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How to partially copy using python an Hdf5 file into a new one keeping the same structure?

Tags:

python

hdf5

h5py

I have a large hdf5 file that looks something like this:

A/B/dataset1, dataset2
A/C/dataset1, dataset2
A/D/dataset1, dataset2
A/E/dataset1, dataset2

...

I want to create a new file with only that: A/B/dataset1, dataset2 A/C/dataset1, dataset2

What is the easiest way in python?

I did:

fs = h5py.File('source.h5', 'r')
fd = h5py.File('dest.h5', 'w')
fs.copy('group B', fd)

the problem is that I get for dest.h5:

B/dataset1, dataset2

and that I am missing part of the arborescence.

like image 273
graham Avatar asked Jul 01 '14 12:07

graham


1 Answers

fs.copy('A/B', fd) doesn't copy the path /A/B/ into fd, it only copies the group B (as you've found out!). So you first need to create the rest of the path:

fd.create_group('A')
fs.copy('A/B', fd['/A'])

or, if you will be using the group a lot:

fd_A = fd.create_group('A')
fs.copy('A/B', fd_A)

This copies the group B from fs['/A/B'] into fd['/A']:

In [1]: fd['A/B'].keys()
Out[1]: [u'dataset1', u'dataset2']

Here's an automatic way of doing this:

# Get the name of the parent for the group we want to copy
group_path = fs['/A/B'].parent.name

# Check that this group exists in the destination file; if it doesn't, create it
# This will create the parents too, if they don't exist
group_id = fd.require_group(group_path)

# Copy fs:/A/B/ to fd:/A/G
fs.copy('/A/B', group_id, name="G")

print(fd['/A/G'].keys())
# [u'dataset1', u'dataset2']
like image 137
Yossarian Avatar answered Oct 11 '22 13:10

Yossarian