I had some directory, with large number of files. Every time I tried to access the list of files within it, I was not able to do that or there was significant delay. I was trying to use ls command within command-line on Linux and web interface from my hosting provider did not help also.
The problem is, that when I just do ls, it takes significant amount of time to even start displaying something. Thus, ls | wc -l would not help also.
After some research I came up with this code (in this example it counts number of new emails on some server):
print sum([len(files) for (root, dirs, files) in walk('/home/myname/Maildir/new')])
The above code is written in Python. I used Python's command-line tool and it worked pretty fast (returned result instantly).
I am interested in the answer to the following question: is it possible to count files in a directory (without subdirectories) faster? What is the fastest way to do that?
ls does a stat(2) call for every file. Other tools, like find(1) and the shell wildcard expansion, may avoid this call and just do readdir. One shell command combination that might work is find dir -maxdepth 1|wc -l, but it will gladly list the directory itself and miscount any filename with a newline in it.
From Python, the straight forward way to get just these names is os.listdir(directory). Unlike os.walk and os.path.walk, it does not need to recurse, check file types, or make further Python function calls.
Addendum: It seems ls doesn't always stat. At least on my GNU system, it can do only a getdents call when further information (such as which names are directories) is not requested. getdents is the underlying system call used to implement readdir in GNU/Linux.
Addition 2: One reason for a delay before ls outputs results is that it sorts and tabulates. ls -U1 may avoid this.