8
votes

Is there any way I can recursively get all keys in h5 file using python library h5py? I tried using the code below

import h5py

h5_data = h5py.File(h5_file_location, 'r')
print(h5_data.keys())

but it only print the top level keys of the h5 file.

3
The group class has visit and visititems methods that can be used to walk the group hierarchy. - hpaulj
This really should be part of the HDF5 API. If I save f["a/b/c"] = data I expect to find "a/b/c"in f.keys(). - TheMechanic

3 Answers

8
votes

Some of the keys returned by keys() on a Group may be Datasets some may be sub Groups. In order to find all keys you need to recurse the Groups. Here is a simple script to do that:

import h5py

def allkeys(obj):
    "Recursively find all keys in an h5py.Group."
    keys = (obj.name,)
    if isinstance(obj, h5py.Group):
        for key, value in obj.items():
            if isinstance(value, h5py.Group):
                keys = keys + allkeys(value)
            else:
                keys = keys + (value.name,)
    return keys

h5 = h5py.File('/dev/null', 'w')
h5.create_group('g1')
h5.create_group('g2')
h5.create_dataset('d1', (10,), 'i')
h5.create_dataset('d2', (10, 10,), 'f')
h5['g1'].create_group('g1')
h5['g1'].create_dataset('d1', (10,), 'i')
h5['g1'].create_dataset('d2', (10,), 'f')
h5['g1/g1'].attrs['a'] = 'b'
print(allkeys(h5))

Gives:

('/', '/d1', '/d2', '/g1', '/g1/d1', '/g1/d2', '/g1/g1', '/g2')
1
votes

You may want to loop through the key to return their values. Below is a simple function that does that.

import h5py

h5_file_location = '../../..'
h5_data = h5py.File(h5_file_location, 'r')

def keys(f):
    return [key for key in f.keys()]
print(keys(h5_data))
0
votes

Based on the top answer I wrote this function to return all bottom-level keys only:

def allbottomkeys(obj):
bottomkeys=[]
def allkeys(obj):
    keys = (obj.name,)
    if isinstance(obj, h5py.Group):
        for key, value in obj.items():
            if isinstance(value, h5py.Group):
                keys = keys + allkeys(value)
            else:
                keys = keys + (value.name,)
                bottomkeys.append(value.name)
    return keys
allkeys(obj)
return bottomkeys