The HDF5 file uses C storage conventions, which is why the matrix is
transposed.
See http://www.hdfgroup.org/HDF5/doc/UG/UG_frame12Dataspaces.html
section: 7.3.2.5. C versus Fortran Dataspaces
On 2013-02-27 06:32, Pradeep Jha wrote:
Hello,
I am trying to convert a unformatted data file created by fortran
into the *h5 format. To do this I am using this program [1] provided
on the website as the base.
I am dealing with a data of size Nx by Ny by Nz. So I changed the
"dims" declaration in the code to something like
-------------------------------------------
**** code ****
INTEGER :: Nx, Ny, Nz
INTEGER(HSIZE_T), DIMENSION(3) :: dims
**** code ****
dims(1) = Nx
dims(2) = Ny
dims(3) = Nz
**** code ****
--------------------------------------------
This code is working perfectly fine. But when I am doing a "h5dump
-H" on the output file, the output is:
HDF5 "output_file" {
GROUP "/" {
DATASET "variable" {
DATATYPE H5T_IEEE_F32LE
DATASPACE SIMPLE { ( Nz,Ny,Nx ) / ( Nz,Ny,Nx) }
}
}
}
Can you please explain is this how the order of Nz,Ny and Nx should
be? Does this represent a cube of size Nx by Ny by Nz? Or does
dims(1)
actually represent the z dimension and I should assign it the value
Nz?
Thank you,
Pradeep
Links:
------
[1]
http://www.hdfgroup.org/ftp/HDF5/examples/introductory/F90/h5_rdwt.f90
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