What is the best way to count "find" results?

Question

My current solution would be find <expr> -exec printf '.' \; | wc -c, but this takes far too long when there are more than 10000 results. Is there no faster/better way to do this?

Answer 1

Why not

find <expr> | wc -l

as a simple portable solution? Your original solution is spawning a new process printf for every individual file found, and that's very expensive (as you've just found).

Note that this will overcount if you have filenames with newlines embedded, but if you have that then I suspect your problems run a little deeper.

Answer 2

Try this instead (require find's -printf support):

find <expr> -type f -printf '.' | wc -c

It will be more reliable and faster than counting the lines.

Note that I use the find's printf, not an external command.

---

Let's bench a bit :

$ ls -1
a
e
l
ll.sh
r
t
y
z

My snippet benchmark :

$ time find -type f -printf '.' | wc -c
8

real    0m0.004s
user    0m0.000s
sys     0m0.007s

With full lines :

$ time find -type f | wc -l
8

real    0m0.006s
user    0m0.003s
sys     0m0.000s

So my solution is faster =) (the important part is the real line)

Answer 3

This solution is certainly slower than some of the other find -> wc solutions here, but if you were inclined to do something else with the file names in addition to counting them, you could read from the find output.

n=0
while read -r -d ''; do
    ((n++)) # count
    # maybe perform another act on file
done < <(find <expr> -print0)
echo $n

It is just a modification of a solution found in BashGuide that properly handles files with nonstandard names by making the find output delimiter a NUL byte using print0, and reading from it using '' (NUL byte) as the loop delimiter.