问题描述
我发现用 Python 读取二进制文件特别困难.你能帮我个忙吗?我需要阅读这个文件，它在 Fortran 90 中很容易被阅读
I find particularly difficult reading binary file with Python. Can you give me a hand?I need to read this file, which in Fortran 90 is easily read by
int*4 n_particles, n_groups
real*4 group_id(n_particles)
read (*) n_particles, n_groups
read (*) (group_id(j),j=1,n_particles)
具体来说，文件格式是:
In detail, the file format is:
Bytes 1-4 -- The integer 8.
Bytes 5-8 -- The number of particles, N.
Bytes 9-12 -- The number of groups.
Bytes 13-16 -- The integer 8.
Bytes 17-20 -- The integer 4*N.
Next many bytes -- The group ID numbers for all the particles.
Last 4 bytes -- The integer 4*N.
如何使用 Python 阅读此内容?我尝试了一切，但从未奏效.我有没有机会在 python 中使用 f90 程序，读取这个二进制文件，然后保存我需要使用的数据?
How can I read this with Python? I tried everything but it never worked. Is there any chance I might use a f90 program in python, reading this binary file and then save the data that I need to use?
推荐答案
读取二进制文件内容如下:
Read the binary file content like this:
with open(fileName, mode='rb') as file: # b is important -> binary
fileContent = file.read()
然后使用struct.unpack解压"二进制数据:
then "unpack" binary data using struct.unpack:
起始字节:struct.unpack("iiiiii", fileContent[:20])
正文:忽略标题字节和尾随字节(= 24)；剩下的部分构成正文，要知道正文中的字节数，进行整数除以 4；得到的商乘以字符串 'i' 为 unpack 方法创建正确的格式:
The body: ignore the heading bytes and the trailing byte (= 24); The remaining part forms the body, to know the number of bytes in the body do an integer division by 4; The obtained quotient is multiplied by the string 'i' to create the correct format for the unpack method:
struct.unpack("i" * ((len(fileContent) -24) // 4), fileContent[20:-4])
结束字节:struct.unpack("i", fileContent[-4:])
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