Qustion about struct.unpack
Steven D'Aprano
steve at REMOVEME.cybersource.com.au
Mon Apr 30 04:41:33 EDT 2007
On Mon, 30 Apr 2007 00:45:22 -0700, OhKyu Yoon wrote:
> Hi!
> I have a really long binary file that I want to read.
> The way I am doing it now is:
>
> for i in xrange(N): # N is about 10,000,000
> time = struct.unpack('=HHHH', infile.read(8))
> # do something
> tdc = struct.unpack('=LiLiLiLi',self.lmf.read(32))
I assume that is supposed to be infile.read()
> # do something
>
> Each loop takes about 0.2 ms in my computer, which means the whole for loop
> takes 2000 seconds.
You're reading 400 million bytes, or 400MB, in about half an hour. Whether
that's fast or slow depends on what the "do something" lines are doing.
> I would like it to run faster.
> Do you have any suggestions?
Disk I/O is slow, so don't read from files in tiny little chunks. Read a
bunch of records into memory, then process them.
# UNTESTED!
rsize = 8 + 32 # record size
for i in xrange(N//1000):
buffer = infile.read(rsize*1000) # read 1000 records at once
for j in xrange(1000): # process each record
offset = j*rsize
time = struct.unpack('=HHHH', buffer[offset:offset+8])
# do something
tdc = struct.unpack('=LiLiLiLi', buffer[offset+8:offset+rsize])
# do something
(Now I'm just waiting for somebody to tell me that file.read() already
buffers reads...)
--
Steven D'Aprano
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