How do i replace [] brackets using SED

Question

I have a string that i am want to remove punctuation from.

I started with

sed \'s/[[:punct:]]/ /g\'

But i had problems on HP-UX n

Answer 1

You need to place the brackets early in the expression:

sed 's/[][=+...-]/ /g'

By placing the ']' as the first character immediately after the opening bracket, it is interpreted as a member of the character set rather than a closing bracket. Placing a '[' anywhere inside the brackets makes it a member of the set.

For this particular character set, you also need to deal with - specially, since you are not trying to build a range of characters between [ and =. So put the - at the end of the class.

Answer 2

You can do it manually:

sed 's/[][\/$*.^|@#{}~&()_:;%+"='\'',`><?!-]/ /g'

This remove the 32 punctuation character, the order of some characters is important:

• - should be at the end like this -]
• [] should be like that [][other characters]
• ' should be escaped like that '\''
• not begin with ^ like in [^
• not begin with [. [= [: and end with .] =] :]
• not end with $]

here you can have explication of why all that http://pubs.opengroup.org/onlinepubs/9699919799/basedefs/V1_chap09.html#tag_09_03_03

Answer 3

You can also specify the characters you want to keep [with inversion]:

sed 's/[^a-zA-Z0-9]/ /g'

Answer 4

Can be handled using the regex capture technique too (Eg: here below) :

echo "narrowPeak_SP1[FLAG]" | sed -e 's/\[\([a-zA-Z0-9]*\)\]/_\1/g'
> narrowPeak_SP1_FLAG

\[ : literal match to open square bracket, since [] is a valid regex
\] : literal match to square close bracket
\(...\) : capture group
\1 : represents the capture group within the square brackets

Answer 5

Here is the final code I ended up with

`echo "$string" | sed 's/[^a-zA-Z0-9]/ /g'`

I had to put = and - at the very end.