python - 使用 h5py 随机播放 HDF5 数据集

Question

我有一个很大的 HDF5 文件 (~30GB)，我需要打乱每个数据集中的条目(沿 0 轴)。通过查看 h5py 文档，我无法找到 randomAccess 或 shuffle 功能，但我希望我错过了一些东西。

是否有人对 HDF5 足够熟悉，可以想出一种快速随机打乱数据的方法？

以下是我将以我有限的知识实现的伪代码:

for dataset in datasets:
    unshuffled = range(dataset.dims[0])
    while unshuffled.length != 0:
        if unshuffled.length <= 100:
            dataset[:unshuffled.length/2], dataset[unshuffled.length/2:] = dataset[unshuffled.length/2:], dataset[:unshuffled.length/2]
            break
        else:
            randomIndex1 = rand(unshuffled.length - 100)
            randomIndex2 = rand(unshuffled.length - 100)

            unshuffled.removeRange(randomIndex1..<randomIndex1+100)
            unshuffled.removeRange(randomIndex2..<randomIndex2+100)

            dataset[randomIndex1:randomIndex1 + 100], dataset[randomIndex2:randomIndex2 + 100] = dataset[randomIndex2:randomIndex2 + 100], dataset[randomIndex1:randomIndex1 + 100]

Answer 1

您可以使用random.shuffle(dataset)。对于配备 Core i5 处理器、8 GB RAM 和 256 GB SSD 的笔记本电脑，这需要 11 分钟多一点的时间。请参阅以下内容:

>>> import os
>>> import random
>>> import time
>>> import h5py
>>> import numpy as np
>>>
>>> h5f = h5py.File('example.h5', 'w')
>>> h5f.create_dataset('example', (40000, 256, 256, 3), dtype='float32')
>>> # set all values of each instance equal to its index
... for i, instance in enumerate(h5f['example']):
...     h5f['example'][i, ...] = \
...             np.ones(instance.shape, dtype='float32') * i
...
>>> # get file size in bytes
... file_size = os.path.getsize('example.h5')
>>> print('Size of example.h5: {:.3f} GB'.format(file_size/2.0**30))
Size of example.h5: 29.297 GB
>>> def shuffle_time():
...     t1 = time.time()
...     random.shuffle(h5f['example'])
...     t2 = time.time()
...     print('Time to shuffle: {:.3f} seconds'.format(str(t2 - t1)))
...
>>> print('Value of first 5 instances:\n{}'
...       ''.format(str(h5f['example'][:10, 0, 0, 0])))
Value of first 5 instances:
[ 0.  1.  2.  3.  4.]
>>> shuffle_time()
Time to shuffle: 673.848 seconds
>>> print('Value of first 5 instances after '
...       'shuffling:\n{}'.format(str(h5f['example'][:10, 0, 0, 0])))
Value of first 5 instances after shuffling:
[ 15733.  28530.   4234. 14869.  10267.]
>>> h5f.close()

洗牌几个较小的数据集的性能应该不会比这差。