Remove one-character words

Question

I'm looking for a regexp to remove one character words. I don't mind whether using perl, awk, sed or bash built-ins.

Test case:

$ echo "a b c d e f g h ijkl m n opqrst u v" | $COMMAND

Desired output:

ijkl opqrst

What I've tried so far:

$ echo "a b c d e f g h ijkl m n opqrst u v" | sed 's/ . //g'
acegijkln opqrstv

I'm guessing that:

• the a isn't removed because there is no white space before it
• the c remains because once the b has been removed, there is no more whitespace before it
• and so on...

Attempt #2:

$ echo "a b c d e f g h ijkl m n opqrst u v" | sed 's/\w.\w//g'
     s v

Here I don't get at all what's happening.

Any help + explanations are welcome, I want to learn.

Answer 1

You have to use word boundary \b (or) \< and \> respectively match the empty string at the beginning and end of a word.

echo "a b c d e f g h ijkl m n opqrst u v" | sed 's/\b\w\b \?//g'

(OR)

echo "a b c d e f g h ijkl m n opqrst u v" | sed 's/\<.\> \?//g'

Answer 2

You could simply use grep:

echo "a b c d e f g h ijkl m n opqrst u v"  | grep -o '[a-z]\{2,\}'

where the regex is matching any word composed with at least 2 characters.

The -o option in grep prints the matching pattern (and not the entire line).

Answer 3

Albeit, Awk is not the most efficient of ways to do this, answering only because it is tagged awk, using its length() string function. It is POSIX compliant, so no issues on portability.

echo "a b c d e f g h ijkl m n opqrst u v" | \
  awk '{for(i=1;i<=NF;i++) {if (length($i)>1) { printf "%s ", $i }} }'
ijkl opqrst

Answer 4

Perl solution: just filter elements on length

echo "a b c d e f g h ijkl m n opqrst u v" | perl -lanE \
  'say join " ", grep {length($_) > 1} @F'

Answer 5

Just for fun, another option: translate spaces to newlines and look for lines with at least 2 characters

$ echo "a b c d e f g h ijkl m n opqrst u v" | tr ' ' '\n' | grep .. | paste -sd " "
ijkl opqrst

Answer 6

Not being familiar with any linux sprung tools, this is somewhat of a guess, but I think the (a) regex you want is

(?:\s\w\b|\b\w\s)

like

$ echo "a b c d e f g h ijkl m n opqrst u v" | sed 's/(?:\s\w\b|\b\w\s)//g'

This would replace any single character either preceded by, or foolowed by, a space with nothing.

Check the regex out here at regex101.

Answer 7

Another in awk. A non-space ([^ ]) is considered a word. Feel free to replace it with your definition of a word.

$ awk '{while(sub(/^[^ ] | [^ ]$/,"")||sub(/ [^ ] /," "));}1'

Using sub it replaces [a space][non-space][a space] tuples with a space and removes from the beginning and end of record the single characters and leading / trailing space. It's in a while so it keeps doing it until there are no hits left. To test it:

$ echo "a b c d e f g h ijkl m n opqrst u v"|awk '{while(sub(/^[^ ] | [^ ]$/,"")||sub(/ [^ ] /," "));}1'
ijkl opqrst

Answer 8

echo "a b c d e f g h ijkl m n opqrst u v"  | grep -wo "\b[a-z][a-z]\+\b"