How to cut a line at a specific character

Question

I have a bunch of lines in a file that look like this

word_word_word 0 word_word
word_word_word 1 wordwordword
word word word word 0 word word word word
word 2 word_word_word word word
word word_word 3 word

I want to cat the file and get an output that looks like this:

word_word_word 0
word_word_word 1
etc...

How to I cut/awk/or whatever the line and display everything from the first byte through the first single-digit number?

Thanks!

Answer 1

Using grep:

grep -oP '^.*?\s[0-9](\s|$)' file

The -o option tells grep to print only the match.

The pattern is looking for:

• The beginning of the line ^
• Followed by literally anything repeating (non-greedy) .*?
• Followed by whitespace \s
• Followed by a single digit between 0 and 9 [0-9]
• Finally that single digit must be followed by whitespace or the end of the line (\s|$)

Answer 2

For loop with awk:

awk '{
  for(i=1;i<=NF;i++){
    if($i ~ /[0-9]/){
      printf "%s\n",$i;
      break
    }
    printf "%s ",$i
  }
}' file

Will get what you want.

Answer 3

If this (using any sed in any shell on every UNIX box) isn't all you need then edit your question to provide a better example including lines that this doesn't work for:

$ sed 's/\( [0-9]\) .*/\1/' file
word_word_word 0
word_word_word 1
word word word word 0
word 2
word word_word 3