从二进制文件创建Numpy数组的有效方法

Question

我有非常大的数据集存储在硬盘上的二进制文件中.以下是文件结构的示例:

文件头

149 Byte ASCII Header

记录开始

4 Byte Int - Record Timestamp

样品开始

2 Byte Int - Data Stream 1 Sample
2 Byte Int - Data Stream 2 Sample
2 Byte Int - Data Stream 3 Sample
2 Byte Int - Data Stream 4 Sample

样品结束

每条记录有122,880个样本,每个文件有713个记录.这样产生的总大小为700,910,521字节.采样率和记录数有时会有所不同,因此我必须编码以检测每个文件的数量.

目前我用来将这些数据导入数组的代码如下:

from time import clock
from numpy import zeros , int16 , int32 , hstack , array , savez
from struct import unpack
from os.path import getsize

start_time = clock()
file_size = getsize(input_file)

with open(input_file,'rb') as openfile:
  input_data = openfile.read()

header = input_data[:149]
record_size = int(header[23:31])
number_of_records = ( file_size - 149 ) / record_size
sample_rate = ( ( record_size - 4 ) / 4 ) / 2

time_series = zeros(0,dtype=int32)
t_series = zeros(0,dtype=int16)
x_series = zeros(0,dtype=int16)
y_series = zeros(0,dtype=int16)
z_series = zeros(0,dtype=int16)

for record in xrange(number_of_records):

  time_stamp = array( unpack( '<l' , input_data[ 149 + (record * record_size) : 149 + (record * record_size) + 4 ] ) , dtype = int32 )
  unpacked_record = unpack( '<' + str(sample_rate * 4) + 'h' , input_data[ 149 + (record * record_size) + 4 : 149 + ( (record + 1) * record_size ) ] ) 

  record_t = zeros(sample_rate , dtype=int16)
  record_x = zeros(sample_rate , dtype=int16)
  record_y = zeros(sample_rate , dtype=int16)
  record_z = zeros(sample_rate , dtype=int16)

  for sample in xrange(sample_rate):

    record_t[sample] = unpacked_record[ ( sample * 4 ) + 0 ]
    record_x[sample] = unpacked_record[ ( sample * 4 ) + 1 ]
    record_y[sample] = unpacked_record[ ( sample * 4 ) + 2 ]
    record_z[sample] = unpacked_record[ ( sample * 4 ) + 3 ]

  time_series = hstack ( ( time_series , time_stamp ) )
  t_series = hstack ( ( t_series , record_t ) )
  x_series = hstack ( ( x_series , record_x ) )
  y_series = hstack ( ( y_series , record_y ) )
  z_series = hstack ( ( z_series , record_z ) )

savez(output_file, t=t_series , x=x_series ,y=y_series, z=z_series, time=time_series)
end_time = clock()
print 'Total Time',end_time - start_time,'seconds'

这目前每700 MB文件大约需要250秒,这对我来说似乎非常高.有没有更有效的方法可以做到这一点？

最终解决方案

使用带有自定义dtype的numpy fromfile方法将运行时间缩短为9秒,比上面的原始代码快27倍.最终代码如下.

from numpy import savez, dtype , fromfile 
from os.path import getsize
from time import clock

start_time = clock()
file_size = getsize(input_file)

openfile = open(input_file,'rb')
header = openfile.read(149)
record_size = int(header[23:31])
number_of_records = ( file_size - 149 ) / record_size
sample_rate = ( ( record_size - 4 ) / 4 ) / 2

record_dtype = dtype( [ ( 'timestamp' , '<i4' ) , ( 'samples' , '<i2' , ( sample_rate , 4 ) ) ] )

data = fromfile(openfile , dtype = record_dtype , count = number_of_records )
time_series = data['timestamp']
t_series = data['samples'][:,:,0].ravel()
x_series = data['samples'][:,:,1].ravel()
y_series = data['samples'][:,:,2].ravel()
z_series = data['samples'][:,:,3].ravel()

savez(output_file, t=t_series , x=x_series ,y=y_series, z=z_series, fid=time_series)

end_time = clock()

print 'It took',end_time - start_time,'seconds'

Answer 1

一些提示:

• 不要使用struct模块.相反,使用Numpy的结构化数据类型和fromfile.点击此处:http://scipy-lectures.github.com/advanced/advanced_numpy/index.html#example-reading-wav-files

• 您可以通过传入合适的count =来一次读取所有记录fromfile.

像这样的东西(未经测试,但你明白了):

import numpy as np

file = open(input_file, 'rb')
header = file.read(149)

# ... parse the header as you did ...

record_dtype = np.dtype([
    ('timestamp', '<i4'), 
    ('samples', '<i2', (sample_rate, 4))
])

data = np.fromfile(file, dtype=record_dtype, count=number_of_records)
# NB: count can be omitted -- it just reads the whole file then

time_series = data['timestamp']
t_series = data['samples'][:,:,0].ravel()
x_series = data['samples'][:,:,1].ravel()
y_series = data['samples'][:,:,2].ravel()
z_series = data['samples'][:,:,3].ravel()