How to replace text in file between known start and stop positions with a command line utility like sed or awk? - bash

Question

I have been tinkering with this for a while but can't quite figure it out. A sample line within the file looks like this:
"...~236 characters of data...Y YYY. Y...many more characters of data"
How would I use sed or awk to replace spaces with a B character only between positions 236 and 246? In that example string it starts at character 29 and ends at character 39 within the string. I would want to preserve all the text preceding and following the target chunk of data within the line.
For clarification based on the comments, it should be applied to all lines in the file and expected output would be:
"...~236 characters of data...YBBYYY.BBY...many more characters of data"

Answer 1

With GNU awk:
$ awk -v FIELDWIDTHS='29 10 *' -v OFS= '{gsub(/ /, "B", $2)} 1' ip.txt
...~236 characters of data...YBBYYY.BBY...many more characters of data
FIELDWIDTHS='29 10 *' means 29 characters for first field, next 10 characters for second field and the rest for third field. OFS is set to empty, otherwise you'll get space added between the fields.
With perl:
$ perl -pe 's/^.{29}\K.{10}/$&=~tr| |B|r/e' ip.txt
...~236 characters of data...YBBYYY.BBY...many more characters of data
^.{29}\K match and ignore first 29 characters
.{10} match 10 characters
e flag to allow Perl code instead of string in replacement section
$&=~tr| |B|r convert space to B for the matched portion

Answer 2

Use this Perl one-liner with substr and tr. Note that this uses the fact that you can assign to substr, which changes the original string:
perl -lpe 'BEGIN { $from = 29; $to = 39; } (substr $_, ( $from - 1 ), ( $to - $from + 1 ) ) =~ tr/ /B/;' in_file > out_file
To change the file in-place, use:
perl -i.bak -lpe 'BEGIN { $from = 29; $to = 39; } (substr $_, ( $from - 1 ), ( $to - $from + 1 ) ) =~ tr/ /B/;' in_file
The Perl one-liner uses these command line flags:
-e : Tells Perl to look for code in-line, instead of in a file.
-p : Loop over the input one line at a time, assigning it to $_ by default. Add print $_ after each loop iteration.
-l : Strip the input line separator ("\n" on *NIX by default) before executing the code in-line, and append it when printing.
-i.bak : Edit input files in-place (overwrite the input file). Before overwriting, save a backup copy of the original file by appending to its name the extension .bak.

Answer 3

I would use GNU AWK following way, for simplicity sake say we have file.txt content
S o m e s t r i n g
and want to change spaces from 5 (inclusive) to 10 (inclusive) position then
awk 'BEGIN{FPAT=".";OFS=""}{for(i=5;i<=10;i+=1)$i=($i==" "?"B":$i);print}' file.txt
output is
S o mBeBsBt r i n g
Explanation: I set field pattern (FPAT) to any single character and output field seperator (OFS) to empty string, thus every field is populated by single characters and I do not get superfluous space when print-ing. I use for loop to access desired fields and for every one I check if it is space, if it is I assign B here otherwise I assign original value, finally I print whole changed line.

Answer 4

Using GNU awk:
awk -v strt=29 -v end=39 '{ ram=substr($0,strt,(end-strt));gsub(" ","B",ram);print substr($0,1,(strt-1)) ram substr($0,(end)) }' file
Explanation:
awk -v strt=29 -v end=39 '{ # Pass the start and end character positions as strt and end respectively
ram=substr($0,strt,(end-strt)); # Extract the 29th to the 39th characters of the line and read into variable ram
gsub(" ","B",ram); # Replace spaces with B in ram
print substr($0,1,(strt-1)) ram substr($0,(end)) # Rebuild the line incorporating raw and printing the result
}'file

Answer 5

This is certainly a suitable task for perl, and saddens me that my perl has become so rusty that this is the best I can come up with at the moment:
perl -e 'local $/=\1;while(<>) { s/ /B/ if $. >= 236 && $. <= 246; print }' input;

Answer 6

Another awk but using FS="":
$ awk 'BEGIN{FS=OFS=""}{for(i=29;i<=39;i++)sub(/ /,"B",$i)}1' file
Output:
"...~236 characters of data...YBBYYY.BBY...many more characters of data"
Explained:
$ awk ' # yes awk yes
BEGIN {
FS=OFS="" # set empty field delimiters
}
{
for(i=29;i<=39;i++) # between desired indexes
sub(/ /,"B",$i) # replace space with B
# if($i==" ") # couldve taken this route, too
# $i="B"
}1' file # implicit output

Answer 7

With sed :
sed '
H
s/\(.\{236\}\)\(.\{11\}\).*/\2/
s/ /B/g
H
g
s/\n//g
s/\(.\{236\}\)\(.\{11\}\)\(.*\)\(.\{11\}\)/\1\4\3/
x
s/.*//
x' infile

Answer 8

When you have an input string without \r, you can use:
sed -r 's/(.{236})(.{10})(.*)/\1\r\2\r\3/;:a;s/(\r.*) (.*\r)/\1B\2/;ta;s/\r//g' input
Explanation:
First put \r around the area that you want to change.
Next introduce a label to jump back to.
Next replace a space between 2 markers.
Repeat until all spaces are replaced.
Remove the markers.
In your case, where the length doesn't change, you can do without the markers.
Replace a space after 236..245 characters and try again when it succeeds.
sed -r ':a; s/^(.{236})([^ ]{0,9}) /\1\2B/;ta' input

Answer 9

This might work for you (GNU sed):
sed -E 's/./&\n/245;s//\n&/236/;h;y/ /B/;H;g;s/\n.*\n(.*)\n.*\n(.*)\n.*/\2\1/' file
Divide the problem into 2 lines, one with spaces and one with B's where there were spaces.
Then using pattern matching make a composite line from the two lines.
N.B. The newline can be used as a delimiter as it is guaranteed not to be in seds pattern space.