I am writing a program where you need to input a title for a file. If you want to delete the file, the command ss "rm name_of_file".
Here Is My Code:
import os
title = raw_input("What Will Your Title Be? ")
os.system("rm", title)
As you can probably imagine, that is only a very small part of the program I am writing.
The Error I Am Getting Is:
File "./texts.py", line 1446, in <module>
os.system("rm", title)
TypeError: system() takes exactly 1 argument (2 given)
I am probably just wording this wrong, and some feedback would be helpful :)
The comma separates it into two arguments, so you are getting that error because that function only takes one argument. Change it so you are adding to the same string thus submitting just one argument to make it work:
os.system("rm "+title)
This is going to be exceedingly hard to implement safely. Consider the case when someone types -rf / as the filename to delete. I'd strongly suggest using the os.unlink function instead:
>>> import os
>>> os.unlink('-rf /')
...
OSError: [Errno 2] No such file or directory: '-rf /'
This is a dangerous operation to attempt on your own. Let the standard library do the heavy lifting for you.
Why don't you just do
if os.path.isfile(title):
os.remove(title)
Related
I have seen multiple posts on passing the string but not able to find good solution on reading the string passed to python script from batch file. Here is my problem.
I am calling python script from batch file and passing the argument.
string_var = "123_Asdf"
bat 'testscript.py %string_var%'
I have following in my python code.
import sys
passed_var = sys.argv[1]
When I run the above code I always see below error.
passed_var = sys.argv[1]
IndexError: list index out of range
Has anyone seen this issue before? I am only passing string and expect it to be read as part of the first argument I am passing to the script.
Try this:
import sys
for x,parameter in enumerate(sys.argv):
print(x, parameter)
If I have read your question and its formatting correctly, I think your .bat file should read:
Set string_var="123_Asdf"
"D:\BuildTools\tools\python27\python.exe" testscript.py %string_var%
Or better still:
Set "string_var=123_Asdf"
"D:\BuildTools\tools\python27\python.exe" testscript.py "%string_var%"
Where %string_var% can be passed with or without its enclosing doublequotes.
Your batch file should be a bit simpler, make sure you have your PATH set correctly or else this won't work.
python testscript.py [argument]
I hope that I can ask this in a clear way, im very much a beginner to python and forums in general so I apologise if i've got anything wrong from the start!
My issue is that I am currently trying to use os.system() to enable a program to run on every file within a directory (this is a directory of ASCII tables which I am crossing with a series of other tables to find matches.
import os
for filename in os.listdir('.'):
os.system('stilts tmatch2 ifmt1=ascii ifmt2=ascii in1=intern in2= %s matcher=2d values1='col1 col2' values2='col1 col2' params=5 out= %s-table.fits'%(filename,filename))
So what im hoping this would do is for every 'filename' it would operate this program known as stilts. Im guessing this gets interrupted/doesn't work because of the presence of apostrophes ' in the line of code itself, which must disrupt the syntax? (please correct me if I am wrong)
I then replaced the ' in os.system() with "" instead. This, however, stops me using the %s notation to refer to filenames throughout the code (at least I am pretty sure anyway).
import os
for filename in os.listdir('.'):
os.system("stilts tmatch2 ifmt1=ascii ifmt2=ascii in1=intern in2= %s matcher=2d values1='col1 col2' values2='col1 col2' params=5 out= %s-table.fits"%(filename,filename))
This now runs but obviously doesn't work, as it inteferes with the %s input.
Any ideas how I can go about fixing this? are there any alternative ways to refer to all of the other files given by 'filename' without using %s?
Thanks in advance and again, sorry for my inexperience with both coding and using this forum!
I am not familiar with os.system() but maybe if you try do some changes about the string you are sending to that method before it could behave differently.
You must know that in python you can "sum" strings so you can save your commands in a variable and add the filenames as in:
os.system(commands+filename+othercommands+filename)
other problem that could be working is that when using:
for file in os.listdir()
you may be recievin file types instead of the strings of their names. Try using a method such as filename.name to check if this is a different type of thing.
Sorry I cant test my answers for you but the computer I am using is too slow for me to try downloading python.
I'm trying to run an external program from a Python script.
After searching and reading multiple post here I came to what seemed to be the solution.
First, I used subprocess.call function.
If I build the command this way:
hmmer1=subprocess.call("D:\Python_Scripts\HMMer3\hmmsearch.exe --tblout hmmTestTab.out SDHA.hmm Test.fasta")
The external program D:\Python_Scripts\HMMer3\hmmsearch.exe is run taking hmmTestTab.out as file name for the output and SDHA.hmm and Test.fasta as input files.
Nevertheless, if I try to replace the file names with the variables outfile, hmmprofile and fastafile (I intend to receive those variables as arguments for the Python script and use them to build the external program call),
hmmer2=subprocess.call("D:\Python_Scripts\HMMer3\hmmsearch.exe --tblout outfile hmmprofile fastafile")
Python prints an error about being unable to open the input files.
I also used "Popen" function with analogous results:
This call works
hmmer3=Popen(['D:\Python_Scripts\HMMer3\hmmsearch.exe', '--tblout','hmmTestTab.out', 'SDHA.hmm','Test.fasta'])
and this one doesn't
hmmer4=Popen(['D:\Python_Scripts\HMMer3\hmmsearch.exe', '--tblout','outfile', 'hmmprofile','fastafile'])
As result of this, I presume I need to understand which is process to follow to interpolate the variables into the call, because it seems that the problem is there.
Would any of you help me with this issue?
Thanks in advance
You have:
hmmer4=Popen(['D:\Python_Scripts\HMMer3\hmmsearch.exe', '--tblout','outfile', 'hmmprofile','fastafile'])
But that's not passing the variable outfile. It's passing a string, 'outfile'.
You want:
hmmer4=Popen(['D:\Python_Scripts\HMMer3\hmmsearch.exe', '--tblout', outfile, hmmprofile, fastafile])
And the other answer is correct, though it addresses a different problem; you should double the backslashes, or use r'' raw strings.
Try to change this:
hmmer1=subprocess.call("D:\Python_Scripts\HMMer3\hmmsearch.exe"
to
hmmer1=subprocess.call('D:\\Python_Scripts\\HMMer3\\hmmsearch.exe'
Edit
argv = ' --tblout outfile hmmprofile fastafile' # your arguments
program = [r'"D:\\Python_Scripts\\HMMer3\\hmmsearch.exe"', argv]
subprocess.call(program)
I'm trying to unzip a file from an FTP site. I've tried it using 7z in a subprocess as well as using 7z in the older os.system format. I get closest however when I'm using the zipfile module in python so I've decided to stick with that. No matter how I edit this I seem to get one of two errors so here are both of them so y'all can see where I'm banging my head against the wall:
z = zipfile.ZipFile(r"\\svr-dc\ftp site\%s\daily\data1.zip" % item)
z.extractall()
NotImplementedError: compression type 6 (implode)
(I think this one is totally wrong, but figured I'd include.)
I seem to get the closest with the following:
z = zipfile.ZipFile(r"\\svr-dc\ftp site\%s\daily\data1.zip" % item)
z.extractall(r"\\svr-dc\ftp site\%s\daily\data1.zip" % item)
IOError: [Errno 2] No such file or directory: '\\\\svr-dc...'
The catch with this is that it is actually giving me the first file name in the zip. I can see the file AJ07242013.PRN at the end of the error so I feel closer because it's at least getting to the point of reading the contents of the zip file.
Pretty much any iteration of this that I try gets me one of those two errors, or a syntax error but that's easily addressed and not my primary concern.
Sorry for being so long winded. I'd love to get this working, so let me know what you think I need to do.
EDIT:
So 7z has finally been added to the path and is running through without any errors with both the subprocess as well as os.system. However, I still can't seem to get anything to unpack. It looks to me, from all I've read in the python documentation that I should be using the subprocess.communicate() module to extract this file but it just won't unpack. When I use os.system it keeps telling me that it cannot find the archive.
import subprocess
cmd = ['7z', 'e']
sp = subprocess.Popen(cmd, stderr=subprocess.STDOUT, stdout=subprocess.PIPE)
sp.communicate('r"\C:\Users\boster\Desktop\Data1.zip"')
I don't think that sp.communicate is right but if I add anything else to it I have too many arguments.
python's zipfile doesn't support compression type 6 (imploded) so its simply not going to work. In the first case, that's obvious from the error. In the second case, things are worse. The parameter for extractfile is an alternate unzip directory. Since you gave it the name of your zip file, a directory of the same name can't be found and zipfile gives up before getting to the not-supported problem.
Make sure you can do this with 7z on the command line, try implementing subprocess again and ask for help on that technique if you need it.
Here's a script that will look for 7z in the usual places:
import os
import sys
import subprocess
from glob import glob
print 'python version:', sys.version
subprocess.call('ver', shell=True)
print
if os.path.exists(r'C:\Program Files\7-Zip'):
print 'have standard 7z install'
if '7-zip' in os.environ['PATH'].lower():
print '...and its in the path'
else:
print '...but its not in the path'
print
print 'find in path...'
found = 0
for p in os.environ['PATH'].split(os.path.pathsep):
candidate = os.path.join(p, '7z.*')
for fn in glob(candidate):
print ' found', fn
found += 1
print
if found:
print '7z located, attempt run'
subprocess.call(['7z'])
else:
print '7z not found'
Accoring to the ZipFile documentation, you might be better off copying the zip first to your working directory. (http://docs.python.org/2/library/zipfile#zipfile.ZipFile.extract)
If you have problems copying, you might want to store the zip in a path with no spaces or protect your code against spaces by using os.path.
I made a small test in which I used os.path.abspath to make sure I had the proper path to my zip and it worked properly.
Also make sure that for extractall the path that you specify is the path where the zip content will be extracted. (If a folder that is specified is not created, it will be created automatically) Your files will be extracted in your current working directory (CWD) if no parameter is passed to extractall.
Cheers!
Managed to get this to work without using the PIPE functionality as subprocess.communicate wouldn't unpack the files. Here was the solution using subprocess.call. Hope this can help someone in the future.
def extract_data_one():
for item in sites:
os.chdir(r"\\svr-dc\ftp site\%s\Daily" % item)
subprocess.call(['7z', 'e', 'data1.zip', '*.*'])
I'm using Python 2.7 on Windows XP.
My script relies on tempfile.mkstemp and tempfile.mkdtemp to create a lot of files and directories with the following pattern:
_,_tmp = mkstemp(prefix=section,dir=indir,text=True)
<do something with file>
os.close(_)
Running the script always incurs the following error (although the exact line number changes, etc.). The actual file that the script is attempting to open varies.
OSError: [Errno 24] Too many open files: 'path\\to\\most\\recent\\attempt\\to\\open\\file'
Any thoughts on how I might debug this? Also, let me know if you would like additional information. Thanks!
EDIT:
Here's an example of use:
out = os.fdopen(_,'w')
out.write("Something")
out.close()
with open(_) as p:
p.read()
You probably don't have the same value stored in _ at the time you call os.close(_) as at the time you created the temp file. Try assigning to a named variable instead of _.
If would help you and us if you could provide a very small code snippet that demonstrates the error.
why not use tempfile.NamedTemporaryFile with delete=False? This allows you to work with python file objects which is one bonus. Also, it can be used as a context manager (which should take care of all the details making sure the file is properly closed):
with tempfile.NamedTemporaryFile('w',prefix=section,dir=indir,delete=False) as f:
pass #Do something with the file here.