计算bash数组中不以特定字符开头的元素

Question

I'm trying to write a bash script that counts all elements in an array that don't begin with the character '-' (it's part of a shell-completion script that counts the arguments and not the options in the words array). A Python equivalent to what I'm trying to write would be:

len([word for word in words if not word.startswith("-")])

I wrote some code that works fine but since I'm very new to bash scripting, I'm sure that some improvements can be made, and I'm wondering if I missed some better way to implement this that doesn't involve so many lines of code, maybe something that looks more like the Python implementation?

This is what I have:

words=('a' 'b' '-c' 'd' '--foo' 'e')
argcount=0
for word in ${words[@]}
do
    if [[ $word =~ ^[^-].*$ ]] ; then
        ((argcount++))
    fi
done
echo $argcount

Any improvement is welcome.

Answer 1

I'd use

for word in "${words[@]}"

which should work even if some argument contain whitespace. Also, instead of a regular expression, you can use a normal pattern matching:

if [[ $word != -* ]] ; then

Other than that, I do not see anything. Bash is not as concise as Python or Perl.

Answer 2

You can also use grep to count the lines for you:

$ words=('a' 'b' '-c' 'd' '--foo' 'e')
$ argcount=$( printf "%s\n" "${words[@]}" | grep -vc "^-" )

(Assuming no word in words can contain an embedded newline, which seems like a safe enough assumption.)

If you must use pure bash , and don't want to use patterns as choroba suggests, you can negate the result of the regular expression match:

if ! [[ $word =~ ^- ]]; then
    (( argcount++ ))
fi