I have the following problem in my simulation.
A is an array 24 x 2. I am going to split it and get 4 or 12 array. It means that I group 6 or 2 array. It will be ok, if I use even "split" coefficient. If it is odd, I can"t split A.[ I can't group 5 or 7, because of 24/5=4*5 + 4 ( or 5*5 -1) or 24/7=7*3+3.
That's why I going to do the following:
If I have 24 x 2 and need group every 5 together:
block 1 : A(1,:), A(2,:),A(3,:),A(4,:),A(5,:)
block 2 : A(6,:), A(7,:),A(8,:),A(9,:),A(10,:)
block 3 : A(11,:), A(12,:),A(13,:),A(14,:),A(15,:)
block 4 : A(16,:), A(17,:),A(18,:),A(19,:),A(20,:)
block 5 : A(21,:), A(22,:),A(23,:),A(24,:), ?
As you can see the 5th block is not full, Matlab gives me an error. My idea is to create A(25,:)=0. For my simulation it will be ok.
I am going to simulate it as function:
A=rand(m,n)
w- # number of a vector that i would like group together ( in ex., it is `5`)
if mod(w,2)==0
if mod(m,2)==0
% do....
else
% remainder = 0
end
else
if mod(m,2)==0
% remainder = 0
else
%do...
end
I was going to simulate like above, but then I have noticed that it doesn't work. Because 24/10 = 2*10+4. So I should write something else
I can find the reminder as r = rem(24,5). As an example above, MatLab gives me r=4. then I can find a difference c= w-r =1 and after that, I don't know how to do that.
Could you suggest to me how to simulate such a calculation?
Determine the number of blocks needed, calculate the virtual amount of rows needed to fill these blocks, and add as many zero rows to A as the difference between the virtual and actual amount of rows. Since you didn't mention, what the actual output should look like (array, cell array, ...), I chose a reshaped array.
Here's the code:
m = 24;
n = 2;
w = 5;
A = rand(m, n)
% Determine number of blocks
n_blocks = ceil(m / w);
% Add zero rows to A
A(m+1:w*n_blocks, :) = 0
% Reshape A into desired format
A = reshape(A.', size(A, 1) / n_blocks * n, n_blocks).'
The output (shortened):
A =
0.9164959 0.1373036
0.5588065 0.1303052
0.4913387 0.6540321
0.5711623 0.1937039
0.7231415 0.8142444
0.9348675 0.8623844
[...]
0.8372621 0.4571067
0.5531564 0.9138423
A =
0.91650 0.13730
0.55881 0.13031
0.49134 0.65403
0.57116 0.19370
0.72314 0.81424
0.93487 0.86238
[...]
0.83726 0.45711
0.55316 0.91384
0.00000 0.00000
A =
0.91650 0.13730 0.55881 0.13031 0.49134 0.65403 0.57116 0.19370 0.72314 0.81424
0.93487 0.86238 0.61128 0.15006 0.43861 0.07667 0.94387 0.85875 0.43247 0.03105
0.48887 0.67998 0.42381 0.77707 0.93337 0.96875 0.88552 0.43617 0.06198 0.80826
0.08087 0.48928 0.46514 0.69252 0.84122 0.77548 0.90480 0.16924 0.82599 0.82780
0.49048 0.00514 0.99615 0.42366 0.83726 0.45711 0.55316 0.91384 0.00000 0.00000
Hope that helps!
Related
I have 3 arrays of size 803500*1 with the following details:
Rid: It can contain any number
RidID: It contains elements from 1 to 184 in random order. Each element appears multiple times.
r: It contains elements 0,1,2,...12. All elements (except zero) appear nearly 3400 to 3700 times at random indices in this array.
Following may be useful for generating sample data:
Rid = rand(803500,1);
RidID = randi(184,803500,1);
r = randi(13,803500,1)-1; %This may not be a good sample for r as per previously mentioned details?
What I want to do?
I want to calculate the sum of those entries of Rid which correspond to each positive unique entry of r and each unique entry of RidID.
This may be clearer with the code which I wrote for this problem:
RNum = numel(unique(RidID));
RSum = ones(RNum,12); %Preallocating for better speed
for i=1:12
RperM = r ==i;
for j = 1:RNum
RSum(j,i) = sum(Rid(RperM & (RidID==j)));
end
end
Issue:
My code works but it takes 5 seconds on average on my computer and I have to do this calculation nearly a thousand times. If this time be reduced from 5 seconds to atleast half of it, I'll be very happy. But how do I optimize this? I don't mind if it is made better with vectorization or any better written loop.
I am using MATLAB R2017b.
You can use accumarray :
u = unique(RidID);
A = accumarray([RidID r+1], Rid);
RSum = A(u, 2:13);
This is slower than accumarray as suggested by rahnema, but using findgroups and splitapply may save memory.
In your example, there may be thousands of zero-valued elements in the resulting matrix, where a combination of RidID and r does not occur. In this case a stacked result would be more memory efficient, like so:
RidID | r | Rid_sum
-------------------------
1 | 1 | 100
2 | 1 | 200
4 | 2 | 85
...
This can be achieved with the following code:
[ID, rn, RidIDn] = findgroups(r,RidID); % Get unique combo ID for 'r' and 'RidID'
RSum = splitapply( #sum, Rid, ID ); % Sum for each ID
output = table( RidIDn, rn, RSum ); % Nicely formatted table output
% Get rid of elements where r == 0
output( output.rn == 0, : ) = [];
You could convert this to the same output as the accumarray method, but it's already a slower method...
% Convert to 'unstacked' 2D matrix (optional)
RSum = full( sparse( 1:numel(Ridn), 1:numel(rn), RSum ) );
I'm newbie college student for programming studies,
so recently i have task to calculate matrix from outside files for Gauss Jordan Numeric Method, in the txt file i provide has 10 (x) and (y) data, and declare with do functions to calculate the 10 data from the txt file each for x^2, x^3, x^4, xy, x^2y
my question is : how to SUM (calculate total) each x^2, x^3 ... that was calculated by program ? i try do sum file in below and still got errors (the first argument of sum must not a scalar.)
the Fortran apps i use was Plato cc from Silverfrost.
I apologize if my english bad and my pogram looks funny.
i have 10 data in my txt looks like these :
(x) (y)
12 10
5 6
28 8
9 11
20 17
6 24
32 9
2 7
1 30
26 22
in program below i open these files and want each x and y i provide read and calculate to get x^2, x^3, x^4, xy, x^2y
Program Gauss_Jordan
Real x(10),y(10),xj,yj,xj2,xj3,xj4,xjyj,xj2yj
Open (10, file='Data.txt')
Do j = 1,10
Read(10,*) x(j), y(j)
xj2 = x(j)**2
xj3 = x(j)**3
xj4 = x(j)**4
xjyj = x(j)*y(j)
xj2yj = (x(j)**2)*y(j)
Do k = 1,10
T(xj2) = SUM( xj2, dim=1)
T(xj3) = SUM (xj3, dim=1)
T(xj4) = SUM (xj4, dim=1)
T(xjyj) = SUM (xjyj, dim=1)
T(xj2yj) = SUM (xj2yj, dim=1)
End Do
End Do
Close(10)
End
for T(xj2) I want to get one result scalar result from SUM the all xj^2 that program has been calculated.
Like in excel was expected :
(A) is 1st xj^2 value that has been calculated
.
.
.
until (J) is 10th xj^2 value that has been calculated
sxj^2 = SUM(Xj^2)
SUM (A-J)
The 'sum' intrinsic needs an array argument, which we can compute from the input arrays without using a loop, so your program could be:
Program Gauss_Jordan
Real x(10), y(10), x2(10), x3(10), x4(10), xy(10), x2y(10)
Open(10, file='Data.txt')
Do j = 1, 10
Read (10, *) x(j), y(j)
End Do
Close(10)
x2 = x**2
x3 = x**3
x4 = x**4
xy = x*y
x2y = (x**2)*y
sx2 = SUM(x2)
sx3 = SUM(x3)
sx4 = SUM(x4)
sxy = SUM(xy)
sx2y = SUM(x2y)
End
From what I see I think you are misunderstanding what the SUM intrinsic does. Since your example isn't storing xj2, xj3 etc. in arrays, SUM isn't going to be useful to you. Instead you could declare totals as scalars (as you described you wanted) and simply add the individual xj2 variables in a loop as in the example below.
Also, you should get in the habit of using the implicit none declaration. It will save you from unexpected errors due to spelling mistakes.
Program Gauss_Jordan
implicit none
Real x(10),y(10),xj,yj,xj2,xj3,xj4,xjyj,xj2yj
real :: Txj2,Txj3,Txj4,Txjyj,Txj2yj
integer :: j
Txj2 = 0
Txj3 = 0
Txj4 = 0
Txjyj= 0
Txj2yj= 0
Open (10, file='Data.txt')
Do j = 1,10
Read(10,*) x(j), y(j)
xj2 = x(j)**2
xj3 = x(j)**3
xj4 = x(j)**4
xjyj = x(j)*y(j)
xj2yj = (x(j)**2)*y(j)
Txj2 = Txj2 + xj2
Txj3 = Txj3 + xj3
Txj4 = Txj4 + xj4
Txjyj = Txjyj + xjyj
Txj2yj = Txj2yj + xj2yj
End Do
print *, 'Txj2 = ', Txj2
Close(10)
End
When I ran this I got the output below which is what I believe you intended:
3175
I have an array with a set of chronological serial numbers and another source array with random serial numbers associated with a numeric value. The code creates a new cell array in MATLAB with the perfectly chronological serial numbers in one column and in the next column it inserts the associated numeric value if the serial numbers match in both original source arrays. If they don't the code simply copies the previous associated value until there is a new match.
j = 1;
A = {random{1:end,1}};
B = cell2mat(A);
value = random{1,2};
data = cell(length(serial), 1);
data(:,1) = serial(:,1);
h = waitbar(0,'Please Wait...');
steps = length(serial);
for k = 1:length(serial)
[row1, col1, vec1] = find(B == serial{k,1});
tf1 = isempty(vec1);
if (tf1 == 0)
prices = random{col1,2};
data(j,2) = num2cell(value);
j = j + 1;
else
data(j,2) = num2cell(value);
j = j + 1;
end
waitbar(k/steps,h,['Please Wait... ' num2str(k/steps*100) ' %'])
end
close(h);
Right now, the run-time for the code is approximately 4 hours. I would like to make this code run faster. Please suggest any methods to do so.
UPDATE
source input (serial)
1
2
3
4
5
6
7
source input (random)
1 100
2 105
4 106
7 107
desired output (data)
SR No Value
1 100
2 105
3 105
4 106
5 106
6 106
7 107
Firstly, run the MATLAB profiler (see 'doc profile') and see where the bulk of the execution time is occuring.
Secondly, don't update the waitbar on every iteration> Particularly if serial contains a large (> 100) number of elements.
Do something like:
if (mod(k, 100)==0) % update on every 100th iteration
waitbar(k/steps,h,['Please Wait... ' num2str(k/steps*100) ' %'])
end
Some points:
Firstly it would help a lot if you gave us some sample input and output data.
Why do you initialize data as one column and then fill it's second in the loop? Rather initialize it as 2 columns upfront: data = cell(length(serial), 2);
Is j ever different from k, they look identical to me and you could just drop both the j = j + 1 lines.
tf1 = isempty(vec1); if (tf1 == 0)... is the same as the single line: if (!isempty(vec1)) or even better if(isempty(vec1)) and then swap the code from your else and your if.
But I think you can probably find a fast vecotrized solution if you provide some (short) sample input and output data.
I have a struct mpc with the following structure:
num type col3 col4 ...
mpc.bus = 1 2 ... ...
2 2 ... ...
3 1 ... ...
4 3 ... ...
5 1 ... ...
10 2 ... ...
99 1 ... ...
to from col3 col4 ...
mpc.branch = 1 2 ... ...
1 3 ... ...
2 4 ... ...
10 5 ... ...
10 99 ... ...
What I need to do is:
1: Re-order the rows of mpc.bus, such that all rows of type 1 are first, followed by 2 and at last, 3. There is only one element of type 3, and no other types (4 / 5 etc.).
2: Make the numbering (column 1 of mpc.bus, consecutive, starting at 1.
3: Change the numbers in the to-from columns of mpc.branch, to correspond to the new numbering in mpc.bus.
4: After running simulations, reverse the steps above to turn up with the same order and numbering as above.
It is easy to update mpc.bus using find.
type_1 = find(mpc.bus(:,2) == 1);
type_2 = find(mpc.bus(:,2) == 2);
type_3 = find(mpc.bus(:,2) == 3);
mpc.bus(:,:) = mpc.bus([type1; type2; type3],:);
mpc.bus(:,1) = 1:nb % Where nb is the number of rows of mpc.bus
The numbers in the to/from columns in mpc.branch corresponds to the numbers in column 1 in mpc.bus.
It's OK to update the numbers on the to, from columns of mpc.branch as well.
However, I'm not able to find a non-messy way of retracing my steps. Can I update the numbering using some simple commands?
For the record: I have deliberately not included my code for re-numbering mpc.branch, since I'm sure someone has a smarter, simpler solution (that will make it easier to redo when the simulations are finished).
Edit: It might be easier to create normal arrays (to avoid woriking with structs):
bus = mpc.bus;
branch = mpc.branch;
Edit #2: The order of things:
Re-order and re-number.
Columns (3:end) of bus and branch are changed. (Not part of this question)
Restore original order and indices.
Thanks!
I'm proposing this solution. It generates a n x 2 matrix, where n corresponds to the number of rows in mpc.bus and a temporary copy of mpc.branch:
function [mpc_1, mpc_2, mpc_3] = minimal_example
mpc.bus = [ 1 2;...
2 2;...
3 1;...
4 3;...
5 1;...
10 2;...
99 1];
mpc.branch = [ 1 2;...
1 3;...
2 4;...
10 5;...
10 99];
mpc.bus = sortrows(mpc.bus,2);
mpc_1 = mpc;
mpc_tmp = mpc.branch;
for I=1:size(mpc.bus,1)
PAIRS(I,1) = I;
PAIRS(I,2) = mpc.bus(I,1);
mpc.branch(mpc_tmp(:,1:2)==mpc.bus(I,1)) = I;
mpc.bus(I,1) = I;
end
mpc_2 = mpc;
% (a) the following mpc_tmp is only needed if you want to truly reverse the operation
mpc_tmp = mpc.branch;
%
% do some stuff
%
for I=1:size(mpc.bus,1)
% (b) you can decide not to use the following line, then comment the line below (a)
mpc.branch(mpc_tmp(:,1:2)==mpc.bus(I,1)) = PAIRS(I,2);
mpc.bus(I,1) = PAIRS(I,2);
end
% uncomment the following line, if you commented (a) and (b) above:
% mpc.branch = mpc_tmp;
mpc.bus = sortrows(mpc.bus,1);
mpc_3 = mpc;
The minimal example above can be executed as is. The three outputs (mpc_1, mpc_2 & mpc_3) are just in place to demonstrate the workings of the code but are otherwise not necessary.
1.) mpc.bus is ordered using sortrows, simplifying the approach and not using find three times. It targets the second column of mpc.bus and sorts the remaining matrix accordingly.
2.) The original contents of mpc.branch are stored.
3.) A loop is used to replace the entries in the first column of mpc.bus with ascending numbers while at the same time replacing them correspondingly in mpc.branch. Here, the reference to mpc_tmp is necessary so ensure a correct replacement of the elements.
4.) Afterwards, mpc.branch can be reverted analogously to (3.) - here, one might argue, that if the original mpc.branch was stored earlier on, one could just copy the matrix. Also, the original values of mpc.bus are re-assigned.
5.) Now, sortrows is applied to mpc.bus again, this time with the first column as reference to restore the original format.
I have the following code
B=[1 2 3; 10 20 30 ; 100 200 300 ; 500 600 800];
A=[100; 500; 300 ; 425];
SA = sum(A);
V={}; % number of rows for cell V = num of combinations -- column = 1
n = 1;
arr = zeros(1,length(B))
for k = 1:length(B)
for idx = nchoosek(1:numel(B), k)'
rows = mod(idx, length(B))
if ~isequal(rows, unique(rows)) %if rows not equal to unique(rows)
continue %combination possibility valid
end %Ignore the combination if there are two elements from the same row
B_subset = B(idx)
if (SA + sum(B_subset) <= 3000) %if sum of A + (combination) < 3000
rows( rows==0 )=4
arr(rows) = B_subset(:)
V(n,1) = {arr}
n = n + 1
arr = zeros(1,length(B))
end
end
end
The combinations are supposed to be valid if the sum of A and some values of B are less than 3000.
Problem with my code is, the last value of B, B(3,3), is only accounted for once in the code.
If you run the code, you will notice one cell of V containing [0;0;0;800] at row 12. But there are other combinations possible as well such as [1;0;0;800]. Where SA + (1 + 800) < 3000 , however the code does not print this possibility.
I cannot figure out why, can someone please help me debug this and find out why some combinations are skipped ? especially B(3,3) ?
I suspect that this line is not doing exactly what you intended:
if ~isequal(rows, unique(rows))
Instead, try this:
if ~isequal(length(rows), length(unique(rows)))