Count whitespace in directory BASH

Question

I just want to count whitespace char in directory but don't find how to do with AWK command. Currently I have this code

res=0
IFS="
"
cd DirPath
res2=0
res3=0
ws=0

for f in `find . ! -type d`; 
do
  let " res += `wc -l $f | awk '{print $1}'`  "
  let " res2 += `wc -c $f | awk '{print $1}'`  "
  let " res3 += `wc -w $f | awk '{print $1}'`  "
  let " ws += `wc -c $f | awk -v RS='[[:space:]]' 'END{print NR}'` "
done

but the result is very low (176) whereas I have 1300000 char in this directory

Answer 1

To count the number of newline characters (-l), words (-w, which for wc are sequences of non-whitespace characters, so words are delimited by either whitespace or non-characters), characters (-m) and bytes (-c), you can do:

find . -type f -exec cat {} + | wc -lwmc

However note that because cat concatenates the files, it could give incorrect results for the word and character count if there are files that don't end in a whitespace character (text files should end in a newline character, which is a whitespace character) as that could end-up joining two bytes into one valid characters for instance, or joining two words together.

Example:

$ od -tx1 a
0000000 c3
0000001
$ od -tx1 b
0000000 a9
0000001
$ wc -m a b
0 a
0 b
0 total
$ cat a b | wc -m
1

$ printf foo > a
$ printf bar > b
$ wc -w a b
1 a
1 b
2 total
$ cat a b | wc -w
1

To count the whitespace characters, POSIXly, you could do:

find . -type f -exec cat {} + | tr -cd '[:space:]' | wc -m

(with the same caveat about joining bytes into characters), but note that with GNU tr, that only works for single-byte characters (so not UTF-8 encoded non-ASCII characters for instance).

On GNU systems, you could resort to GNU grep and use:

grep -rzo '[[:space:]]' . | LC_ALL=C tr -cd '\0' | wc -c

Though note that because with -z, grep works on NUL-delimited records, that would end up slurping whole text files in memory (as text files typically don't contain NUL bytes).

Answer 2

cd DIRPATH
res2=0
res3=0
ws=0
let " ws += `grep -rzo '[[:space:]]' . | LC_ALL=C tr -cd '\0' | wc -c` "

for f in `find . ! -type d`; 
do
let " res += `wc -l $f | awk '{print $1}'`  "
let " res2 += `wc -c $f | awk '{print $1}'`  "
let " res3 += `wc -w $f | awk '{print $1}'`  "
done

The previous code give you number of whitespace char in directory on ws var

Answer 3

Try with GNU grep:

grep -ro '[[:space:]]' | wc -l