I was trying to replace the first word in a line with the last word in a line using a sed command i=but i can't seem to figure it out.

  • In every line, or a specific line. If so which line? – Jesse_b Nov 28 '18 at 21:29

Using awk:

awk '{ t=$1; $1=$NF; $NF=t; print}'

This will:

  • t=$1 - set t to the first word
  • $1=$NF - set the first word to the last word
  • $NF=t - set the last word to the first word
  • print - print the new line.

$ echo 'one two three four five six' | awk '{ f=$1; l=$NF; $1=l; $NF=f; print}'
six two three four five one
  • @Kusalananda: Thanks. Didn't think of it. – Jesse_b Nov 28 '18 at 21:41
  • 1
    Slightly shorter : awk '{ t=$1; $1=$NF; $NF=t}1' – steve Nov 28 '18 at 22:37

Using Perl, and assuming whitespace delimited input and space-delimited output:

perl -ape '($F[0],$F[-1])=($F[-1],$F[0]);$_="@F\n"'


$ printf 'Cleanse Fold and Manipulate\n' | perl -ape '($F[0],$F[-1])=($F[-1],$F[0]);$_="@F\n"'
Manipulate Fold and Cleanse

The Perl code, using -a to split the input on whitespace into the array @F, simply swaps the two elements at the start and end of that array before joining the resulting list with spaces, adding a newline at the end.

A shorter Perl variant that matches the first and last words and swaps them in a substitution (this assumes that there are no flanking whitespace in the input though):

perl -pe 's/^(\w*)(.*?)(\w*)$/$3$2$1/'

The middle bit, .*?, matches the middle of the string non-greedily. We couldn't have done this this easily with sed as there is no non-greedy modifier like that ? after .*.

  • 1
    Or, allowing for leading trailing whitespace: perl -pe 's/^(\s*)(\S+)(.+?)(\S+)(\s*)$/$1$4$3$2$5/' -- note the middle bit has at least 1 char to account for separate first/last words. – glenn jackman Nov 28 '18 at 22:24

Semi-serious answer: it's not you, you are fine. The problem is totally in sed's s/// command's verbosity (compare this with the alternative answers):

$ echo "Hello some good world!" |
sed 's/\(^[^[:space:]]\+\)\([[:space:]].*[[:space:]]\|[[:space:]]\+\)\([^[:space:]]\+$\)/\3\2\1/'
world! some good Hello

We may also want to swap the first and the last words even if we have space characters before the first and/or after the latter (thanks to comments and other answers):

$ echo "  Hello some   good world!  " |
sed 's/^\([[:space:]]*\)\([^[:space:]]\+\)\([[:space:]].*[[:space:]]\|[[:space:]]\+\)\([^[:space:]]\+\)\([[:space:]]*\)$/\1\4\3\2\5/'
  world! some   good Hello  

However, these commands use some non-POSIX GNU extensions to the BRE - Basic Regular Expression - syntax (namely, + and |).
A (more portable) command that satisfies the POSIX standard while keeping the convenience of alternation (|) would require Extended regular expressions. For example, using GNU sed with the --posix option (which disables GNU extensions - not actually required):

$ echo "  Hello some   good world!  " |
sed --posix -E 's/^([[:space:]]*)([^[:space:]]{1,})([[:space:]].*[[:space:]]|[[:space:]]{1,})([^[:space:]]{1,})([[:space:]]*)$/\1\4\3\2\5/'
  world! some   good Hello  

Note, however, that POSIX ERE syntax does not include backreferences. This command, too, will require an implementation with extensions to succeed.

  • 3
    Will not destroy the original whitespace. +1 – glenn jackman Nov 28 '18 at 22:22

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