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我有下一个任务：我需要从文件中读取结构数组。读取一个结构没有问题：

structFmt = "=64s 2L 3d"    # char[ 64 ] long[ 2 ] double [ 3 ]
structLen = struct.calcsize( structFmt )
f = open( "path/to/file", "rb" )
structBytes = f.read( structLen )
s = struct.unpack( structFmt, structBytes )

读取“简单”类型的数组也没有问题：

f = open( "path/to/file", "rb" )
a = array.array( 'i' )
a.fromfile( f, 1024 )

但是structFmt从文件中读取 1024 个结构是有问题的（当然对我来说）。我认为，读取 1024 次 struct 并将其附加到列表中是一种开销。我不想使用像numpy.

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2 回答 2

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我会查看对文件进行映射，然后使用 ctypes 类方法 from_buffer() 调用。这将映射 ctypes 定义的结构数组http://docs.python.org/library/ctypes#ctypes-arrays。

这将结构映射到 mmap 文件上，而无需显式读取/转换和复制内容。

我不知道最终结果是否合适。

只是为了好玩，这里是一个使用 mmap 的快速示例。（我使用 dd 创建了一个文件dd if=/dev/zero of=./test.dat bs=96 count=10240

from ctypes import Structure
from ctypes import c_char, c_long, c_double
import mmap
import timeit


class StructFMT(Structure):
     _fields_ = [('ch',c_char * 64),('lo',c_long *2),('db',c_double * 3)]

d_array = StructFMT * 1024

def doit():
    f = open('test.dat','r+b')
    m = mmap.mmap(f.fileno(),0)
    data = d_array.from_buffer(m)

    for i in data:
        i.ch, i.lo[0]*10 ,i.db[2]*1.0   # just access each row and bit of the struct and do something, with the data.

    m.close()
    f.close()

if __name__ == '__main__':
    from timeit import Timer
    t = Timer("doit()", "from __main__ import doit")
    print t.timeit(number=10)
于 2012-08-08T07:05:05.080 回答
1

Alas, there is no analog for array that holds complex structs.

The usual technique is to make many calls to struct.unpack and append the results to a list.

structFmt = "=64s 2L 3d"    # char[ 64 ] long[ 2 ] double [ 3 ]
structLen = struct.calcsize( structFmt )
results = []
with open( "path/to/file", "rb" ) as f:
    structBytes = f.read( structLen )
    s = struct.unpack( structFmt, structBytes )
    results.append(s)

If you're concerned about being efficient, know that struct.unpack caches the parsed structure between successive calls.

于 2012-08-08T04:37:54.870 回答