sed replace last line matching pattern

Question

Given a file like this:

a
b
a
b

I\'d like to be able to use sed to replace just the last line that contains an instance of \"a

Answer 1

A two-pass solution for when buffering the entire input is intolerable:

sed "$(sed -n /a/= file | sed -n '$s/$/ s,a,c,/p' )" file

(the earlier version of this hit a bug with history expansion encountered on a redhat bash-4.1 install, this way avoids a $!d that was being mistakenly expanded.)

A one-pass solution that buffers as little as possible:

sed '/a/!{1h;1!H};/a/{x;1!p};$!d;g;s/a/c/'

Simplest:

tac | sed '0,/a/ s/a/c/' | tac

Answer 2

Here's the command:

sed '$s/.*/a/' filename.txt

And here it is in action:

> echo "a
> b
> a
> b" > /tmp/file.txt

> sed '$s/.*/a/' /tmp/file.txt
a
b
a
a

Answer 3

Not quite sed only:

tac file | sed '/a/ {s//c/; :loop; n; b loop}' | tac

testing

% printf "%s\n" a b a b a b | tac | sed '/a/ {s//c/; :loop; n; b loop}' | tac
a
b
a
b
c
b

Reverse the file, then for the first match, make the substitution and then unconditionally slurp up the rest of the file. Then re-reverse the file.

Note, an empty regex (here as s//c/) means re-use the previous regex (/a/)

I'm not a huge sed fan, beyond very simple programs. I would use awk:

tac file | awk '/a/ && !seen {sub(/a/, "c"); seen=1} 1' | tac

Answer 4

This might work for you (GNU sed):

sed -r '/^PATTERN/!b;:a;$!{N;/^(.*)\n(PATTERN.*)/{h;s//\1/p;g;s//\2/};ba};s/^PATTERN/REPLACEMENT/' file

or another way:

sed '/^PATTERN/{x;/./p;x;h;$ba;d};x;/./{x;H;$ba;d};x;b;:a;x;/./{s/^PATTERN/REPLACEMENT/p;d};x' file

or if you like:

sed -r ':a;$!{N;ba};s/^(.*\n?)PATTERN/\1REPLACEMENT/' file

On reflection, this solution may replace the first two:

sed  '/a/,$!b;/a/{x;/./p;x;h};/a/!H;$!d;x;s/^a$/c/M' file

If the regexp is no where to found in the file, the file will pass through unchanged. Once the regex matches, all lines will be stored in the hold space and will be printed when one or both conditions are met. If a subsequent regex is encountered, the contents of the hold space is printed and the latest regex replaces it. At the end of file the first line of the hold space will hold the last matching regex and this can be replaced.

Answer 5

Here is all done in one single awk

awk 'FNR==NR {if ($0~/a/) f=NR;next} FNR==f {$0="c"} 1' file file
a
b
c
b

This reads the file twice. First run to find last a, second run to change it.

Answer 6

Another one:

tr '\n' ' ' | sed 's/\(.*\)a/\1c/' | tr ' ' '\n'

in action:

$ printf "%s\n" a b a b a b | tr '\n' ' ' | sed 's/\(.*\)a/\1c/' | tr ' ' '\n'
a
b
a
b
c
b