sed: replace letter between square brackets - sed

Question

I have the following string:
signal[i]
signal[bg]
output [10:0]
input [i:1]
what I want is to replace the letters between square brackets (by underscore for example) and to keep the other strings that represents table declaration:
signal[_]
signal[__]
output [10:0]
input [i:1]
thanks

Answer 1

try:
awk '{gsub(/\[[a-zA-Z]+\]/,"[_]")} 1' Input_file
Globally substituting the (bracket)alphabets till their longest match then with [_]. Mentioning 1 will print the lines(edited or without edited ones).
EDIT: Above will substitute all alphabets with one single _, so to get as many underscores as many characters are there following may help in same.
awk '{match($0,/\[[a-zA-Z]+\]/);VAL=substr($0,RSTART+1,RLENGTH-2);if(VAL){len=length(VAL);;while(i<len){q=q?q"_":"_";i++}};gsub(/\[[a-zA-Z]+\]/,"["q"]")}1' Input_file
OR
awk '{
match($0,/\[[a-zA-Z]+\]/);
VAL=substr($0,RSTART+1,RLENGTH-2);
if(VAL){
len=length(VAL);
while(i<len){
q=q?q"_":"_";
i++
}
};
gsub(/\[[a-zA-Z]+\]/,"["q"]")
}
1
' Input_file
Will add explanation soon.
EDIT2: Following is the one with explanation purposes for OP and users.
awk '{
match($0,/\[[a-zA-Z]+\]/); #### using match awk's built-in utility to match the [alphabets] as per OP's requirement.
VAL=substr($0,RSTART+1,RLENGTH-2); #### Creating a variable named VAL which has substr($0,RSTART+1,RLENGTH-2); which will have substring value, whose starting point is RSTART+1 and ending point is RLENGTH-2.
RSTART and RLENGTH are the variables out of the box which will be having values only when awk finds any match while using match.
if(VAL){ #### Checking if value of VAL variable is NOT NULL. Then perform following actions.
len=length(VAL); #### creating a variable named len which will have length of variable VAL in it.
while(i<len){ #### Starting a while loop which will run till the value of VAL from i(null value).
q=q?q"_":"_"; #### creating a variable named q whose value will be concatenated it itself with "_".
i++ #### incrementing the value of variable i with 1 each time.
}
};
gsub(/\[[a-zA-Z]+\]/,"["q"]") #### Now globally substituting the value of [ alphabets ] with [ value of q(which have all underscores in it) then ].
}
1 #### Mentioning 1 will print (edited or non-edited) lines here.
' Input_file #### Mentioning the Input_file here.

Answer 2

Alternative gawk solution:
awk -F'\\[|\\]' '$2!~/^[0-9]+:[0-9]$/{ gsub(/./,"_",$2); $2="["$2"]" }1' OFS= file
The output:
signal[_]
signal[__]
output [10:0]
-F'\\[|\\]' - treating [ and ] as field separators
$2!~/^[0-9]+:[0-9]$/ - performing action if the 2nd field does not represent table declaration
gsub(/./,"_",$2) - replace each character with _

Answer 3

This might work for you (GNU sed);
sed ':a;s/\(\[_*\)[[:alpha:]]\([[:alpha:]]*\]\)/\1_\2/;ta' file
Match on opening and closing square brackets with any number of _'s and at least one alpha character and replace said character by an underscore and repeat.

Answer 4

awk '{sub(/\[i\]/,"[_]")sub(/\[bg\]/,"[__]")}1' file
signal[_]
signal[__]
output [10:0]
input [i:1]
The explanation is as follows: Since bracket is as special character it has to be escaped to be handled literally then it becomes easy use sub.