I am having an issue with the following section of Python code:
# Open/Create the output file
with open(sys.argv[1] + '/Concatenated.csv', 'w+') as outfile:
try:
with open(sys.argv[1] + '/MatrixHeader.csv') as headerfile:
for line in headerfile:
outfile.write(line + '\n')
except:
print 'No Header File'
Specifically the error is as follows:
Traceback (most recent call last): File "ConcatenateFiles.py", line 12, in <module> with open(sys.argv[1] + 'Concatenated.csv', 'w+') as outfile:
IndexError: list index out of range
I've done some research and it seems that the sys.argv
might require an argument at the command line when running the script, but I'm not sure what to add or what the issue might be! I've also searched the site, but all of the solutions I've found either have no comments and/or don't include the open function as mine does.
Any help is greatly appreciated.
sys.argv
represents the command line options you execute a script with.
sys.argv[0]
is the name of the script you are running. All additional options are contained in sys.argv[1:]
.
You are attempting to open a file that uses sys.argv[1]
(the first argument) as what looks to be the directory.
Try running something like this:
python ConcatenateFiles.py /tmp
sys.argv
is the list of command line arguments passed to a Python script, where sys.argv[0]
is the script name itself.
It is erroring out because you are not passing any commandline argument, and thus sys.argv
has length 1 and so sys.argv[1]
is out of bounds.
To "fix", just make sure to pass a commandline argument when you run the script, e.g.
python ConcatenateFiles.py /the/path/to/the/directory
However, you likely wanted to use some default directory so it will still work when you don't pass in a directory:
cur_dir = sys.argv[1] if len(sys.argv) > 1 else '.'
with open(cur_dir + '/Concatenated.csv', 'w+') as outfile:
try:
with open(cur_dir + '/MatrixHeader.csv') as headerfile:
for line in headerfile:
outfile.write(line + '\n')
except:
print 'No Header File'
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