15 20]
It is really faster than for_loop and you do not need other big functions to slow your program.
Hey guys. I have this question to ask. In C programming, if we want to store several values in an array, we implement that using loops like this:
j=0; //initialize
for (idx=1,idx less than a constant; idex++)
{
slope[j]=(y2-y1)/(x2-x1);
j++;
}
My question is in Matlab do we have any simpler way to get the same array 'slope' without manually increasing j? Something like:
for idx=1:constant
slope[]=(y2-y1)/(x2-x1);
Thank you!
Such operations can usually be done without looping.
For example, if the slope is the same for all entries, you can write
slope = ones(numRows,numCols) * (y2-y1)/(x2-x1);
where numRows and numCols are the size of the array slope.
If you have a list of y-values and x-values, and you want the slope at every point, you can call
slope = (y(2:end)-y(1:end-1))./(x(2:end)-x(1:end-1)
and get everything in one go. Note that y(2:end) are all elements from the second to the last, and y(1:end-1) are all elements from the first to the second to last. Thus, the first element of the slope is calculated from the difference between the second and the first element of y. Also, note the ./ instead of /. The dot makes it an element-wise operation, meaning that I divide the first element of the array in the numerator by the first element of the array in the denominator, etc.