I need to output a matrix with FORTRAN. I have a working code that calculates the values, but instead of a matrix, I get single a column. The matrix is huge, ixj = ~2000x2000.
Here is my sample code:
open(19, file="results1.txt", status="old", position="rewind",
& action="write")
do j=0,p
do i=0,o
write(19,*) mat_user_yield_surface(d, eps(i), deps(j), 200.0d0)
end do
end do
close(19)
Use an implied do loop:
do j=0,p
write(19,'(2000g22.14)') (mat_user_yield_surface(d, eps(i), deps(j),200.0d0),i=0,o)
end do
I suggest not using "o" as a variable name, since it is easily confused with zero.
This "write(19,'(2000g22.14)')" worked perfectly! Thanks. So the final code is:
open(19, file="results1.txt", status="old", position="rewind",
& action="write")
do j=0,p
write(19,'(2000g22.14)') (mat_user_yield_surface(d, eps(i),
& deps(j), 200.0d0), i=0,o)
end do
close(19)
Related
My code seems to run very slowly and I can't think of any way to make it faster. All my arrays have been preallocated. S is a large number of element (say 10000 element, for example). I know my code runs slowly because of the "for k=1:S" but i cant think of another way to perform this loop at a relatively fast speed. Can i please get help because it takes hours to run.
[M,~] = size(Sample2000_X);
[N,~] = size(Sample2000_Y);
[S,~] = size(Prediction_Point);
% Speed Preallocation
Distance = zeros(M,N);
Distance_Prediction = zeros(M,1);
for k=1:S
for i=1:M
for j=1:N
Distance(i,j) = sqrt(power((Sample2000_X(i)-Sample2000_X(j)),2)+power((Sample2000_Y(i)-Sample2000_Y(j)),2));
end
Distance_Prediction(i,1) = sqrt(power((Prediction_Point(k,1)-Sample2000_X(i)),2)+power((Prediction_Point(k,2)-Sample2000_Y(i)),2));
end
end
Thanks.
I realized the major problem was organization of my code. I was performing calculation in a loop where it was absolutely unnecessary. So i seperated the code in two blocks and it Works much faster.
for i=1:M
for j=1:N
Distance(i,j) = sqrt(power((Sample2000_X(i)-Sample2000_X(j)),2)+power((Sample2000_Y(i)-Sample2000_Y(j)),2));
end
end
for k=1:S
for i=1:M
Distance_Prediction(i,1) = sqrt(power((Prediction_Point(k,1)-Sample2000_X(i)),2)+power((Prediction_Point(k,2)-Sample2000_Y(i)),2));
end
end
Thanks to the community for the help.
Your matrix Distance does not depend on k, so you can easily calculate it outside the main for-loop, for instance using:
d = sqrt((repmat(Sample2000_X, [1,M]) - repmat(Sample2000_X', [M,1])).^2 + (repmat(Sample2000_Y, [1,N]) - repmat(Sample2000_Y', [N,1])).^2);
I assume M=N, because elsewise your code won't work. Next, you can calculate your Distance_Prediction matrix. It is rather strange that you calculate this inside the for-loop over k, because the matrix will be changed in every iteration without using it. Anyway, this will do exactly the same as your code:
for k=1:S
Distance_Prediction = sqrt((Sample2000_X - Prediction_Point(k,1)).^2 + (Sample2000_Y - Prediction_Point(k,1)).^2);
end
I am relatively new to Modelica (Dymola-environment) and I am getting very desperate/upset that I cannot solve such a simple problem as a random number generation in Modelica and I hope that you can help me out.
The simple function random produces a random number between 0 and 1 with an input seed seedIn[3] and produces the output seed seedOut[3] for the next time step or event. The call
(z,seedOut) = random(seedIn);
works perfectly fine.
The problem is that I cannot find a way in Modelica to compute this assignment over time by using the seedOut[3] as the next seedIn[3], which is very frustrating.
My simple program looks like this:
*model Randomgenerator
Real z;
Integer seedIn[3]( start={1,23,131},fixed=true), seedOut[3];
equation
(z,seedOut) = random(seedIn);
algorithm
seedIn := seedOut;
end Randomgenerator;*
I have tried nearly all possibilities with algorithm assignments, initial conditions and equations but none of them works. I just simply want to use seedOut in the next time step. One problem seems to be that when entering into the algorithm section, neither the initial conditions nor the values from the equation section are used.
Using the 'sample' and 'reinit' functions the code below will calculate a new random number at the frequency specified in 'sample'. Note the way of defining the "start value" of seedIn.
model Randomgenerator
Real seedIn[3] = {1,23,131};
Real z;
Real[3] seedOut;
equation
(z,seedOut) = random(seedIn);
when sample(1,1) then
reinit(seedIn,pre(seedOut));
end when;
end Randomgenerator;
The 'pre' function allows the use of the previous value of the variable. If this was not used, the output 'z' would have returned a constant value. Two things regarding the 'reinint' function, it requires use of 'when' and requires 'Real' variables/expressions hence seedIn and seedOut are now defined as 'Real'.
The simple "random" generator I used was:
function random
input Real[3] seedIn;
output Real z;
output Real[3] seedOut;
algorithm
seedOut[1] :=seedIn[1] + 1;
seedOut[2] :=seedIn[2] + 5;
seedOut[3] :=seedIn[3] + 10;
z :=(0.1*seedIn[1] + 0.2*seedIn[2] + 0.3*seedIn[3])/(0.5*sum(seedIn));
end random;
Surely there are other ways depending on the application to perform this operation. At least this will give you something to start with. Hope it helps.
I am writing a simple code in matlab which has the purpose of creating the histogram of a grayscale image without using the function hist. I am stuck at the point in which mathlab displays the error "Subscript indices must either be real positive integers or logicals." Can you help me finding where is the wrong indices?
indirizzo='file.jpg';
immagine=imread(indirizzo);
immaginebn=rgb2gray(immagine);
n=zerps(0,255);
for x=0:255;
numeroennesimo=sum(sum(immaginebn==x));
n(x)=numeroennesimo;
end
plot(x,n)
you cant use 0 as index. Either make n(x+1) or for x = 1:256 and substract the 1 in your comparison. And there is a typo, I guess it means zeros instead of zerps, which also doesnt work with a 0. And one more, your plot will also not work as the x has only a size of 1 while n is an array of 266 and for a histogram I would use a barplot instead.
indirizzo='file.jpg';
immagine=imread(indirizzo);
immaginebn=rgb2gray(immagine);
n=zeros(1,256);
for x=0:255;
numeroennesimo=sum(sum(immaginebn==x-1));
n(x+1)=numeroennesimo;
end
bar(0:255,n)
or
indirizzo='file.jpg';
immagine=imread(indirizzo);
immaginebn=rgb2gray(immagine);
n=zeros(1,256);
xplot=zeros(1,256);
for x=1:256;
numeroennesimo=sum(sum(immaginebn==x-1));
n(x)=numeroennesimo;
xplot(x) = x-1;
end
plot(xplot,n)
I'm trying to create a random number generator in Lua. I found out that I can just use math.random(1,100) to randomize a number between 1 and 100 and that should be sufficient.
But I don't really understand how to use the randomize number as variables in the script.
Tried this but of course it didn't work.
$randomCorr = math.random(1,100);
http.request_batch({
{"POST", "https://store.thestore.com/priceAndOrder/selectProduct", headers={["Content-Type"]="application/json;charset=UTF-8"}, data="{\"ChoosenPhoneModelId\":4,\"PricePlanId\":\"phone\",\"CorrelationId\":\"$randomCorr\",\"DeliveryTime\":\"1 vecka\",\"$$hashKey\":\"006\"},\"ChoosenAmortization\":{\"AmortizationLength\":0,\"ChoosenDataPackage\":{\"Description\":\"6 GB\",\"PricePerMountInKr\":245,\"DataAmountInGb\":6,\"$$hashKey\":\"00W\"},\"ChoosenPriceplan\":{\"IsPostpaid\":true,\"Title\":\"Fastpris\",\"Description\":\"Fasta kostnader till fast pris\",\"MonthlyAmount\":0,\"AvailiableDataPackages\":null,\"SubscriptionBinding\":0,\"$$hashKey\":\"00K\"}}", auto_decompress=true},
{"GET", "https://store.thestore.com/api/checkout/getproduct?correlationId=$randomCorr", auto_decompress=true},
})
In Lua, you can not start a variable name with $. This is where your main issue is at. Once the $ is removed from your code, we can easily see how to refer to variables in Lua.
randomCorr = math.random(100)
print("The random number:", randomCorr)
randomCorr = math.random(100)
print("New Random Number:", randomCorr)
Also, concatenation does not work the way you are implying it into your Http array. You have to concatenate the value in using .. in Lua
Take a look at the following example:
ran = math.random(100)
data = "{\""..ran.."\"}"
print(data)
--{"14"}
The same logic can be implied into your code:
data="{\"ChoosenPhoneModelId\":4,\"PricePlanId\":\"phone\",\"CorrelationId\":\""..randomCorr.."\",\"DeliveryTime\":\"1 vecka\",\"$$hashKey\":\"006\"},\"ChoosenAmortization\":{\"AmortizationLength\":0,\"ChoosenDataPackage\":{\"Description\":\"6 GB\",\"PricePerMountInKr\":245,\"DataAmountInGb\":6,\"$$hashKey\":\"00W\"},\"ChoosenPriceplan\":{\"IsPostpaid\":true,\"Title\":\"Fastpris\",\"Description\":\"Fasta kostnader till fast pris\",\"MonthlyAmount\":0,\"AvailiableDataPackages\":null,\"SubscriptionBinding\":0,\"$$hashKey\":\"00K\"}}"
Or you can format the value in using one of the methods provided by the string library
Take a look at the following example:
ran = math.random(100)
data = "{%q}"
print(string.format(data,ran))
--{"59"}
The %q specifier will take whatever you put as input, and safely surround it with quotations
The same logic can be applied to your Http Data.
Here is a corrected version of the code snippet:
local randomCorr = math.random(1,100)
http.request_batch({
{"POST", "https://store.thestore.com/priceAndOrder/selectProduct", headers={["Content-Type"]="application/json;charset=UTF-8"}, data="{\"ChoosenPhoneModelId\":4,\"PricePlanId\":\"phone\",\"CorrelationId\":\"" .. randomCorr .. "\",\"DeliveryTime\":\"1 vecka\",\"$$hashKey\":\"006\"},\"ChoosenAmortization\":{\"AmortizationLength\":0,\"ChoosenDataPackage\":{\"Description\":\"6 GB\",\"PricePerMountInKr\":245,\"DataAmountInGb\":6,\"$$hashKey\":\"00W\"},\"ChoosenPriceplan\":{\"IsPostpaid\":true,\"Title\":\"Fastpris\",\"Description\":\"Fasta kostnader till fast pris\",\"MonthlyAmount\":0,\"AvailiableDataPackages\":null,\"SubscriptionBinding\":0,\"$$hashKey\":\"00K\"}}", auto_decompress=true},
{"GET", "https://store.thestore.com/api/checkout/getproduct?correlationId=" .. randomCorr, auto_decompress=true},
})
There is something called $$hashKey also, in the quoted string. Not sure if that is supposed to be referencing a variable or not. If it is, it also needs to be concatenated into the resulting string, using the .. operator (just like with the randomCorr variable).
I'm trying to speed up my code using parfor. The purpose of the code is to slide a 3D square window on a 3D image and for each block of mxmxm apply a function.
I wrote this code:
function [ o_image ] = SlidingWindow( i_image, i_padSize, i_fun, i_options )
%SLIDINGWINDOW Summary of this function goes here
% Detailed explanation goes here
o_image = zeros(size(i_image,1),size(i_image,2),size(i_image,3));
i_image = padarray(i_image,i_padSize,'symmetric');
i_padSize = num2cell(i_padSize);
[m,n,p] = deal(i_padSize{:});
[row,col,depth] = size(i_image);
windowShape = i_options.windowShape;
mask = i_options.mask;
parfor (i = m+1:row-m,i_options.cores)
temp = i_image(i-m:i+m,:,:);
for j = n+1:col-n
for h = p+1:depth-p
ii = i-m;
jj = j-n;
hh = h-p;
temp = temp(:,j-n:j+n, h-p:h+p);
o_image(ii,jj,hh) = parfeval(i_fun, temp, windowShape, mask);
end
end
end
end
I get one warning and one error that I don't understand how to solve.
The warning says:
the entire array or structure 'i_image' is a broadcast variable.
The error says:
the PARFOR loop can not run due to the way variable 'o_image' is used.
I don't understand how to fix these two things. Any help is greatly appreciated!
As far as I understand, parfeval takes care of running your function on the available number of workers, which is why it doesn't need to be surrounded by parfor. Assuming you already have an active parpool, changing the external parfor into for eliminates both problems.
Unfortunately, I can't support my answer with a benchmark or suggest a more fitting solution because your inputs are unknown.
It seems to me that the code can be optimized in other ways, mainly by vectorization. I would suggest you looked into the following resources:
This question, for additional info on parfeval.
Examples on how to use bsxfun and permute and benchmarks thereof: ex1, ex2, ex3.
P.S.: The 2nd part of (i = m+1:row-m,i_options.cores) seems out of place...