How can I open an xyz format file in MATLAB?

Hello all, I have a xyz format file. I got it from my simulation program by LAMMPS. In this file I have some statements that are not useful for me but I can't delete them. My useful data in each timestep are the numbers that come after "ITEM: ATOMS id type z" , other statements aren't important. I have lots of data in each timestep in my program. My file contains lots of timesteps that come after each other continuously . For more detail I put 2 timesteps of my data here (timestep 0; timestep 10000):

ITEM: TIMESTEP

0

ITEM: NUMBER OF ATOMS

8553

ITEM: BOX BOUNDS pp pp pp

-21.97 21.963

-22.261 22.261

-121.7 68.7

ITEM: ATOMS id type z

1912 5 3.499

1913 4 4.234

8149 5 4.2

8150 4 3.56

8151 4 4.415

143 4 4.754

346 5 5.053

ITEM: TIMESTEP

10000

ITEM: NUMBER OF ATOMS

8553

ITEM: BOX BOUNDS pp pp pp

-21.97 21.963

-22.261 22.261

-121.7 68.7

ITEM: ATOMS id type z

2894 4 -107.919

2893 5 -107.733

2895 4 -108.452

1001 4 -112.385

1000 5 -111.436

1002 4 -110.96

3963 4 -101.516

I want to know how can I open and read this file by MATLAB? I want to read the numbers in each timestep and then apply some functions, also the same for other steps till end. How can I do this? Would you please guide me? Many thanks

[dpb]

Do you know how many items are in the section of interest? Is it given by the value after the line

ITEM: NUMBER OF ATOMS

8553 here? If so, that'll make it simpler.

Also, I presume the extra linefeed is a fignewton of the posting and not included in the file?

[Samaneh]

My file contains 100 timesteps and each timestep includes 8553 line value.I just put 7 line in each timestep here as an example. Other statements exist in my file anyway, because my program produces them but I don't need them. How can I read this file?

Answer 1

[Image Analyst]

Ask the publisher of your simulation program if they have a MATLAB reader for their data files. They very well may have a reader, and if so, that is by far the easiest way to go.

[dpb]

Good point, IA...

+1

[Samaneh]

yeah it's a good idea. Thanks

[Tina-Nefeli Vartziotis]

Did you find anything regarding that? I need the exact same code!

Thanks

[Image Analyst]

@Tina-Nefeli Vartziotis, did you contact the publishers of "simulation program by LAMMPS" like I suggested?

If the answer is yes, and (1) they were unwilling to help you, or (2) or you tried but couldn't find them (like they went out of business), then attach the xyz file (zipped up) with the paperclip icon.

If the answer is no, then try them first.

[hamza ashar]

@Image Analyst

please will you help me to code this First part i have coded sucessfully but second part for repetiotion code i am not able to do it. Plz help me to do it i am in urgent of today.

Image transmission over Binary Channel.

We wish to study how a degraded image looks through binary channels. Starts with a binary image (black and white image) of your choice.

I. Assume that this image is transmitted through a BSC with p=0.01

Use the repetition code(n=3) for error correction and then show the improvement in the above case.

i have asked question here also but no help got

https://in.mathworks.com/matlabcentral/answers/819695-image-transmission-over-binary-channel?s_tid=srchtitle

x=imread("cameraman.tif");

disp(size(x))
x=im2double(x);
T=dctmtx(8);
dct= @(block_struct)T*block_struct.data*T';
B=blockproc(x,[8 8],dct);
mask= [1   1   1   1   0   0   0   0
      1   1   1   0   0   0   0   0
      1   1   0   0   0   0   0   0
      1   0   0   0   0   0   0   0
      0   0   0   0   0   0   0   0
      0   0   0   0   0   0   0   0
      0   0   0   0   0   0   0   0
      0   0   0   0   0   0   0   0];
B2=blockproc(B,[8 8],@(block_struct)mask.*block_struct.data);
invdct= @(block_struct)T' *block_struct.data*T;
I2=blockproc(B2,[8 8],invdct);
figure(1)
imshow(I2)
% BW = im2bw(x,0.4);
BW = imbinarize(x);
figure(2)
imshow(BW); 
ndata = bsc(BW,0.01);
figure(3)
imshow(ndata);
ndata=im2double(ndata);
BW=im2double(BW);
peaksnr = psnr(ndata,BW);
disp(peaksnr);
out=repmat(ndata,3,1);
figure (4)
imshow(out)

[Image Analyst]

Not here @hamza ashar, because it has nothing at all to do with @Samaneh's original question.

Answer 2

[Image Analyst]

Try this. It took me about half an hour to write a reader for this type of file.

It works for the data you mentioned (attached, because you forgot to attach it).

% Demo by Image Analyst.
clc;    % Clear the command window.
fprintf('Beginning to run %s.m ...\n', mfilename);
close all;  % Close all figures (except those of imtool.)
clearvars;
workspace;  % Make sure the workspace panel is showing.
format short g;
format compact;
fontSize = 16;

% One record will look like this:
% ITEM: TIMESTEP
% 0
% ITEM: NUMBER OF ATOMS
% 8553
% ITEM: BOX BOUNDS pp pp pp
% -21.97 21.963
% -22.261 22.261
% -121.7 68.7
% ITEM: ATOMS id type z
% 1912 5 3.499
% 1913 4 4.234
% 8149 5 4.2
% 8150 4 3.56
% 8151 4 4.415
% 143 4 4.754
% 346 5 5.053
records.TimeStep = 0;
records.NumberOfAtoms = 0;
records.BoxBounds = zeros(3, 2);
records.Atoms = zeros(7, 3);

fullFileName = fullfile(pwd, 'LAMMPS.XYZ');
% Open the file for reading in text mode.
fileID = fopen(fullFileName, 'rt');
recordNumber = 1;
textLine = 'Start';
while ischar(textLine)
	% Get the time step.
	% Read the next line, if we haven't already.
	if ~contains(textLine, 'ITEM')
		textLine = fgetl(fileID); % Should be "ITEM: TIMESTEP"
	end
	if textLine == -1
		break; % End of file
	end
	% Get the next line which should be a number.
	textLine = fgetl(fileID); % Should be a number
	records(recordNumber).TimeStep = str2double(textLine);
	
	% Get the number of atoms.
	% Read the next line.
	textLine = fgetl(fileID); % ITEM: NUMBER OF ATOMS
	textLine = fgetl(fileID); % Should be a number
	% Get the next line which should be a number.
	records(recordNumber).NumberOfAtoms = str2double(textLine);
	numberOfAtoms = records(recordNumber).NumberOfAtoms;
	
	% Get the box bounds.
	% Read the next line.
	textLine = fgetl(fileID); % BOX BOUNDS pp pp pp
	bb = zeros(3, 2);
	for k = 1 : size(bb, 1) % Read 3 rows
		textLine = fgetl(fileID); % Should be two numbers.
		if textLine == -1
			% Did not have all the lines.  Started with the next record early for some reason.
			bb = bb(1 : k-1, :); % Crop matrix so that it's shorter.
			% Break if it's the end of the file.
			break;
		end
		if contains(textLine, 'ITEM', 'IgnoreCase', true)
			% Did not have all the lines.  Started with the next record early for some reason.
			bb = bb(1 : k-1, :); % Crop matrix so that it's shorter.
			break;
	I	end
		twoNumbers = sscanf(textLine, '%f %f')';
		bb(k, :) = twoNumbers;
	end
	% Get the next line which should be a number.
	records(recordNumber).BoxBounds = bb;

	% Get the Atom numbers.
	% Read the next line, if we haven't already.
	if ~contains(textLine, 'ITEM')
		textLine = fgetl(fileID); % ATOMS id type z
	end
	atomsMatrix = zeros(numberOfAtoms, 3);
	for k = 1 : numberOfAtoms % Read all rows
		textLine = fgetl(fileID); % Should be three numbers.
		if textLine == -1
			% Did not have all the lines.  Started with the next record early for some reason.
			atomsMatrix = atomsMatrix(1 : k-1, :); % Crop matrix so that it's shorter.
			% Break if it's the end of the file.
			break;
		end
		if contains(textLine, 'ITEM', 'IgnoreCase', true)
			% Did not have all the lines.  Started with the next record early for some reason.
			atomsMatrix = atomsMatrix(1 : k-1, :); % Crop matrix so that it's shorter.
			break;
		end
		threeNumbers = sscanf(textLine, '%f %f %f')';
		atomsMatrix(k, :) = threeNumbers;
	end
	% Get the next line which should be a number.
	records(recordNumber).Atoms = atomsMatrix;

	recordNumber = recordNumber + 1;
end
% All done reading all lines, so close the file.
fclose(fileID);

% Display in command window.
records
fprintf('Done running %s.m\n', mfilename);