I am writing a Pokemon game and I want to make folders to save different types of Pokemon, as well as other sorts of information, in. I want to use folders because it would be really messy if I were to save all my data into a single file.
Is it possible to create folders with a Python program? This would make it easier and cleaner for me when I try to import the Pokemon data from external websites.
You can use open with the a mode, which opens a file in append mode, and creates it if it does not exist:
my_file = open('file.txt', 'a')
# Optionally: write stuff to my_file, using my_file.write('stuff')
my_file.close()
You can run any command that you want with python by doing:
import os
os.popen("mkdir random_name") # This creates a directory called "random_name"
os.popen("touch rando_name.txt") # This creates a file called "random_name"
You can run any command that you would usually run in terminal inside the popen.
You can use the popen() command in both UNIX(Linux, macOS) as well as Windows OS.
You can learn more regarding this by taking a look in the python documentation.
https://docs.python.org/2/library/subprocess.html
If you'd like to create folders (or directories), you need os.mkdir:
import os
os.mkdir("folder_name")
To create some deep folders at once, use os.makedirs:
os.makedirs("path/to/something")
Then all that structure of three folders will be created.
Tutorialspoint has a short tutorial about os.mkdir.
You can use with statement like this:
with open('some_file_100.txt', 'a') as f:
pass
The above will just create an empty file, if you want to write something to the created file, you can try:
with open('some_file_100.txt', 'a') as f:
f.write('some text')
When you use with statement you do not need to close your files explicitly because they are closed autmatically at the end of the block.
Related
First and foremost, I am recently new to Unix and I have tried to find a solution to my question online, but I could not find a solution.
So I am running Python through my Unix terminal, and I have a program that parses xml files and inputs the results into a .dat file.
My program works, but I have to input every single xml file (which number over 50) individually.
For example:
clamshell: python3 my_parser2.py 'items-0.xml' 'items-1.xml' 'items-2.xml' 'items-3.xml' .....`
So I was wondering if it is possible to read from the directory, which contains all of my files into my program? Rather than typing all the xml file names individually and running the program that way.
Any help on this is greatly appreciated.
import glob
listOffiles = glob.glob('directory/*.xml')
The shell itself can expand wildcards so, if you don't care about the order of the input files, just use:
python3 my_parser2.py items-*.xml
If the numeric order is important (you want 0..9, 10-99 and so on in that order, you may have to adjust the wildcard arguments slightly to guarantee this, such as with:
python3 my_parser2.py items-[0-9].xml items-[1-9][0-9].xml items-[1-9][0-9][0-9].xml
python3 my_parser2.py *.xml should work.
Other than the command line option, you could just use glob from within your script and bypass the need for command arguments:
import glob
filenames = glob.glob("*.xml")
This will return all .xml files (as filenames) in the directory from which you are running the script.
Then, if needed you can simply iterate through all the files with a basic loop:
for file in filenames:
with open(file, 'r') as f:
# do stuff to f.
With the code open, I can create a new file in the computer. How do i decide which folder it goes? I need to put them in a certain folder when i am creating them. Should I put sth in the brackets?
e.g. open("apple juice. txt", "a")
If you don't specify a path, then the file will be created in the current directory. Where exactly that is depends on how you started the interpreter. For example, when you start Python 3.4 from the Windows Start Menu, then the file will be saved in C:\Python34\.
If you want to specify a certain path, then do so:
f = open(r"C:\Users\David\Python Files\apple juice.txt", "a")
Give the full path:
with open("path_where/to_save/apple_juice.txt", "a") as f:
# do work
with will automatically close your file.
If you are looking for a place for temp files, use the module tempfile.
You can use the function tempfile.gettempdir() to get a path to a folder directory.
You can use tempfile.TemporaryFile() to generate a full path to the place where the temp files are usually stored on your OS.
The method used in either case to generate the temp path is explained here.
So I'm a newb :) Python question
I have a list of files and I'm looking to open/read these files using an I/O method
I understand if I explicitly go through each test file I've created and opening them one by one would be fine but how about if I have an unknown file and I tell it to be open/read, how would this be done?
Logically thinking, it sounds like I need to create a variable and assign it to a list of files and from there tell it open all the files in the list. So a for loop perhaps?
You can do it as follows:
import os
for fl in os.listdir(os.getcwd()):
with open(fl) as f:
#do stuff
Alternatively, if your files are not in the same directory as your script, you can do:
for fl in os.listdir('custom/path/to/files'):
I have the python code which will download the .war file and put it in a path which is specified by the variable path.
Now I wish to extract a specific file from that war to a specific folder.
But I got struck up here :
os.system(jar -xvf /*how to give the path varible here*/ js/pay.js)
I'm not sure how to pass on the variable path to os.system command.
I'm very new to python, kindly help me out.
If you really want to use os.system, the shell command line is passed as a string, and you can pass any string you want. So:
os.system('jar -xvf "' + pathvariable + '" js/pay.js)
Or you can use {} or %s formatting, etc.
However, you probably do not want to use os.system.
First, if you want to run other programs, it's almost always better to use the subprocess module. For example:
subprocess.check_call(['jar', '-xvf', pathvariable, 'js/pay.js'])
As you can see, you can pass a list of arguments instead of trying to work out how to put a string together (and deal with escaping and quoting and all that mess). And there are lots of other advantages, mostly described in the documentation itself.
However, you probably don't want to run the war tool at all. As jimhark says, a WAR file is just a special kind of JAR file, which is just a special kind of ZIP file. For creating them, you generally want to use JAR/WAR-specific tools (you need to verify the layout, make sure the manifest is the first entry in the ZIP directory, take care of the package signature, etc.), but for expanding them, any ZIP tool will work. And Python has ZIP support built in. What you want to do is probably as simple as this:
import zipfile
with zipfile.ZipFile(pathvariable, 'r') as zf:
zf.extract('js/pay.js', destinationpathvariable)
IIRC, you can only directly use ZipFile in a with statement in 2.7 and 3.2+, so if you're on, say, 2.6 or 3.1, you have to do it indirectly:
from contextlib import closing
import zipfile
with closing(zipfile.ZipFile(pathvariable, 'r')) as zf:
zf.extract('js/pay.js', destinationpathvariable)
Or, if this is just a quick&dirty script that quits as soon as it's done, you can get away with:
import zipfile
zf = zipfile.ZipFile(pathvariable, 'r')
zf.extract('js/pay.js', destinationpathvariable)
But I try to always use with statements whenever possible, because it's a good habit to have.
Isn't a war file a type of zip file? Python has zipfile support (click link for docs page).
You can use os.environ, it holds all environment variables on it.
It is a dict, so you can just use it like:
pypath = os.environ['PYTHONPATH']
now if you mean it's a common python variable, just use it like:
var1 = 'pause'
os.system('#echo & %s' % var1)
I'm not sure even where to start.
I have a list of output files from a program, lets call them foo. They are numbered outputs like foo_1.out
I'd like to make a directory for each file, move the file to its directory, run a bash script within that directory, take the output from each script, copy it to the root directory as a concatenated single file.
I understand that this is not a forum for "hey, do my work for me", I'm honestly trying to learn. Any suggestions on where to look are sincerely appreciated!
Thanks!
You should probably look up the documentation for the python modules os - specifically os.path and a couple of others - and subprocess which can be found here and here respectively.
Without wanting to do it all for you as you stated - you'll be wanting to do something like:
for f in filelist:
[pth, ext] = os.path.splitext(f)
os.mkdir(pth)
out = subprocess.Popen(SCRIPTNAME, stdout=...)
# and so on...
To get a list of all files in a directory or make folders, check out the os module. Specifically, try os.listdir and os.mkdir
To copy files, you could either manually open each file, copy the contents to a string, and rewrite it to a different file. Alternatively, look at the shutil module
To run bash scripts, use the subprocess library.
All three of those should be a part of python's standard library.