Last n words from a string in bash - string

Question

I have a string containing many (the total number varies) words, and I need to get last 10 of them. How do I do it? I'm looking at awk, grep and cut but nothing really comes to mind.
An example (although it seems to me that the question is clear):
aaa bda fdkfj fds fsd ... dsad dsas dsad zrthd shshh
I want the last 10 words of this string.
Again, the total number of words in the initial string isn't defined.

Answer 1

Just play with tr, tail and xargs:
$ echo "1 2 3 4 5 6 7 8 9 10" | tr ' ' '\n' | tail -5 | xargs -n5
6 7 8 9 10
This prints the words one in every line, so that tail gets the desired amount of them. Then, xargs "remerges" them in the same line.
You can also set awk's NF to the value you want after reversing the text:
$ echo "1 2 3 4 5 6 7 8 9 10" | rev | awk '{NF=5}1' | rev
6 7 8 9 10

Answer 2

When you're trying to find words or characters at the last, it's better to use end of the line anchor $ in your regex.
$ echo "aaa bda fdkfj fds fsd bar dsad dsas dsad zrthd shshh" | grep -o '[^[:space:]]\+\([[:space:]]\+[^[:space:]]\+\)\{9\} *$'
bda fdkfj fds fsd bar dsad dsas dsad zrthd shshh
You could use the same regex in sed also.
OR
$ echo "aaa bda fdkfj fds fsd bar dsad dsas dsad zrthd shshh" | grep -oP '\S+(?:\s+\S+){9} *$'
bda fdkfj fds fsd bar dsad dsas dsad zrthd shshh

Answer 3

In awk, the builtin variable NF is set to the number of fields (which are by default words) on each line. So you can:
echo "${STRING}" | awk '{
for (i = NF - 9; i <= NF; i++) {printf "%s ", $i}
printf "\n"
}'
assuming that you always have at least 10 words on the line. If not, you can add extra checks for that. And do something more if you don't want the extra space at the end of the line.

Answer 4

The canonical, pure Bash way of doing this is to use read:
string='one two three four five six seven eight nine ten eleven twelve thirteen fourteen fifteen sixteen seventeen eighteen nineteen forty two'
read -r -d '' -a array < <(printf '%s\0' "$string")
# Print only ten last words:
printf '%s\n' "${array[*]: -10}"
If there are less than 10 words, the last expansion fails, but this can be easily fixed:
printf '%s\n' "${array[*]:${#array[#]}<10?0:-10}"

Answer 5

You want shell? This is pure shell. No awk, no cut, no sed, no perl. You can't get more sell than this. (Okay, I do use wc which is a utility and not part of Bash shell, but everything else is part of Bash).
FOO="one two three four five six seven eight nine ten eleven twelve thirteen"
set $FOO
((shift=$(wc -w<<<$FOO)-10))
shift $shift
echo $*
The set sets the positional parameters. (The $1, $2, etc. in command line arguments).
The $(wc -w<<<$FOO) finds the number of parameters.
I subtract that number from 10, and get the number of parameters greater than ten. I set this to $shift
I then shift $shift parameters. This leaves the last ten parameters which I echo.
You don't really need wc. $# expands to the number of positional parameters set. – gniourf_gniourf
Oh, I forgot about that. Now, we have a pure Bash answer:
FOO='one two three four five six seven eight nine ten eleven twelve thirteen'
set $FOO
((shift=$#-10))
shift $shift
echo $*

Answer 6

echo $string | perl -lanE 'say join " ", #F[-10..-1]'

Answer 7

Your string:
string="Lorem ipsum dolor sit amet"
The last four words by using a pure Bash/Shell one-liner:
echo ${string/${string% * * * *} /}
Repeat or remove  * to fetch more or less words.
Explanation
We use the Shell Parameter Expansion ${parameter/pattern/string} to replace x words by nothing. And as the pattern ${str% * * * *}  returns everything in front of the last 4 words it removes the leading Lorem from our string.