I have a python script that should run 7z.exe with the command: "x" and switch " -o" using subprocess.run(). The script is as follows:
import subprocess as sb
zipperpath = "C:\\Program Files\\7-zip\\7z.exe"
dirname ="C:\\Users\\ajain\\Desktop\\TempData"
archivename="UnprocessedData_v3.7z"
outputfilename="foo"
sb.run([zipperpath,"x",os.path.join(dirname,archivename)," -o",os.path.join(dirname,outputfilename)])
Output is:
Although the return code is 0, the zip never gets unzipped.
Try with this :
import subprocess
# variable cmd is is your command line , like you put in your console
cmd='7z.exe x UnprocessedData_v3.7z'
process = subprocess.Popen(cmd,shell=True,stdin=None,stdout=subprocess.PIPE,stderr=subprocess.PIPE)
# The output from your shell command in an array
result=process.stdout.readlines()
if len(result) >= 1:
for line in result:
print(line)
Related
I wrote a script to run mafft module from the terminal:
import subprocess
def linsi_MSA(sequnces_file_path):
cmd = ' mafft --maxiterate 1000 --localpair {seqs} > {out}'.format(seqs=sequnces_file_path, out=sequnces_file_path)
subprocess.call(cmd.split(), shell=True)
if __name__ == '__main__':
import logging
logger = logging.getLogger('main')
from sys import argv
if len(argv) < 2:
logger.error('Usage: MSA <sequnces_file_path> ')
exit()
else:
linsi_MSA(*argv[1:])
for some reason when trying to run the script from the terminal using:
python ./MSA.py ./sample.fa
I get the mafft interactive version opening directly in the trminal (asking for input ..output etc..)
when i'm trying to write the cmd directly in the terminal using:
mafft --maxiterate 1000 --localpair sample.fa > sample.fa
its working as expected and perfoming the command line version as without opening the interactive version.
I want my script to be able to perform the cmd line version on the terminal. what seems to be the problem?
thanks!
If you use shell=True you should pass one string as argument, not a list, e.g.:
subprocess.call("ls > outfile", shell=True)
It's not explained in the docs, but I suspect it has to do with what low-level library function is ultimately called:
call(["ls", "-l"]) --> execlp("ls", "-l")
^^^^^^^^^^ ^^^^^^^^^^
call("ls -l", shell=True) --> execlp("sh", "-c", "ls -l")
^^^^^^^ ^^^^^^^
call(["ls", "-l"], shell=True) --> execlp("sh", "-c", "ls", "-l")
# which can be tried from command line:
sh -c ls -l
# result is a list of files without details, -l was ignored.
# see sh(1) man page for -c string syntax and what happens to further arguments.
I am trying to run python subprocess.run function to execute following command:
pdftoppm -jpeg -f 1 -scale-to 200 data/andromeda.pdf and-page
pdftoppm - is part of poppler utility and it generates images from pdf files.
File data/andromeda.pdf exists. Folder data is on same level with python script and/or where I run command from.
Command basically will generate a jpeg file, from page 1 (-f 1) 200px wide (-scale-to) from given file of and-page-1.jpeg format (so called ppmtroot).
Long story short: from command line it works as expected i.e. if I call the above command either from zsh or bash shell, manually - it generates thumbnail as expected. However if I run it from python subprocess module - it fails it returns 99 error code!
Following is python code (file name is sc_02_thumbnails.py):
import subprocess
import sys
def main(filename, ppmroot):
cmd = [
'pdftoppm',
'-f 1',
'-scale-to 200',
'-jpeg',
filename,
ppmroot
]
result = subprocess.run(
cmd,
stdout=subprocess.PIPE,
stderr=subprocess.PIPE
)
if result.returncode:
print("Failed to generate thumbnail. Return code: {}. stderr: {}".format(
result.returncode,
result.stderr
))
print("Used cmd: {}".format(' '.join(cmd)))
sys.exit(1)
else:
print("Success!")
if __name__ == "__main__":
if len(sys.argv) > 2:
filename = sys.argv[1]
ppmroot = sys.argv[2]
else:
print("Usage: {} <pdffile> <ppmroot>".format(sys.argv[0]))
sys.exit(1)
main(filename, ppmroot)
And here is repo which includes data/andromeda.pdf file as well.
I call my script with as (from zsh):
$ chmod +x ./sc_02_thumbnauils.py
$ ./sc_02_thumbnails.py data/andromeda.pdf and-page
and ... thumbnail generating fails!
I have tried executing python script from both, from zsh and bash shells :(
What I am doing wrong?
The quoting is wrong, you should have '-f', '1', etc
I want to redirect o/p of shell commands to file using variable "path" but it is not working
import os, socket, shutil, subprocess
host = os.popen("hostname -s").read().strip()
path = "/root/" + host
if os.path.exists(path):
print(path, "Already exists")
else:
os.mkdir("Directory", path , "Created")
os.system("uname -a" > path/'uname') # I want to redirect o/p of shell commands to file using varibale "path" but it is not working
os.system("df -hP"> path/'df')
I think the problem is the bare > and / symbols in the os.system command...
Here is a python2.7 example with os.system that does what you want
import os
path="./test_dir"
command_str="uname -a > {}/uname".format(path)
os.system(command_str)
Here's a very minimal example using subprocess.run. Also, search StackOverflow for "python shell redirect", and you'll get this result right away:
Calling an external command in Python
import subprocess
def run(filename, command):
with open(filename, 'wb') as stdout_file:
process = subprocess.run(command, stdout=subprocess.PIPE, shell=True)
stdout_file.write(process.stdout)
return process.returncode
run('test_out.txt', 'ls')
I am trying to run the shell command which tees the output to some file within python script but when I see the log file of tee output it's empty. The code looks like this:
val_home = os.environ["xyz"]
pwd = os.getcwd()
os.chdir(val_home)
os.chdir("..")
avk_home = os.environ["abc"]
del os.environ['abc']
command = "source %s/somefile/ %s/inputparamater | tee config.log" %(val_home,val_home)
execute(command)#function written to execute shell command ,no issue with this
assert search_string("config.log","Success") ,"Target config failed"
rm_file(['config.log'])
os.environ["AVKRUN_HOME"] = avk_home
os.chdir(pwd)
The file config.out is empty but when I run the command manually I see the tee file output.
I have an .R file saved locally at the following path:
Rfilepath = "C:\\python\\buyback_parse_guide.r"
The command for RScript.exe is:
RScriptCmd = "C:\\Program Files\\R\\R-2.15.2\\bin\\Rscript.exe --vanilla"
I tried running:
subprocess.call([RScriptCmd,Rfilepath],shell=True)
But it returns 1 -- and the .R script did not run successfully. What am I doing wrong? I'm new to Python so this is probably a simple syntax error... I also tried these, but they all return 1:
subprocess.call('"C:\Program Files\R\R-2.15.2\bin\Rscript.exe"',shell=True)
subprocess.call('"C:\\Program Files\\R\\R-2.15.2\\bin\\Rscript.exe"',shell=True)
subprocess.call('C:\Program Files\R\R-2.15.2\bin\Rscript.exe',shell=True)
subprocess.call('C:\\Program Files\\R\\R-2.15.2\\bin\\Rscript.exe',shell=True)
Thanks!
The RScriptCmd needs to be just the executable, no command line arguments. So:
RScriptCmd = "\"C:\\Program Files\\R\\R-2.15.2\\bin\\Rscript.exe\""
Then the Rfilepath can actually be all of the arguments - and renamed:
RArguments = "--vanilla \"C:\\python\\buyback_parse_guide.r\""
It looks like you have a similar problem to mine. I had to reinstall RScript to a path which has no spaces.
See: Running Rscript via Python using os.system() or subprocess()
This is how I worked out the communication between Python and Rscript:
part in Python:
from subprocess import PIPE,Popen,call
p = subprocess.Popen([ path/to/RScript.exe, path/to/Script.R, Arg1], stdout=subprocess.PIPE, stderr=subprocess.PIPE, stdin=subprocess.PIPE)
out = p.communicate()
outValue = out[0]
outValue contains the output-Value after executing the Script.R
part in the R-Script:
args <- commandArgs(TRUE)
argument1 <- as.character(args[1])
...
write(output, stdout())
output is the variable to send to Python