filter lines featuring some character on sed

Question

I have two unsorted lists of words, and I want to produce a new list excluding the words featuring some selected characters, and in order. I tried the following command:

 cat roa7.lst gr7.usl | sed -e 's/\s/\n/g' | sed -n 's/[^hlrw]/&/p' | sort -u | less -N

However, the last sed command, which tries to exclude the lines containing the set of characters [hlrw] fails to exclude these words. Somehow I'm missing the pragmatics of the caret within the delimiters. How can I fix this pipe?

Answer 1

The right syntax for the last sed command is

sed -n '/[hlrw]/!p'

or

sed '/[hlrw]/d'

which are roughly equivalent. The corresponding awk program would be

awk '!/[hlrw]/'

The wrong command in the original example means "contains any character not in the set hlrw" (which is true, unless the line consists wholly of characters h,l,r,w or is empty) while the second one is "find the lines containing any character in the set hlrw", and then, do not print them; or "find the lines including the set hlrw", and delete them from the input.

Answer 2

Here, you don't need sed, though. You could just do:

grep -v '[hlrw]'

The first (non-standard) sed -e 's/\s/\n/g' could also be replaced with:

tr -s '[:space:]' '[\n*]'

which transliterates and squeezes all sequences of white spaces into a newline character.

That would actually justify the use of cat (which is not needed for sed, as sed can take filenames as arguments and read from them directly).

Thus, the alternative syntax would be

cat roa7.lst gr7.usl | tr -s '[:space:]' '[\n*]' | grep -v '[hlrw]' | sort -u | less -N

which avoids completely the use of sed.