Read and process a Fortran90 binary file
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Hi,
I have a Fortran90 code where a variable called 'stream' is defined such that it stores certain variable every nstep:
open(19,file='stream',form='UNFORMATTED',position='append')
if(mod(NSTEP,100).eq.0)then
write(19)NSTEP,U,W,zeta
endif
How can I properly read the file in Matlab for postprocessing?
FYI, the fortran code is built with double precision.
Thank you
3 个评论
James Tursa
2024-11-21
编辑：James Tursa
2024-11-21
Unformatted Fortran can contain beginning and end of line markers in the file. You need to give us a small sample file to work with to verify this, and also tell us the type and dimensions of the variables NSTEP, U, W, and zeta.
Walter Roberson
2024-11-21
The FORTRAN runtime system embeds the record boundaries in the data by inserting an INTEGER*4 byte count at the beginning and end of each unformatted sequential record during an unformatted sequential WRITE. The trailing byte count enables BACKSPACE to operate on records.
Carola Forlini
2024-11-21
采纳的回答
In light of the additional variable size information and sample file provided by the OP, and the information from Walter about Fortran inserting record boundary byte counts at the beginning and end of each record written with an unformatted sequential WRITE, here is a fresh answer.
Note that this makes an assumption about the order the data is written (column major order). If this assumption is not correct, the inner for loop will need to be modified accordingly.
Hopefully this is close enough to get you where you need to go.
filenames = unzip('stream.zip')
fid = fopen(filenames{1}, 'rb');
% start/stop byte counts (4 bytes each) + NSTEP (4 bytes) + U, W, zeta (3*32*512*4 bytes)
recordSize = 2*4 + 4 + 3*32*512*4;
fseek(fid, 0, 'eof'); % seek to the end of the file
nBytes = ftell(fid); % find number of bytes in the file
frewind(fid); % go back to the beginning of the file
nRecords = nBytes / recordSize;
% preallocate
nStep = zeros(nRecords, 1); % column vector
U = zeros(3, 512, nRecords); % 3D arrays
W = zeros(3, 512, nRecords);
zeta = zeros(3, 512, nRecords);
for iRecord = 1:nRecords
fseek(fid, 4, 'cof'); % skip begin record byte count field
nStep(iRecord) = fread(fid, 1, 'integer*4');
for iRow = 1:3
U(iRow, :, iRecord) = fread(fid, 512, 'real*4');
W(iRow, :, iRecord) = fread(fid, 512, 'real*4');
zeta(iRow, :, iRecord) = fread(fid, 512, 'real*4');
end
fseek(fid, 4, 'cof'); % skip end record byte count field
end
fclose(fid); % close the file
whos nStep U W zeta % check variable sizes
plot(nStep); % make a plot see if nStep makes sense
grid on
Well, I don't know what your NSTEP should look like, but I would have expected a monotonically increasing count. So, either I made a mistake, or there is some additional information needed. It does appear that the size information you gave is either correct, or just coincidentally came out to be an integer divisor of the file size.
4 个评论
Reasonable approach, but I'm not sure why you assume 3-by-512 matrices and read them one row at a time. If instead 32x512 is used and each read at once, then the nStep comes out correct (i.e., [0 100 200 ... 2000]).
filenames = unzip('stream.zip')
fid = fopen(filenames{1}, 'rb');
% start/stop byte counts (4 bytes each) + NSTEP (4 bytes) + U, W, zeta (3*32*512*4 bytes)
recordSize = 2*4 + 4 + 3*32*512*4;
fseek(fid, 0, 'eof'); % seek to the end of the file
nBytes = ftell(fid); % find number of bytes in the file
frewind(fid); % go back to the beginning of the file
nRecords = nBytes / recordSize;
% preallocate
nStep = zeros(nRecords, 1); % column vector
U = zeros(32, 512, nRecords); % 3D arrays
W = zeros(32, 512, nRecords);
zeta = zeros(32, 512, nRecords);
for iRecord = 1:nRecords
fseek(fid, 4, 'cof'); % skip begin record byte count field
nStep(iRecord) = fread(fid, 1, 'uint32');
U(:, :, iRecord) = fread(fid, [32 512], 'single');
W(:, :, iRecord) = fread(fid, [32 512], 'single');
zeta(:, :, iRecord) = fread(fid, [32 512], 'single');
fseek(fid, 4, 'cof'); % skip end record byte count field
end
% make sure the entire file was read
assert(isequal(ftell(fid),nBytes))
fclose(fid); % close the file
whos nStep U W zeta % check variable sizes
plot(nStep,'-o'); % make a plot see if nStep makes sense
grid on
Carola Forlini
2024-11-22
Les Beckham
2024-11-22
Good catch, Voss. Thanks. I don't know why I switched from using 32 in the number of records calculation, to using 3 in the section where I did the preallocation and the reads. I should have spent more time reviewing my answer before submitting it, as it was obvious that something was wrong.
James Tursa
2024-11-22
编辑：James Tursa
2024-11-22
"... this makes an assumption about the order the data is written (column major order) ..."
Fortran is column-major storage order, same as MATLAB. When you specify just the variable name in a read/write statement, that is the element order for the read/write.
更多回答（1 个）
Les Beckham
2024-11-21
编辑：Les Beckham
2024-11-21
It would help to have a sample file to experiment with.
Nevertheless, assuming that the write call that you have shown is the only thing that writes to this file in your Fortran code, I would suggest something like this using fread and reshape:
fid = fopen('your_file_name.ext', 'rb');
A = fread(fid);
data = reshape(A, 4, numel(A)/4)';
nstep = A(:,1);
U = A(:,2);
W = A(:,3);
zeta = A(:,4);
Note that you might run into endian-ness issues if your Fortran code runs on a different architecture than your Matlab is running on. See the documentation for the machinefmt option to fread if necessary.
5 个评论
Walter Roberson
2024-11-21
form='UNFORMATTED' has record length markers before and after each record.
Walter Roberson
2024-11-21
If you start with fread(fid, 1, 'integer*4') and discard that, and thereafter fread() with a skip field of 8 then it should be properly positioned to fread() more.
Unfortunately we are not told the size of NSTEP, U, W, zeta
Carola Forlini
2024-11-22
Les Beckham
2024-11-22
Thanks for adding the example.
Can you clarify what you mean by this?
NSTEP is an integer = 2000
Especially the "= 2000" part. Is that just an example of a possible value? What is the size of this integer (number of bits)?
Also, are U, W, and zeta double precision (64 bit) floating point 2-d arrays with the given size?
Carola Forlini
2024-11-22
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