I need to read a text file with the os module as such:
t = os.open('te.txt', os.O_RDONLY)
r = os.read(t, 20)
rs = r.decode('utf-8')
print(rs)
What if I don't know the byte size of the file. I could put a very large number instead of 20 as a value seems to be required, but perhaps there is a more pythonic way.
The second argument isn't supposed to hold the size of the file in bytes; it's only supposed to hold the maximum amount of content you're prepared to read at a time (which should typically be divisible by both your operating system's block size and page size; 64kb is not a bad default).
The "why" of this is because memory has to be allocated in userspace before the kernel can be instructed to write content into that memory. This isn't the kind of detail that Python developers need to think about often, but you're using a low-level interface built for use from C; it accordingly has implementation details leaking out of that underlying layer.
The operating system is free to give you less than the number of bytes you indicate as a maximum (for example, if it gets interrupted, or the filesystem driver isn't written to provide that much data at a time), so no matter what, you need to be prepared to call it repeatedly; only when it returns an empty string (as opposed to throwing an exception or returning a shorter-than-requested string) are you certain to have reached the end of the file.
os.read() isn't a Pythonic interface, and it isn't supposed to be. It's a thin wrapper around the syscall provided by the operating system kernel. If you want a Pythonic interface, don't use os.read(), but instead use Python's native file objects.
If you wanted to load the whole file and you have to use os, you could use os.stat(filename).st_size or os.path.getsize(filename) to get the size of the file in bytes.
filename = 'te.txt'
t = os.open(filename, os.O_RDONLY)
b = os.stat(filename).st_size
r = os.read(t, b)
rs = r.decode('utf-8')
print(rs)
I am looking to do, in Python 3.8, the equivalent of:
xz --decompress --stdout < hugefile.xz > hugefile.out
where neither the input nor output might fit well in memory.
As I read the documentation at https://docs.python.org/3/library/lzma.html#lzma.LZMADecompressor
I could use LZMADecompressor to process incrementally available input, and I could use its decompress() function to produce output incrementally.
However it seems that LZMADecompressor puts its entire decompressed output into a single memory buffer, and decompress() reads its entire compressed input from a single input memory buffer.
Granted, the documentation confuses me as to when the input and/or output can be incremental.
So I figure I will have to spawn a separate child process to execute the "xz" binary.
Is there anyway of using the lzma Python module for this task?
Instead of using the low-level LZMADecompressor, use lzma.open to get a file object. Then, you can copy data into an other file object with the shutil module:
import lzma
import shutil
with lzma.open("hugefile.xz", "rb") as fsrc:
with open("hugefile.out", "wb") as fdst:
shutil.copyfileobj(fsrc, fdst)
Internally, shutils.copyfileobj reads and write data in chunks, and the LZMA decompression is done on the fly. This avoids decompressing the whole data into memory.
How can I quickly create a large file on a Linux (Red Hat Linux) system?
dd will do the job, but reading from /dev/zero and writing to the drive can take a long time when you need a file several hundreds of GBs in size for testing... If you need to do that repeatedly, the time really adds up.
I don't care about the contents of the file, I just want it to be created quickly. How can this be done?
Using a sparse file won't work for this. I need the file to be allocated disk space.
dd from the other answers is a good solution, but it is slow for this purpose. In Linux (and other POSIX systems), we have fallocate, which uses the desired space without having to actually writing to it, works with most modern disk based file systems, very fast:
For example:
fallocate -l 10G gentoo_root.img
This is a common question -- especially in today's environment of virtual environments. Unfortunately, the answer is not as straight-forward as one might assume.
dd is the obvious first choice, but dd is essentially a copy and that forces you to write every block of data (thus, initializing the file contents)... And that initialization is what takes up so much I/O time. (Want to make it take even longer? Use /dev/random instead of /dev/zero! Then you'll use CPU as well as I/O time!) In the end though, dd is a poor choice (though essentially the default used by the VM "create" GUIs). E.g:
dd if=/dev/zero of=./gentoo_root.img bs=4k iflag=fullblock,count_bytes count=10G
truncate is another choice -- and is likely the fastest... But that is because it creates a "sparse file". Essentially, a sparse file is a section of disk that has a lot of the same data, and the underlying filesystem "cheats" by not really storing all of the data, but just "pretending" that it's all there. Thus, when you use truncate to create a 20 GB drive for your VM, the filesystem doesn't actually allocate 20 GB, but it cheats and says that there are 20 GB of zeros there, even though as little as one track on the disk may actually (really) be in use. E.g.:
truncate -s 10G gentoo_root.img
fallocate is the final -- and best -- choice for use with VM disk allocation, because it essentially "reserves" (or "allocates" all of the space you're seeking, but it doesn't bother to write anything. So, when you use fallocate to create a 20 GB virtual drive space, you really do get a 20 GB file (not a "sparse file", and you won't have bothered to write anything to it -- which means virtually anything could be in there -- kind of like a brand new disk!) E.g.:
fallocate -l 10G gentoo_root.img
Linux & all filesystems
xfs_mkfile 10240m 10Gigfile
Linux & and some filesystems (ext4, xfs, btrfs and ocfs2)
fallocate -l 10G 10Gigfile
OS X, Solaris, SunOS and probably other UNIXes
mkfile 10240m 10Gigfile
HP-UX
prealloc 10Gigfile 10737418240
Explanation
Try mkfile <size> myfile as an alternative of dd. With the -n option the size is noted, but disk blocks aren't allocated until data is written to them. Without the -n option, the space is zero-filled, which means writing to the disk, which means taking time.
mkfile is derived from SunOS and is not available everywhere. Most Linux systems have xfs_mkfile which works exactly the same way, and not just on XFS file systems despite the name. It's included in xfsprogs (for Debian/Ubuntu) or similar named packages.
Most Linux systems also have fallocate, which only works on certain file systems (such as btrfs, ext4, ocfs2, and xfs), but is the fastest, as it allocates all the file space (creates non-holey files) but does not initialize any of it.
truncate -s 10M output.file
will create a 10 M file instantaneously (M stands for 10241024 bytes, MB stands for 10001000 - same with K, KB, G, GB...)
EDIT: as many have pointed out, this will not physically allocate the file on your device. With this you could actually create an arbitrary large file, regardless of the available space on the device, as it creates a "sparse" file.
For e.g. notice no HDD space is consumed with this command:
### BEFORE
$ df -h | grep lvm
/dev/mapper/lvm--raid0-lvm0
7.2T 6.6T 232G 97% /export/lvm-raid0
$ truncate -s 500M 500MB.file
### AFTER
$ df -h | grep lvm
/dev/mapper/lvm--raid0-lvm0
7.2T 6.6T 232G 97% /export/lvm-raid0
So, when doing this, you will be deferring physical allocation until the file is accessed. If you're mapping this file to memory, you may not have the expected performance.
But this is still a useful command to know. For e.g. when benchmarking transfers using files, the specified size of the file will still get moved.
$ rsync -aHAxvP --numeric-ids --delete --info=progress2 \
root#mulder.bub.lan:/export/lvm-raid0/500MB.file \
/export/raid1/
receiving incremental file list
500MB.file
524,288,000 100% 41.40MB/s 0:00:12 (xfr#1, to-chk=0/1)
sent 30 bytes received 524,352,082 bytes 38,840,897.19 bytes/sec
total size is 524,288,000 speedup is 1.00
Where seek is the size of the file you want in bytes - 1.
dd if=/dev/zero of=filename bs=1 count=1 seek=1048575
Examples where seek is the size of the file you want in bytes
#kilobytes
dd if=/dev/zero of=filename bs=1 count=0 seek=200K
#megabytes
dd if=/dev/zero of=filename bs=1 count=0 seek=200M
#gigabytes
dd if=/dev/zero of=filename bs=1 count=0 seek=200G
#terabytes
dd if=/dev/zero of=filename bs=1 count=0 seek=200T
From the dd manpage:
BLOCKS and BYTES may be followed by the following multiplicative suffixes: c=1, w=2, b=512, kB=1000, K=1024, MB=1000*1000, M=1024*1024, GB =1000*1000*1000, G=1024*1024*1024, and so on for T, P, E, Z, Y.
To make a 1 GB file:
dd if=/dev/zero of=filename bs=1G count=1
I don't know a whole lot about Linux, but here's the C Code I wrote to fake huge files on DC Share many years ago.
#include < stdio.h >
#include < stdlib.h >
int main() {
int i;
FILE *fp;
fp=fopen("bigfakefile.txt","w");
for(i=0;i<(1024*1024);i++) {
fseek(fp,(1024*1024),SEEK_CUR);
fprintf(fp,"C");
}
}
You can use "yes" command also. The syntax is fairly simple:
#yes >> myfile
Press "Ctrl + C" to stop this, else it will eat up all your space available.
To clean this file run:
#>myfile
will clean this file.
I don't think you're going to get much faster than dd. The bottleneck is the disk; writing hundreds of GB of data to it is going to take a long time no matter how you do it.
But here's a possibility that might work for your application. If you don't care about the contents of the file, how about creating a "virtual" file whose contents are the dynamic output of a program? Instead of open()ing the file, use popen() to open a pipe to an external program. The external program generates data whenever it's needed. Once the pipe is open, it acts just like a regular file in that the program that opened the pipe can fseek(), rewind(), etc. You'll need to use pclose() instead of close() when you're done with the pipe.
If your application needs the file to be a certain size, it will be up to the external program to keep track of where in the "file" it is and send an eof when the "end" has been reached.
One approach: if you can guarantee unrelated applications won't use the files in a conflicting manner, just create a pool of files of varying sizes in a specific directory, then create links to them when needed.
For example, have a pool of files called:
/home/bigfiles/512M-A
/home/bigfiles/512M-B
/home/bigfiles/1024M-A
/home/bigfiles/1024M-B
Then, if you have an application that needs a 1G file called /home/oracle/logfile, execute a "ln /home/bigfiles/1024M-A /home/oracle/logfile".
If it's on a separate filesystem, you will have to use a symbolic link.
The A/B/etc files can be used to ensure there's no conflicting use between unrelated applications.
The link operation is about as fast as you can get.
The GPL mkfile is just a (ba)sh script wrapper around dd; BSD's mkfile just memsets a buffer with non-zero and writes it repeatedly. I would not expect the former to out-perform dd. The latter might edge out dd if=/dev/zero slightly since it omits the reads, but anything that does significantly better is probably just creating a sparse file.
Absent a system call that actually allocates space for a file without writing data (and Linux and BSD lack this, probably Solaris as well) you might get a small improvement in performance by using ftrunc(2)/truncate(1) to extend the file to the desired size, mmap the file into memory, then write non-zero data to the first bytes of every disk block (use fgetconf to find the disk block size).
This is the fastest I could do (which is not fast) with the following constraints:
The goal of the large file is to fill a disk, so can't be compressible.
Using ext3 filesystem. (fallocate not available)
This is the gist of it...
// include stdlib.h, stdio.h, and stdint.h
int32_t buf[256]; // Block size.
for (int i = 0; i < 256; ++i)
{
buf[i] = rand(); // random to be non-compressible.
}
FILE* file = fopen("/file/on/your/system", "wb");
int blocksToWrite = 1024 * 1024; // 1 GB
for (int i = 0; i < blocksToWrite; ++i)
{
fwrite(buf, sizeof(int32_t), 256, file);
}
In our case this is for an embedded linux system and this works well enough, but would prefer something faster.
FYI the command dd if=/dev/urandom of=outputfile bs=1024 count = XX was so slow as to be unusable.
Shameless plug: OTFFS provides a file system providing arbitrarily large (well, almost. Exabytes is the current limit) files of generated content. It is Linux-only, plain C, and in early alpha.
See https://github.com/s5k6/otffs.
So I wanted to create a large file with repeated ascii strings. "Why?" you may ask. Because I need to use it for some NFS troubleshooting I'm doing. I need the file to be compressible because I'm sharing a tcpdump of a file copy with the vendor of our NAS. I had originally created a 1g file filled with random data from /dev/urandom, but of course since it's random, it means it won't compress at all and I need to send the full 1g of data to the vendor, which is difficult.
So I created a file with all the printable ascii characters, repeated over and over, to a limit of 1g in size. I was worried it would take a long time. It actually went amazingly quickly, IMHO:
cd /dev/shm
date
time yes $(for ((i=32;i<127;i++)) do printf "\\$(printf %03o "$i")"; done) | head -c 1073741824 > ascii1g_file.txt
date
Wed Apr 20 12:30:13 CDT 2022
real 0m0.773s
user 0m0.060s
sys 0m1.195s
Wed Apr 20 12:30:14 CDT 2022
Copying it from an nfs partition to /dev/shm took just as long as with the random file (which one would expect, I know, but I wanted to be sure):
cp ascii1gfile.txt /home/greygnome/
uptime; free -m; sync; echo 1 > /proc/sys/vm/drop_caches; free -m; date; dd if=/home/greygnome/ascii1gfile.txt of=/dev/shm/outfile bs=16384 2>&1; date; rm -f /dev/shm/outfile
But while doing that I ran a simultaneous tcpdump:
tcpdump -i em1 -w /dev/shm/dump.pcap
I was able to compress the pcap file down to 12M in size! Awesomesauce!
Edit: Before you ding me because the OP said, "I don't care about the contents," know that I posted this answer because it's one of the first replies to "how to create a large file linux" in a Google search. And sometimes, disregarding the contents of a file can have unforeseen side effects.
Edit 2: And fallocate seems to be unavailable on a number of filesystems, and creating a 1GB compressible file in 1.2s seems pretty decent to me (aka, "quickly").
You could use https://github.com/flew-software/trash-dump
you can create file that is any size and with random data
heres a command you can run after installing trash-dump (creates a 1GB file)
$ trash-dump --filename="huge" --seed=1232 --noBytes=1000000000
BTW I created it
I have a few text files whose sizes range between 5 gigs and 50 gigs. I am using Python to read them. I have specific anchors in terms of byte offsets, to which I can seek and read the corresponding data from each of these files (using Python's file api).
The issue that I am seeing is that for relatively smaller files (< 5 gigs), this reading approach works well. However, for the much larger files (> 20 gigs) and especially when the file.seek function has to take longer jumps (like a few multi-million bytes at a time), it (sometimes) takes a few hundred milliseconds for it to do so.
My impression was that seek operations within the files are constant time operations. But apparently, they are not. Is there a way around it?
Here is what I am doing:
import time
f = open(filename, 'r+b')
f.seek(209)
current = f.tell()
t1 = time.time()
next = f.seek(current + 1200000000)
t2 = time.time()
line = f.readline()
delta = t2 - t1
The delta variable is varying between few microseconds to few hundreeld milliseconds, intermittently. I also profiled the cpu usage, and didnt see anything busy there as well.
Your code runs consistently in under 10 microseconds on my system (Windows 10, Python 3.7), so there is no obvious error in your code.
NB: You should use time.perf_counter() instead of time.time() for measuring performance. The granularity of time.time() can be very bad ("not all systems provide time with a better precision than 1 second"). When comparing timings with other systems you may get strange results.
My best guess is that the seek triggers some buffering (read-ahead) action, which might be slow, depending on your system.
According to the documentation:
Binary files are buffered in fixed-size chunks; the size of the buffer is chosen using a heuristic trying to determine the underlying device’s “block size” and falling back on io.DEFAULT_BUFFER_SIZE. On many systems, the buffer will typically be 4096 or 8192 bytes long.
You could try to disable buffering by adding the argument buffering=0 to open() and check if that makes a difference:
open(filename, 'r+b', buffering=0)
A good way around that could be combining functions from OS module os.open (with flag os.O_RDONLY in your case), os.lseek, os.read which are at low-level I/O
I use the following method to read binary data from any given offset in the binary file. The binary file I have is huge 10GB, so I usually read portion of it when needed by specifying from which offset I should start_read and how many bytes to read num_to_read. I use Python 3.6.4 :: Anaconda, Inc., platform Darwin-17.6.0-x86_64-i386-64bit and os module:
def read_from_disk(path, start_read, num_to_read, dim):
fd = os.open(path, os.O_RDONLY)
os.lseek(fd, start_read, 0) # Where to (start_read) from the beginning 0
raw_data = os.read(fd, num_to_read) # How many bytes to read
C = np.frombuffer(raw_data, dtype=np.int64).reshape(-1, dim).astype(np.int8)
os.close(fd)
return C
This method works very well when the chunk of data to be read is about less than 2GB. When num_to_read > 2GG, I get this error:
raw_data = os.read(fd, num_to_read) # How many to read (num_to_read)
OSError: [Errno 22] Invalid argument
I am not sure why this issue appears and how to fix it. Any help is highly appreciated.
The os.read function is just a thin wrapper around the platform's read function.
On some platforms, this is an unsigned or signed 32-bit int,1 which means the largest you can read in a single go on these platforms is, respectively, 4GB or 2GB.
So, if you want to read more than that, and you want to be cross-platform, you have to write code to handle this, and to buffer up multiple reads.
This may be a bit of a pain, but you are intentionally using the lowest-level directly-mapping-to-the-OS-APIs function here. If you don't like that:
Use io module objects (Python 3.x) or file objects (2.7) that you get back from open instead.
Just let NumPy read the files—which will have the added advantage that NumPy is smart enough to not try to read the whole thing into memory at once in the first place.
Or, for files this large, you may want to go lower level and use mmap (assuming you're on a 64-bit platform).
The right thing to do here is almost certainly a combination of the first two. In Python 3, it would look like this:
with open(path, 'rb', buffering=0) as f:
f.seek(start_read)
count = num_to_read // 8 # how many int64s to read
return np.fromfile(f, dtype=np.int64, count=count).reshape(-1, dim).astype(np.int8)
1. For Windows, the POSIX-emulation library's _read function uses int for the count argument, which is signed 32-bit. For every other modern platform, see POSIX read, and then look up the definitions of size_t, ssize_t, and off_t, on your platform. Notice that many POSIX platforms have separate 64-bit types, and corresponding functions, instead of changing the meaning of the existing types to 64-bit. Python will use the standard types, not the special 64-bit types.