Filter löschen
Filter löschen

2D Surface Random Number Plot

5 Ansichten (letzte 30 Tage)
Syed Arafun Nabi
Syed Arafun Nabi am 15 Dez. 2022
Kommentiert: Syed Arafun Nabi am 15 Dez. 2022
The 2D square surface has 1m length in each side and splitted into 10 grid(Total grid is 10x10). A ball is hitting the surface 1000 times. The hit is random. The exact position of each hit is determined by random number. How can I find the total hit count in each grid? Each grid will be named as
A0101, A0201 . . . .A1001
A0102, A0202 . . . .A1002
.........................................
A1001, A1002 . . . A1010
Initial codes are as followings -
x = rand;
y = rand;
x1 = linspace(0,1,11)
y1 = linspace(0,1,11)
for i = 1:1000
[x,y];
end

Akzeptierte Antwort

KSSV
KSSV am 15 Dez. 2022
% Make grid
x = linspace(0,1,11) ;
y = linspace(0,1,11) ;
[X,Y] = meshgrid(x,y) ;
% Get mid points of each element
mx = linspace(0.05,0.95,10) ;
my = linspace(0.05,0.95,10) ;
[MX,MY] = meshgrid(mx,my) ;
plot(X,Y,'.r',MX,MY,'.b')
MZ = zeros(size(MX)) ; % Initialize the hit count
for i = 1:1000
hit = rand(1,2) ;
% Check where it is hit
idx = knnsearch([MX(:) MY(:)],hit) ;
% UPdate the count
MZ(idx) = MZ(idx)+1 ;
end
figure
hold on
plot(X,Y,'k',X',Y','k')
scatter(MX(:),MY(:),[],MZ(:),'filled') ;
colormap(jet) ;
cb = colorbar ;
ylabel(cb,'Number of hits')
  3 Kommentare
KSSV
KSSV am 15 Dez. 2022
Z is in matrix already you can plot a contour.
About boundary, we need to change the strategy.
Syed Arafun Nabi
Syed Arafun Nabi am 15 Dez. 2022
If the probablity of hitting the bondary is 20% & inside the surface is 80%. The boundary hit will be consider in the corresponding grid. How can we modify? Can you please assist?
My actual code is more than 250 lines. so this is just a template
a = rand
a = 0.8406
if a<.05
%hit y=0 plane
x=rand
y=0
elseif a<.10
%hit y=0 plane
x=rand
y=1
elseif a<.15
%hit x=0 plane
x=0
y=rand
elseif a<.20
%hit x=1 plane
x=1
y=rand
else
%inside the surface
x=rand
y=rand
end
x = 0.4815
y = 0.4564

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (2)

Steven Lord
Steven Lord am 15 Dez. 2022
I recommend using histogram2.

Fifteen12
Fifteen12 am 15 Dez. 2022
Bearbeitet: Fifteen12 am 15 Dez. 2022
The best way to do this is to use sequential indices, rather than row, column. Then you're just keeping track of a single number. If you do need the x, y coordinates (row, column), you can try this:
hit_count = zeros(10);
hits = randi(numel(hit_count), 1, 1000); %Randomly hit an cell index between 1 and 100, 1000 times
for i = 1:10
for j = 1:10
seq_index = (i-1) * 10 + j;
hit_count(i, j) = sum(hits == seq_index);
end
end
This would be the same as doing the following:
hit_count = zeros(10);
hits = randi(numel(hit_count), 1, 1000); %Randomly hit an cell index between 1 and 100, 1000 times
for i = 1:numel(hit_count)
hit_count(i) = sum(hits == i);
end
As for the name of the cells, I'm not sure how you want to access them. You can easily make up a calling function though that returns the hits per cell based on an inputted name though...
hits_at_3_5 = getNumberOfHitsPerCell('A0305', hit_count);
disp(hits_at_3_5)
8
function [num_hits] = getNumberOfHitsPerCell(cell_name, hit_count)
x = str2num(cell_name(2:3));
y = str2num(cell_name(4:5));
num_hits = hit_count(x, y);
end
You'd probably want to do some more intense string validation if you go this route though. Good luck!

Produkte


Version

R2021b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by