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Finding which files are being read from during a session (python code)

I have a large system written in python. when I run it, it reads all sorts of data from many different files on my filesystem. There are thousands lines of code, and hundreds of files, most of them are not actually being used. I want to see which files are actually being accessed by the system (ubuntu), and hopefully, where in the code they are being opened. Filenames are decided dynamically using variables etc., so the actual filenames cannot be determined just by looking at the code.
I have access to the code, of course, and can change it.
I try to figure how to do this efficiently, with minimal changes in the code:
is there a Linux way to determine which files are accessed, and at what times? this might be useful, although it won't tell me where in the code this happens
is there a simple way to make an "open file" command also log the file name, time, etc... of the open file? hopefully without having to go into the code and change every open command, there are many of them, and some are not being used at runtime.
Thanks

You can trace file accesses without modifying your code, using strace.
Either you start your program with strace, like this
strace -f -e trace=file your_program.py
Otherwise you attach strace to a running program like this
strace -f -e trace=file -p <PID>

For 1 - You can use
ls -la /proc/<PID>/fd`
Replacing <PID> with your process id.
Note that it will give you all the open file descriptors, some of them are stdin stdout stderr, and often other things, such as open websockets (which use a file descriptor), however filtering it for files should be easy.
For 2- See the great solution proposed here -
Override python open function when used with the 'as' keyword to print anything
e.g. overriding the open function with your own, which could include the additional logging.

One possible method is to "overload" the open function. This will have many effects that depend on the code, so I would do that very carefully if needed, but basically here's an example:
>>> _open = open
>>> def open(filename):
... print(filename)
... return _open(filename)
...
>>> open('somefile.txt')
somefile.txt
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "<stdin>", line 3, in open
FileNotFoundError: [Errno 2] No such file or directory: 'somefile.txt'
As you can see my new open function will return the original open (renamed as _open) but will first print out the argument (the filename). This can be done with more sophistication to log the filename if needed, but the most important thing is that this needs to run before any use of open in your code

how to solve python ilock FileNotFoundError?

so i am trying to use ilock in python as a system wide lock, but after few iterations in my code i get the following error, what might cause such an error? and how i can start solving it
FileNotFoundError: [Errno 2] No such file or directory: '/tmp/ilock-bfe0d208735d8d5f20bb2b8abcf8bf67d696f23629b4ee2d4e7304f69063db61.lock'

I'm seeing the same error. Looks to me like it's a bug in the ilock library.
In short, when two ILock objects are created with the same unique name, they'll use the same file as the locking entity (passed to portalocker). They create the file (if it doesn't exist) using open(path, 'w') upon ILock.__enter__, and they call os.unlink(path) upon ILock.__exit__.
However consider the following scenario:
process1: ILock.__enter__ # file is created, lock acquired
process2: ILock.__enter__ # file already exists, lock pending
process1: does its thing under the lock
process1: ILock.__exit__ # file is unlinked, lock released
process2: does its thing under the lock
process2: ILock.__exit__ # Error: cannot unlink, file does not exist
On the surface, it could be that this can be fixed by silently allowing unlink to fail; or perhaps, by recreating the file as necessary after the lock has been acquired. I am not sure though if portalocker would behave nicely in this case.
Perhaps the easiest workaround is to simply NEVER delete the file (get rid of os.unlink altogether).

Lock file for access on windows

Using portalocker we can lock a file for access through the following way:
f=open("M99","r+")
portalocker.lock(f,portalocker.LOCK_EX)
The lock over the file can be removed using
f.close() #or
portalocker.unlock(file) #needs `file` ie reference to file it locked ..pretty obvious too
Can this same thing be done by any other way in python wherein
We can lock the file for access
Restart Python (so no longer have the original Python file object or file number).
Unlock the file for access in the new process.
I cannot save f or file object so can't use pickle or something either. Is there a way using the Python standard library or some win32api call?
Any windows utility will also do...any command line from windows?

It appears you want to lock access to resources where the lock persists between program invocations. You need a different strategy for that.
Create a lock file using exclusive create mode; in Python 2 this requires using the os.open() call (followed by os.fdopen() to produce a Python file object), in Python 3 you can use the 'x' mode when using the built-in open().
In Python 2:
import os
LOCKFILE = r'some\path\to\lockfile'
class AlreadyLocked(Exception):
pass
def lock():
try:
fd = os.open(LOCKFILE, os.O_WRONLY | os.O_CREAT | os.O_EXCL)
except IOError:
# file already exists
raise AlreadyLocked()
with os.fdopen(fd, 'w') as lockfile:
# write the PID of the current process so you can debug
# later if a lockfile can be deleted after a program crash
lockfile.write(os.getpid())
def unlock():
os.remove(LOCKFILE)
In Python 3 the lock() function would be:
def lock():
try:
with open(LOCKFILE, 'x') as lockfile:
# write the PID of the current process so you can debug
# later if a lockfile can be deleted after a program crash
lockfile.write(os.getpid())
except IOError:
# file already exists
raise AlreadyLocked()
You need to use exclusive create mode to avoid race conditions; in exclusive create mode the file can only be created if it doesn't yet exist, a condition checked by the Operating System, rather than by a separate step in Python which would open a window for another program to create the lock as well.
Now you can lock and unlock without tracking the file descriptor. The lockfile is now a signal file; if it is present something has claimed a lock, and deleting the file means something is unlocked.
This does mean that access to the files or directories you are trying to protect is only protected because all your code honours this lock system, not because the OS is enforcing locks on those files or directories.
This all means that this only works if all access to the shared resource is handled by processes that cooperate in this strategy. It cannot be used if another process doesn't honour this scheme. In that case your only option is to use OS level locking and you have to keep your process running for the full duration of the lock.

there is a method in win32api to set file attributes if you have a read of the following:
python SetFileAttributes
MSDN file attributes
these give you the python method to set file attributes:
win32api.SetFileAttributes(file, win32con.FILE_ATTRIBUTE_NORMAL)
where file is the name/path of the file, and not a file object
and the second argument is a attribute mask, is you wanted to set several attributes at once, you can use bitwise xor to add them:
win32con.FILE_ATTRIBUTE_HIDDEN | win32con.FILE_ATTRIBUTE_READONLY
and there are more constants named in the MSDN page.
EDIT:
for file locking you can also look at the win32file.LockFileEx method
i haven't used this before so it may take some playing around, but it appears to need you to pass it a file object (not a path) and then certain constants to set the access permissions, more info on the constants can be found on MSDN

You could use subprocess to open the file in notepad or excel:
import subprocess, time
subprocess.call('start excel.exe "\lockThisFile.txt\"', shell = True)
time.sleep(10) # if you need the file locked before executing the next commands, you may need to sleep it for a few seconds
or
subprocess.call('notepad > lockThisFile.txt', shell = True)
As written you need shell = True, otherwise windows will give you a syntax error.
(subprocess.Popen() works as well)
You can then close the process later using:
subprocess.call('taskkill /f /im notepad.exe') # or excel.exe
Other options include
-write some C++ code and call it from python (https://msdn.microsoft.com/en-us/library/windows/desktop/aa365203(v=vs.85).aspx)
-call 3rd party programs with subprocess.call():
FileLocker http://www.jensscheffler.de/filelocker (https://superuser.com/questions/294826/how-to-purposefully-exclusively-lock-a-file)
Easy File Locker http://www.xoslab.com/efl.html and Dispatch (from win32com.client import Dispatch), although last choice is the most complex

How to open a game from python

I want to open a file and then run a game from in it. I tried:
import os
os.startfile(r'C:/Users/Andy/Desktop/Jnes')
os.startfile("Jnes.exe")
But that only brought up the first file and didn't run the game.

Your first line is superfluous. The parent Python process still has it's default working directory. That line will not change it, so the second line will fail as it will not find the file named Jnes.exe.
So either use:
os.startfile(r'C:/Users/Andy/Desktop/Jnes/Jnes.exe')
This will keep the working directory (for the Python process) unchanged, or, if you want it changed, use:
os.chdir(r'C:/Users/Andy/Desktop/Jnes')
os.startfile("Jnes.exe")

Can't access temporary files created with tempfile

I am using tempfile.NamedTemporaryFile() to store some text until the program ends. On Unix is working without any issues but on Windows the file returned isn't accessible for reading or writing: python gives Errno 13. The only way is to set delete=False and manually delete the file with os.remove(). Why?

This causes the IOError because the file can be opened only once after it is created.
The reason is because NamedTemporaryFile creates the file with FILE_SHARE_DELETE flag on Windows. On Windows when a file has been created/opened with specific share flag all subsequent open operations have to pass this share flag. It's not the case with Python's open function which does not pass FILE_SHARE_DELETE flag. See my answer on How to create a temporary file that can be read by a subprocess? question for more details and a workaround.

Take a look: http://docs.python.org/2/library/tempfile.html
tempfile.NamedTemporaryFile([mode='w+b'[, bufsize=-1[, suffix=''[, prefix='tmp'[, dir=None[, delete=True]]]]]])
This function operates exactly as TemporaryFile() does, except that the file is guaranteed to have a visible name in the file system (on Unix, the directory entry is not unlinked). That name can be retrieved from the name attribute of the file object. Whether the name can be used to open the file a second time, while the named temporary file is still open, varies across platforms (it can be so used on Unix; it cannot on Windows NT or later). If delete is true (the default), the file is deleted as soon as it is closed.

Thanks to #Rnhmjoj here is a working solution:
file = NamedTemporaryFile(delete=False)
file.close()
You have to keep the file with the delete-flag and then close it after creation. This way, Windows will unlock the file and you can do stuff with it!