I wanna write out put of the process which will be run using subprocess into test file. Below is my code.
Proc = "./server.sh >>out.txt"
Subprocess.Popen(proc,stdout=subprocess.Pipe,shell= True)
If I run above code , only first three lines of out put is written, can any tell how to get my process entire output into text file?
You can pass a file handle directly to the Popen() constructor:
with open('/tmp/out.txt', 'w') as tempf:
subprocess.Popen('server.sh', stdout=tempf, shell=True)
Related
Python version: Python 2.7.13
I am trying to write a script that is able to go through a *.txt file and launch a batch file to execute a particular test.
The code below goes through the input file, changes the string from 'N' to 'Y' which allows the particular test to be executed. I am in the process of creating a for loop to go through all the lines within the *.txt file and execute all the test in a sequence. However, my problem is that I do not want to execute the test at the same time (which is what would happen if I just write the test code).
Is there a way to wait until the initial test is finished to launch the next one?
Here is what I have so far:
from subprocess import Popen
import os, glob
path = r'C:/Users/user1/Desktop/MAT'
for fname in os.listdir(path):
if fname.startswith("fort"):
os.remove(os.path.join(path, fname))
with open('RUN_STUDY_CHECKLIST.txt', 'r') as file:
data = file.readlines()
ln = 4
ch = list(data[ln])
ch[48] = 'Y'
data[ln] = "".join(ch)
with open('RUN_STUDY_CHECKLIST.txt', 'w') as file:
file.writelines(data)
matexe = Popen('run.bat', cwd=r"C:/Users/user1/Desktop/MAT")
stdout, stderr = matexe.communicate()
In this particular instance I am changing the 'N' in line 2 of the *.txt file to a 'Y' which will be used as an input for another python script.
I have to mention that I would like to do this task without having to interact with any prompt, I would like to do execute the script and leave it running (since it would take a long time to go through all the tests).
Best regards,
Jorge
After further looking through several websites I managed to get a solution to my question.
I used:
exe1 = subprocess.Popen(['python', 'script.py'])
exe1.wait()
I wanted to post the answer just in case this is helpful to anyone.
I am running the following code in Python 2.7:
values = os.system("bazel build tensorflow/examples/image_retraining:"
"label_image && bazel-bin/tensorflow/examples/image_retraining/label_image "
"--graph=/tmp/output_graph.pb --labels=/tmp/output_labels.txt "
"--output_layer=final_result:0 --image=$HOME/Desktop/Image-3/image1.png")
print values
But for the values variable I am returned a 0. I believe this means that I am not getting any errors. How do I store the output into a variable or into a .txt file?
You can just redirect the output of the system call appending > output.txt to your command.
The output of the command will be in file output.txt in the directory where you invoke the command (likely the very same one you invoke your python script in).
Since I can't readily reproduce your command, I used a simple example - try to switch to Pyopen in the subprocess module:
from subprocess import Popen
proc = Popen(['ls', '-t'], stdout = open('/path/redir.txt', 'w'))
Here you run the command in square brackets and redirect the output from stdout i.e. the terminal to a file redir.txt.
How can I open a file once I am inside Python, that is, once I have typed "python" in the terminal? I know how to open a file by typing something similar to the following in a script, and then running it:
from sys import argv
script, filename = argv
txt = open(filename)
print txt.read()
But I have no idea how to do it once I'm inside the Python interpreter. I've tried to type open (file.txt) and also open ("file.txt"), but I get a long error message either way.
Which is the correct way to do this?
You have to add a mode to the call txt = open('filename.txt', 'r') if you want to read (or w/a for writing or appending). I just tried it, it works :)
I would like to run LIWC (installed in my Mac) within a python 2.7 script.
I have been reading about subprocess (popen and check_output seem the way to go), but I do not get the syntax for:
opening the program;
getting a text file to be analysed;
running the program;
getting the output (analysis) and storing it in a text file.
This is my first approach to subprocess, is this possible?
I appreciate the suggestions.
EDIT
This is the closest to implementing a solution (still does not work):
I can open the application.
subprocess.call(['open', '/file.app'])
But cannot make it process the input file and get an output one.
subprocess.Popen(['/file.app', '-input', 'input.txt', '-output', 'output.txt'])
Nothing comes out of this code.
EDIT 2
After reading dozens of posts, I am still very confused about the syntax for the solution.
Following How do I pipe a subprocess call to a text file?
I came out with this code:
g = open('in_file.txt', 'rb', 0)
f = open('out_file.txt', 'wb')
subprocess.call(['open', "file.app"] stdin=g, stdout=f)
The output file comes out empty.
EDIT 3
Following http://www.cplusplus.com/forum/unices/40680/
When I run the following shell script on the Terminal:
cat input.txt | /Path/LIWC > output.txt
The output txt file is empty.
EDIT 4
When I run:
subprocess.check_call(['/PATH/LIWC', 'PATH/input.txt', 'PATH/output.txt'])
It opens LIWC, does not create an output file and freezes.
EDIT 5
When I run:
subprocess.call(['/PATH/LIWC', 'PATH/input.txt', 'PATH/output.txt'])
It runs LIWC, creates an empty output.txt file and freezes (the process does not end).
The problem with using 'open' in subprocess.call(['open', "file.app"] stdin=g, stdout=f) is that it requests that a file be opened through a service, and doesn't directly attach it to your python process. You'll need to instead use the path to LIWC. I'm not sure that it supports reading from stdin, though, so you might need to even pass in the path to the file you'd like it to open.
I have two .py files. The first file executes the second file, and also needs to be able to send an argument to the second file.
Here's the file1.py:
from subprocess import Popen, PIPE
import sys
file_names = ['one.csv' , 'two.csv']
for f in file_names:
process = Popen([sys.executable , "file2.py"] , stdout = PIPE , stdin = PIPE)
process.communicate(f)
And here's file2.py:
def c(x):
print x
c(f)
The first file successfully executes the second file, but doesn't pass the argument f to the second file. I've also tried using process.stdin.write(f) instead of process.communicate(f) but this doesn't work either, and I'd rather use communicate instead of stdin because multiple instances of file2.py need to be executed at the same time without blocking.
When you do:
process.communicate(f)
you are writing to the standard input of process; the problem is that the latter program must read the data from its input stream in order to make use of it. You could change file2.py to something like:
import sys
def c(x):
print x
for f in sys.stdin:
c(f)
You seem to be thinking that communicate() allows the parent and child processes to share variable bindings; that's not the case. As written in your question, file2.py would generate an error to the effect that name 'f' is not defined.