I'm getting an error in a program that is supposed to run for a long time that too many files are open. Is there any way I can keep track of which files are open so I can print that list out occasionally and see where the problem is?
11 Answers
To list all open files in a cross-platform manner, I would recommend psutil.
#!/usr/bin/env python
import psutil
for proc in psutil.process_iter():
print(proc.open_files())
The original question implicitly restricts the operation to the currently running process, which can be accessed through psutil's Process class.
proc = psutil.Process()
print(proc.open_files())
Lastly, you'll want to run the code using an account with the appropriate permissions to access this information or you may see AccessDenied errors.

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3This only seems to work for disk-based files, not sockets, fifos, etc. – NeilenMarais Aug 29 '14 at 13:14
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@NeilenMarais: it does work for sockets, see examples at https://pypi.python.org/pypi/psutil – LetMeSOThat4U Jan 13 '15 at 10:53
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Doesn't work at all for me. Simple test: `from psutil import Process; proc = Process(); with open('fileincurrentdirectory') as f: print(proc.open_files())` - you'll need to add a line break and indent that final print statement. It'll print out an empty list. And I did use the name of an actual file actually in the current directory. – ArtOfWarfare Jan 18 '18 at 14:57
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I ended up wrapping the built-in file object at the entry point of my program. I found out that I wasn't closing my loggers.
import io
import sys
import builtins
import traceback
from functools import wraps
def opener(old_open):
@wraps(old_open)
def tracking_open(*args, **kw):
file = old_open(*args, **kw)
old_close = file.close
@wraps(old_close)
def close():
old_close()
open_files.remove(file)
file.close = close
file.stack = traceback.extract_stack()
open_files.add(file)
return file
return tracking_open
def print_open_files():
print(f'### {len(open_files)} OPEN FILES: [{", ".join(f.name for f in open_files)}]', file=sys.stderr)
for file in open_files:
print(f'Open file {file.name}:\n{"".join(traceback.format_list(file.stack))}', file=sys.stderr)
open_files = set()
io.open = opener(io.open)
builtins.open = opener(builtins.open)

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@Claudiu - can you please how using close all open files - def closeall(self): print "### CLOSING All files ###" oldfile.close(self,(f.x for f in openfiles)) openfiles.remove(self,(f.x for f in openfiles)) Will work perfectly? – Programmer Dec 13 '14 at 11:52
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3In order not to extend lifetimes of the file objects (and hence prevent auto-closing on reference counted objects in cpython) it's IMO worth using a `weakref.WeakSet` instead of plain `set` for `openfiles`. – coldfix Apr 23 '15 at 03:17
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Thanks, worked brilliantly for me too. Note that I had exceptions when patching a program also using sklearn joblib: the same file could be closed twice by joblib.load, causing an exception in `openfiles.remove(self)`. – massyah Aug 22 '16 at 12:38
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1When I use this snippet I get the next error: AttributeError: 'file' object attribute 'close' is read-only. Any idea how can I fix it? $python --version Python 2.7.10 – fsquirrel Jun 08 '18 at 14:22
On Linux, you can look at the contents of /proc/self/fd
:
$ ls -l /proc/self/fd/
total 0
lrwx------ 1 foo users 64 Jan 7 15:15 0 -> /dev/pts/3
lrwx------ 1 foo users 64 Jan 7 15:15 1 -> /dev/pts/3
lrwx------ 1 foo users 64 Jan 7 15:15 2 -> /dev/pts/3
lr-x------ 1 foo users 64 Jan 7 15:15 3 -> /proc/9527/fd

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Is this just for CPython or all implementations? I remember seeing, I think, that files open in ipython are listed in `/proc/ipython_pid/fd/`. Also, in the list above, how do you know what are files you opened and which are files that Python opened (and which you shouldn't close)? – Chris Jan 26 '12 at 11:13
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3This is for Linux systems which provide the `/proc` filesystem. It's independent of language; any program in any language that can access the "files" in `/proc` can get this information. I haven't messed with ipython, but the basic idea would be to record the contents of `/proc/self/fd` after initialization and then compare the contents later in the run to look for changes. – Mike DeSimone Jan 26 '12 at 14:18
Although the solutions above that wrap opens are useful for one's own code, I was debugging my client to a third party library including some c extension code, so I needed a more direct way. The following routine works under darwin, and (I hope) other unix-like environments:
Python 3:
import os
import shutil
import subprocess
def get_open_fds() -> int:
"""Get the number of open file descriptors for the current process."""
lsof_path = shutil.which("lsof")
if lsof_path is None:
raise NotImplementedError("Didn't handle unavailable lsof.")
raw_procs = subprocess.check_output(
[lsof_path, "-w", "-Ff", "-p", str(os.getpid())]
)
def filter_fds(lsof_entry: str) -> bool:
return lsof_entry.startswith("f") and lsof_entry[1:].isdigit()
fds = list(filter(filter_fds, raw_procs.decode().split(os.linesep)))
return len(fds)
Python 2:
def get_open_fds():
'''
return the number of open file descriptors for current process
.. warning: will only work on UNIX-like os-es.
'''
import subprocess
import os
procs = subprocess.check_output(
['lsof', '-w', '-Ff', '-p', str(os.getpid())]
)
nprocs = len(
filter(
lambda s: s and s[0] == 'f' and s[1:].isdigit(),
procs.split('\n'))
)
return nprocs
If anyone can extend to be portable to windows, I'd be grateful.

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On Linux, you can use lsof
to show all files opened by a process.

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2Has python some internal function for lsof, or I really have to call linux lsof? – sumid Sep 06 '11 at 21:02
As said earlier, you can list fds on Linux in /proc/self/fd, here is a simple method to list them programmatically:
import os
import sys
import errno
def list_fds():
"""List process currently open FDs and their target """
if not sys.platform.startswith('linux'):
raise NotImplementedError('Unsupported platform: %s' % sys.platform)
ret = {}
base = '/proc/self/fd'
for num in os.listdir(base):
path = None
try:
path = os.readlink(os.path.join(base, num))
except OSError as err:
# Last FD is always the "listdir" one (which may be closed)
if err.errno != errno.ENOENT:
raise
ret[int(num)] = path
return ret

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On Windows, you can use Process Explorer to show all file handles owned by a process.

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There are some limitations to the accepted response, in that it does not seem to count pipes. I had a python script that opened many sub-processes, and was failing to properly close standard input, output, and error pipes, which were used for communication. If I use the accepted response, it will fail to count these open pipes as open files, but (at least in Linux) they are open files and count toward the open file limit. The lsof -p
solution suggested by sumid and shunc works in this situation, because it also shows you the open pipes.

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Get a list of all open files. handle.exe
is part of Microsoft's Sysinternals Suite. An alternative is the psutil Python module, but I find 'handle' will print out more files in use.
Here is what I made. Kludgy code warning.
#!/bin/python3
# coding: utf-8
"""Build set of files that are in-use by processes.
Requires 'handle.exe' from Microsoft SysInternals Suite.
This seems to give a more complete list than using the psutil module.
"""
from collections import OrderedDict
import os
import re
import subprocess
# Path to handle executable
handle = "E:/Installers and ZIPs/Utility/Sysinternalssuite/handle.exe"
# Get output string from 'handle'
handle_str = subprocess.check_output([handle]).decode(encoding='ASCII')
""" Build list of lists.
1. Split string output, using '-' * 78 as section breaks.
2. Ignore first section, because it is executable version info.
3. Turn list of strings into a list of lists, ignoring first item (it's empty).
"""
work_list = [x.splitlines()[1:] for x in handle_str.split(sep='-' * 78)[1:]]
""" Build OrderedDict of pid information.
pid_dict['pid_num'] = ['pid_name','open_file_1','open_file_2', ...]
"""
pid_dict = OrderedDict()
re1 = re.compile("(.*?\.exe) pid: ([0-9]+)") # pid name, pid number
re2 = re.compile(".*File.*\s\s\s(.*)") # File name
for x_list in work_list:
key = ''
file_values = []
m1 = re1.match(x_list[0])
if m1:
key = m1.group(2)
# file_values.append(m1.group(1)) # pid name first item in list
for y_strings in x_list:
m2 = re2.match(y_strings)
if m2:
file_values.append(m2.group(1))
pid_dict[key] = file_values
# Make a set of all the open files
values = []
for v in pid_dict.values():
values.extend(v)
files_open = sorted(set(values))
txt_file = os.path.join(os.getenv('TEMP'), 'lsof_handle_files')
with open(txt_file, 'w') as fd:
for a in sorted(files_open):
fd.write(a + '\n')
subprocess.call(['notepad', txt_file])
os.remove(txt_file)

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I'd guess that you are leaking file descriptors. You probably want to look through your code to make sure that you are closing all of the files that you open.

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I figured that's what the problem was. However the code is very complex, and this would be an easy way to immediately spot which files aren't being closed. – Claudiu Jan 07 '10 at 21:06
You can use the following script. It builds on Claudiu's answer. It addresses some of the issues and adds additional features:
- Prints a stack trace of where the file was opened
- Prints on program exit
- Keyword argument support
Here's the code and a link to the gist, which is possibly more up to date.
"""
Collect stacktraces of where files are opened, and prints them out before the
program exits.
Example
========
monitor.py
----------
from filemonitor import FileMonitor
FileMonitor().patch()
f = open('/bin/ls')
# end of monitor.py
$ python monitor.py
----------------------------------------------------------------------------
path = /bin/ls
> File "monitor.py", line 3, in <module>
> f = open('/bin/ls')
----------------------------------------------------------------------------
Solution modified from:
https://stackoverflow.com/questions/2023608/check-what-files-are-open-in-python
"""
from __future__ import print_function
import __builtin__
import traceback
import atexit
import textwrap
class FileMonitor(object):
def __init__(self, print_only_open=True):
self.openfiles = []
self.oldfile = __builtin__.file
self.oldopen = __builtin__.open
self.do_print_only_open = print_only_open
self.in_use = False
class File(self.oldfile):
def __init__(this, *args, **kwargs):
path = args[0]
self.oldfile.__init__(this, *args, **kwargs)
if self.in_use:
return
self.in_use = True
self.openfiles.append((this, path, this._stack_trace()))
self.in_use = False
def close(this):
self.oldfile.close(this)
def _stack_trace(this):
try:
raise RuntimeError()
except RuntimeError as e:
stack = traceback.extract_stack()[:-2]
return traceback.format_list(stack)
self.File = File
def patch(self):
__builtin__.file = self.File
__builtin__.open = self.File
atexit.register(self.exit_handler)
def unpatch(self):
__builtin__.file = self.oldfile
__builtin__.open = self.oldopen
def exit_handler(self):
indent = ' > '
terminal_width = 80
for file, path, trace in self.openfiles:
if file.closed and self.do_print_only_open:
continue
print("-" * terminal_width)
print(" {} = {}".format('path', path))
lines = ''.join(trace).splitlines()
_updated_lines = []
for l in lines:
ul = textwrap.fill(l,
initial_indent=indent,
subsequent_indent=indent,
width=terminal_width)
_updated_lines.append(ul)
lines = _updated_lines
print('\n'.join(lines))
print("-" * terminal_width)
print()
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I recent python, there is no module `__builtin__`, it can be replaced by `builtins`. – EmmaRenauld Apr 05 '23 at 19:39