Thomas Guettler <[EMAIL PROTECTED]> wrote: > bvidinli schrieb: > > is there a way to find out if file open in system ? - > > please write if you know a way other than lsof. because lsof if slow for > > me. > > i need a faster way. > > i deal with thousands of files... so, i need a faster / python way for this. > > thanks. > > On Linux there are symlinks from /proc/PID/fd to the open > files. You could use this: > > #!/usr/bin/env python > import os > pids=os.listdir('/proc') > for pid in sorted(pids): > try: > int(pid) > except ValueError: > continue > fd_dir=os.path.join('/proc', pid, 'fd') > for file in os.listdir(fd_dir): > try: > link=os.readlink(os.path.join(fd_dir, file)) > except OSError: > continue > print pid, link
Unfortunately I think that is pretty much exactly what lsof does so is unlikely to be any faster! However if you have 1000s of files to check you only need to do the above scan once and build a dict with all open files in whereas lsof will do it once per call so that may be a useful speedup. Eg ------------------------------------------------------------ import os pids=os.listdir('/proc') open_files = {} for pid in sorted(pids): try: int(pid) except ValueError: continue fd_dir=os.path.join('/proc', pid, 'fd') try: fds = os.listdir(fd_dir) except OSError: continue for file in fds: try: link=os.readlink(os.path.join(fd_dir, file)) except OSError: continue if not os.path.exists(link): continue open_files.setdefault(link, []).append(pid) for link in sorted(open_files.keys()): print "%s : %s" % (link, ", ".join(map(str, open_files[link]))) ------------------------------------------------------------ You might want to consider http://pyinotify.sourceforge.net/ depending on exactly what you are doing... -- Nick Craig-Wood <[EMAIL PROTECTED]> -- http://www.craig-wood.com/nick -- http://mail.python.org/mailman/listinfo/python-list