Develop Reference

Read and write from pigz and subprocess with python3

I am trying to use the pigz function from linux to speed up file decompression and compression. I managed to open a file with pigz using subprocess.Popen() function but after different tries, I don't manage to read the stream from Popen() make modifications on some lines, and write it directly on a new file using pigz and subprocess as well. In the end, I use the gzip.open() function from the gzip library to write the new file and the process is as slow as reading and writing directly from the gzip.open() function.
Question:
On the following code is there a way to modify the data from the output of subprocess and directly write it to a compressed file using subprocess and pigz in order to speed up the whole operation?
inputFile = "file1.txt.gz"
outputFile = "file2.txt.gz"
def pigzStream2(inputFile, outputFile):
cmd = f'pigz -dkc {inputFile} ' # -dkc: decompress, k: do not delete original file, c:Write all processed output to stdout (won't delete)
if not sys.platform.startswith("win"):
cmd = shlex.split(cmd)
res = Popen(cmd, stdout=PIPE, stdin=PIPE, bufsize=1, text=True)
with res.stdout as f_in:
with gzip.open(outputFile, 'ab') as f_out:
count = 0
while True:
count += 1
line = f_in.readline()
if line.startswith('#'):
line = f"line {count} changed"
if not line:
print(count)
break
f_out.write(line.encode())
return 0```

Unable to read file with python

I'm trying to read the content of a file with python 3.8.5 but the output is empty, I don't understand what I'm doing wrong.
Here is the code:
import subprocess
import os
filename = "ls.out"
ls_command = "ls -la"
file = open(filename, "w")
subprocess.Popen(ls_command, stdout=file, shell=True)
file.close()
# So far, all is ok. The file "ls.out" is correctly created and filled with the output of "ls -la" command"
file = open(filename, "r")
for line in file:
print(line)
file.close()
The output of this script is empty, it doesn't print anything. I'm not able to see the content of ls.out.
What is not correct here ?

Popen creates a new process and launches it but returns immediately. So the end result is that you've forked your code and have both processes running at once. Your python code in executing faster than the start and finish of ls. Thus, you need to wait for the process to finish by adding a call to wait():
import subprocess
import os
filename = "ls.out"
ls_command = "ls -la"
file = open(filename, "w")
proc = subprocess.Popen(ls_command, stdout=file, shell=True)
proc.wait()
file.close()
file = open(filename, "r")
for line in file:
print(line)
file.close()

Popen merely starts the subprocess. Chances are the file is not yet populated when you open it.
If you want to wait for the Popen object to finish, you have to call its wait method, etc; but a much better and simpler solution is to use subprocess.check_call() or one of the other higher-level wrappers.
If the command prints to standard output, why don't you read it drectly?
import subprocess
import shlex
result = subprocess.run(
shlex.split(ls_command), # avoid shell=True
check=True, text=True, capture_output=True)
line = result.stdout

why can't I read the output of this subprocess call in python?

I want to make a subprocess call to get the directory structure of a folder called ORIG.
Here's my code:
import os
from subprocess import call
# copy the directory structure of ORIG into file
f = open("outputFile.txt","r+b")
call(['find', './ORIG', '-type', 'd'], stdout=f)
a = f.read()
print(a)
The call command is working, as I see the contents of the file outputFile.txt when I open it:
./ORIG
./ORIG/child_one
./ORIG/child_one/grandchild_one
./ORIG/child_two
But why can't I read this/print the output?
According to Luke.py's suggestion, I tried the following also:
import os
import re
from subprocess import call, PIPE
# copy the directory structure of ORIG into file
# make directory structure from ORIG file in FINAL folder
process = call(['find', './ORIG', '-type', 'd'], stdout=PIPE, stderr=PIPE)
stdout, stderr = process.communicate()
if stderr:
print stderr
else:
print stdout
This gives me the outut:
Traceback (most recent call last):
File "feeder.py", line 9, in <module>
stdout, stderr = process.communicate()
AttributeError: 'int' object has no attribute 'communicate'

First: There is no need to call external programs. If you want to get the subdirectories of some path, there's the python function os.walk. You could use that and check each entry with os.path.isdir, or e.g. use os.fwalk and just use the directories.
If you really want to call an external program and get its stdout, usually the high level function subprocess.run is the right choice.
You can get the stdout with:
subprocess.run(command, stdout=subprocess.PIPE).stdout
Without the need for a temporary file or low level functions.

try Popen-
You'll need to import PIPE from subprocess.
process = subprocess.Popen(['find', './ORIG', '-type', 'd'], stdout=PIPE, stderr=PIPE)
stdout, stderr = process.communicate()
if stderr:
print stderr
else:
print stdout

If you don't want to close and reopen the file in between writing and reading, you can use the seek command to read the file from the beginning.
import os
from subprocess import call
# copy the directory structure of ORIG into file
f = open("outputFile.txt","r+b")
call(['find', './ORIG', '-type', 'd'], stdout=f)
# move back to the beginning of the file
f.seek(0, 0)
a = f.read()
print(a)

Writing to a file and reading it from a subprocess in python?

I'm creating a text file, and immediately after calling a subprocess that does some computation based on the text file.
When I call the subprocess by itself, it's able to read from the file as expected, but when I try to create a file and write to it immediately before, it is not able to read from the file.
f = open('name_data.txt', 'w')
f.write(name)
f.close()
cmd = ['g2p-seq2seq', '--decode', 'name_data.txt', '--model', 'g2p-seq2seq-cmudict']
process = subprocess.Popen(cmd, stdout=subprocess.PIPE)
process.wait()
#etc....

import subprocess
open("Edited.py", "w").write("Thing To Write")
A = subprocess.Popen('Command you want to call', shell = True, stdout = subprocess.PIPE, stderr = subprocess.PIPE)
print(A.communicate())

Ouputting the results of os.popen()

I am trying to send the results of os.popen() to an output file. Here is the code I have been trying
import os
cmd = 'dir'
fp = os.popen(cmd)
print(fp.read()) --Prints the results to the screen
res = fp.read()
fob = open('popen_output.txt','w')
fob.write(res)
fob.close()
fp.close()
The output file is just blank. The results of the command are however displayed on screen. I have also tried using Popen like this (as per the subprocess management documentation):
import subprocess
fob = Popen('dir',stdout='popen_output.txt',shell=true).stdout
As well as:
import subprocess
subprocess.Popen('dir',stdout='popen_output.txt,shell=true)

Pass a file object to stdout not a the file name as a string, you can also use check_call in place of Popen which will raise a CalledProcessError for a non-zero exit status:
with open('popen_output.txt',"w") as f:
subprocess.check_call('dir',stdout=f)
If you are on windows subprocess.check_call('dir',stdout=f, shell=True), you could also redirect using > using shell=True:
subprocess.check_call('dir > popen_output.txt',shell=True)

Ok. This got it going. Thanks for the help!
fob = open('popen_output.txt','a')
subprocess.Popen('dir',stdout=fob,shell=True)
fob.close()

This seems to be more what you'd like to do. You can process then write to file.
process = subprocess.Popen(cmd, stdout=subprocess.PIPE, shell=True)
for line in process.stdout:
#processing then write line to file...
file.write(line)
If you don't want to process, then you could just do it in your subprocess call.
subprocess.run('dir > popen_output.txt', shell=true)

The problem is you are calling fp.read() twice instead of saving the results of a single fp.read() call as res, printing res, and writing the res to the output file. A file handle is stateful, so if you call read on it twice, the current position after the first call will be at the end of the file/stream, hence your blank file.
Try this (just providing the relevant changes):
fp = os.popen(cmd)
res = fp.read()
print(res)