check what files are open in Python

Question:

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?

Asked By: Claudiu

||

Answers:

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.

Answered By: Adam Crossland

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

Answered By: interjay

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

Answered By: eduffy

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)
Answered By: Claudiu

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
Answered By: Mike DeSimone

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:

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

    pid = os.getpid()
    procs = subprocess.check_output( 
        [ "lsof", '-w', '-Ff', "-p", str( pid ) ] )

    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.

Answered By: shaunc

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.

Answered By: stacksia

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.*sss(.*)")  # 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)
Answered By: Matt-the-Bat

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
Answered By: Romuald Brunet

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.

Answered By: jkhines

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()
Answered By: chriscz
Categories: questions Tags: , , ,
Answers are sorted by their score. The answer accepted by the question owner as the best is marked with
at the top-right corner.