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How to get single character after space?

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.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;

2

0

How can I obtain below output? I want the first field as it is and a single character after space.

echo "Hello world"
Hellow

If it also has a 3rd field than the beginning character of the 3rd field should be in the output.

echo "hello world unix"
hellou

linux awk sed grep cut

share|improve this question

asked Apr 14 at 4:45

BDNBDN

130112

add a comment | 

2

0

How can I obtain below output? I want the first field as it is and a single character after space.

echo "Hello world"
Hellow

If it also has a 3rd field than the beginning character of the 3rd field should be in the output.

echo "hello world unix"
hellou

linux awk sed grep cut

share|improve this question

asked Apr 14 at 4:45

BDNBDN

130112

add a comment | 

How can I obtain below output? I want the first field as it is and a single character after space.

echo "Hello world"
Hellow

If it also has a 3rd field than the beginning character of the 3rd field should be in the output.

echo "hello world unix"
hellou

linux awk sed grep cut

share|improve this question

asked Apr 14 at 4:45

BDNBDN

130112

echo "Hello world"
Hellow

echo "hello world unix"
hellou

share|improve this question

asked Apr 14 at 4:45

BDNBDN

130112

share|improve this question

asked Apr 14 at 4:45

BDNBDN

130112

asked Apr 14 at 4:45

BDNBDN

130112

asked Apr 14 at 4:45

BDNBDN

130112

3 Answers
3

active

oldest

votes

4

Using awk to output the first whitespace-delimited word concatenated with the first character of the last whitespace-delimited word:

awk ' print $1 substr($NF, 1, 1) '

The substr() function returns a number of characters from a given position of a string, and $1 and $NF is the first and last whitespace-delimited word on the current line, respectively.

Testing:

$ echo 'hello world' | awk ' print $1 substr($NF, 1, 1) '
hellow

$ echo 'apple beet carrot' | awk ' print $1 substr($NF, 1, 1) '
applec

share|improve this answer

edited Apr 14 at 6:37

answered Apr 14 at 6:29

Kusalananda♦Kusalananda

148k18279468

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Will not this duplicated the first letter of a word for a line with that single word only?
  
  – αғsнιη
  Apr 14 at 6:41
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @αғsнιη Yes it would. The question does not specify what should happen in the case when there is only a single word.
  
  – Kusalananda♦
  Apr 14 at 6:43
  

  
  
  
  

add a comment | 

6

With sed:

Edit: improved by glenn jackmann, thanks!

$ echo "Hello world" | sed -E 's/(S+).*s(S).*$/12/'
Hellow
$ echo "hello world unix" | sed -E 's/(S+).*s(S).*$/12/'
hellou

Description with "hello world unix" as example:

• 
  s/ substitute the following pattern


• 
  (S+) 1st group, one or more non-space characters: "hello"


• 
  .* the middle part, any characters: " world"


• 
  s space character: " "


• 
  (S) 2nd group, non-space character: "u"


• 
  .*$ any characters up to the end: "nix"


• 
  /12/ replace with 1st and 2nd group: "hellou"

With bash:

$ var="Hello world"
$ var_end=$var##* ;echo $var%% *$var_end:0:1
Hellow

$ var="hello world unix"
$ var_end=$var##* ;echo $var%% *$var_end:0:1
hellou

Description with "hello world unix" as example:

• 
  var_end=$var##* remove matching prefix pattern, longest match,
  
  "hello world ", result: "unix"


• 
  $var%% * remove matching suffix pattern, longest match,
  
  " world unix", result: "hello"


• 
  $var_end:0:1 get the first character: "u"

share|improve this answer

edited Apr 14 at 12:28

answered Apr 14 at 5:31

FreddyFreddy

3,8231417

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Your regex can be a bit simpler: you don't need to capture the 2nd group, and it can contain any characters: ^(S+).*s(S) will do. Also, I believe the perl-like regex means you must use GNU sed.
  
  – glenn jackman
  Apr 14 at 12:03
  

  
  
  
  

add a comment | 

3

Using bash:

text="hello world unix"
if [[ $text =~ ^([^[:space:]]+).*[[:space:]]([^[:space:]]) ]]; then
 declare -p BASH_REMATCH
 echo "$BASH_REMATCH[1]$BASH_REMATCH[2]"
fi

declare -ar BASH_REMATCH='([0]="hello world u" [1]="hello" [2]="u")'
hellou

share|improve this answer

answered Apr 14 at 12:05

glenn jackmanglenn jackman

53.9k674115

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  just curious to understand to know, why is declare -p needed here? The array is already populated right?
  
  – Inian
  Apr 15 at 5:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  It's not needed at all. Merely to show the contents of the array, to see the results of the regex match with capturing parentheses.
  
  – glenn jackman
  Apr 15 at 10:42
  
  

  
  
  
  

add a comment | 

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4

Using awk to output the first whitespace-delimited word concatenated with the first character of the last whitespace-delimited word:

awk ' print $1 substr($NF, 1, 1) '

The substr() function returns a number of characters from a given position of a string, and $1 and $NF is the first and last whitespace-delimited word on the current line, respectively.

Testing:

$ echo 'hello world' | awk ' print $1 substr($NF, 1, 1) '
hellow

$ echo 'apple beet carrot' | awk ' print $1 substr($NF, 1, 1) '
applec

share|improve this answer

edited Apr 14 at 6:37

answered Apr 14 at 6:29

Kusalananda♦Kusalananda

148k18279468

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Will not this duplicated the first letter of a word for a line with that single word only?
  
  – αғsнιη
  Apr 14 at 6:41
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @αғsнιη Yes it would. The question does not specify what should happen in the case when there is only a single word.
  
  – Kusalananda♦
  Apr 14 at 6:43
  

  
  
  
  

add a comment | 

4

Using awk to output the first whitespace-delimited word concatenated with the first character of the last whitespace-delimited word:

awk ' print $1 substr($NF, 1, 1) '

The substr() function returns a number of characters from a given position of a string, and $1 and $NF is the first and last whitespace-delimited word on the current line, respectively.

Testing:

$ echo 'hello world' | awk ' print $1 substr($NF, 1, 1) '
hellow

$ echo 'apple beet carrot' | awk ' print $1 substr($NF, 1, 1) '
applec

share|improve this answer

edited Apr 14 at 6:37

answered Apr 14 at 6:29

Kusalananda♦Kusalananda

148k18279468

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Will not this duplicated the first letter of a word for a line with that single word only?
  
  – αғsнιη
  Apr 14 at 6:41
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @αғsнιη Yes it would. The question does not specify what should happen in the case when there is only a single word.
  
  – Kusalananda♦
  Apr 14 at 6:43
  

  
  
  
  

add a comment | 

Using awk to output the first whitespace-delimited word concatenated with the first character of the last whitespace-delimited word:

awk ' print $1 substr($NF, 1, 1) '

The substr() function returns a number of characters from a given position of a string, and $1 and $NF is the first and last whitespace-delimited word on the current line, respectively.

Testing:

$ echo 'hello world' | awk ' print $1 substr($NF, 1, 1) '
hellow

$ echo 'apple beet carrot' | awk ' print $1 substr($NF, 1, 1) '
applec

share|improve this answer

edited Apr 14 at 6:37

answered Apr 14 at 6:29

Kusalananda♦Kusalananda

148k18279468

awk ' print $1 substr($NF, 1, 1) '

$ echo 'hello world' | awk ' print $1 substr($NF, 1, 1) '
hellow

$ echo 'apple beet carrot' | awk ' print $1 substr($NF, 1, 1) '
applec

share|improve this answer

edited Apr 14 at 6:37

answered Apr 14 at 6:29

Kusalananda♦Kusalananda

148k18279468

answered Apr 14 at 6:29

Kusalananda♦Kusalananda

148k18279468

answered Apr 14 at 6:29

Kusalananda♦Kusalananda

148k18279468

6

With sed:

Edit: improved by glenn jackmann, thanks!

$ echo "Hello world" | sed -E 's/(S+).*s(S).*$/12/'
Hellow
$ echo "hello world unix" | sed -E 's/(S+).*s(S).*$/12/'
hellou

Description with "hello world unix" as example:

• 
  s/ substitute the following pattern


• 
  (S+) 1st group, one or more non-space characters: "hello"


• 
  .* the middle part, any characters: " world"


• 
  s space character: " "


• 
  (S) 2nd group, non-space character: "u"


• 
  .*$ any characters up to the end: "nix"


• 
  /12/ replace with 1st and 2nd group: "hellou"

With bash:

$ var="Hello world"
$ var_end=$var##* ;echo $var%% *$var_end:0:1
Hellow

$ var="hello world unix"
$ var_end=$var##* ;echo $var%% *$var_end:0:1
hellou

Description with "hello world unix" as example:

• 
  var_end=$var##* remove matching prefix pattern, longest match,
  
  "hello world ", result: "unix"


• 
  $var%% * remove matching suffix pattern, longest match,
  
  " world unix", result: "hello"


• 
  $var_end:0:1 get the first character: "u"

share|improve this answer

edited Apr 14 at 12:28

answered Apr 14 at 5:31

FreddyFreddy

3,8231417

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Your regex can be a bit simpler: you don't need to capture the 2nd group, and it can contain any characters: ^(S+).*s(S) will do. Also, I believe the perl-like regex means you must use GNU sed.
  
  – glenn jackman
  Apr 14 at 12:03
  

  
  
  
  

add a comment | 

6

With sed:

Edit: improved by glenn jackmann, thanks!

$ echo "Hello world" | sed -E 's/(S+).*s(S).*$/12/'
Hellow
$ echo "hello world unix" | sed -E 's/(S+).*s(S).*$/12/'
hellou

Description with "hello world unix" as example:

• 
  s/ substitute the following pattern


• 
  (S+) 1st group, one or more non-space characters: "hello"


• 
  .* the middle part, any characters: " world"


• 
  s space character: " "


• 
  (S) 2nd group, non-space character: "u"


• 
  .*$ any characters up to the end: "nix"


• 
  /12/ replace with 1st and 2nd group: "hellou"

With bash:

$ var="Hello world"
$ var_end=$var##* ;echo $var%% *$var_end:0:1
Hellow

$ var="hello world unix"
$ var_end=$var##* ;echo $var%% *$var_end:0:1
hellou

Description with "hello world unix" as example:

• 
  var_end=$var##* remove matching prefix pattern, longest match,
  
  "hello world ", result: "unix"


• 
  $var%% * remove matching suffix pattern, longest match,
  
  " world unix", result: "hello"


• 
  $var_end:0:1 get the first character: "u"

share|improve this answer

edited Apr 14 at 12:28

answered Apr 14 at 5:31

FreddyFreddy

3,8231417

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Your regex can be a bit simpler: you don't need to capture the 2nd group, and it can contain any characters: ^(S+).*s(S) will do. Also, I believe the perl-like regex means you must use GNU sed.
  
  – glenn jackman
  Apr 14 at 12:03
  

  
  
  
  

add a comment | 

With sed:

Edit: improved by glenn jackmann, thanks!

$ echo "Hello world" | sed -E 's/(S+).*s(S).*$/12/'
Hellow
$ echo "hello world unix" | sed -E 's/(S+).*s(S).*$/12/'
hellou

Description with "hello world unix" as example:

• 
  s/ substitute the following pattern


• 
  (S+) 1st group, one or more non-space characters: "hello"


• 
  .* the middle part, any characters: " world"


• 
  s space character: " "


• 
  (S) 2nd group, non-space character: "u"


• 
  .*$ any characters up to the end: "nix"


• 
  /12/ replace with 1st and 2nd group: "hellou"

With bash:

$ var="Hello world"
$ var_end=$var##* ;echo $var%% *$var_end:0:1
Hellow

$ var="hello world unix"
$ var_end=$var##* ;echo $var%% *$var_end:0:1
hellou

Description with "hello world unix" as example:

• 
  var_end=$var##* remove matching prefix pattern, longest match,
  
  "hello world ", result: "unix"


• 
  $var%% * remove matching suffix pattern, longest match,
  
  " world unix", result: "hello"


• 
  $var_end:0:1 get the first character: "u"

share|improve this answer

edited Apr 14 at 12:28

answered Apr 14 at 5:31

FreddyFreddy

3,8231417

$ echo "Hello world" | sed -E 's/(S+).*s(S).*$/12/'
Hellow
$ echo "hello world unix" | sed -E 's/(S+).*s(S).*$/12/'
hellou

$ var="Hello world"
$ var_end=$var##* ;echo $var%% *$var_end:0:1
Hellow

$ var="hello world unix"
$ var_end=$var##* ;echo $var%% *$var_end:0:1
hellou

share|improve this answer

edited Apr 14 at 12:28

answered Apr 14 at 5:31

FreddyFreddy

3,8231417

answered Apr 14 at 5:31

FreddyFreddy

3,8231417

answered Apr 14 at 5:31

FreddyFreddy

3,8231417

3

Using bash:

text="hello world unix"
if [[ $text =~ ^([^[:space:]]+).*[[:space:]]([^[:space:]]) ]]; then
 declare -p BASH_REMATCH
 echo "$BASH_REMATCH[1]$BASH_REMATCH[2]"
fi

declare -ar BASH_REMATCH='([0]="hello world u" [1]="hello" [2]="u")'
hellou

share|improve this answer

answered Apr 14 at 12:05

glenn jackmanglenn jackman

53.9k674115

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  just curious to understand to know, why is declare -p needed here? The array is already populated right?
  
  – Inian
  Apr 15 at 5:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  It's not needed at all. Merely to show the contents of the array, to see the results of the regex match with capturing parentheses.
  
  – glenn jackman
  Apr 15 at 10:42
  
  

  
  
  
  

add a comment | 

3

Using bash:

text="hello world unix"
if [[ $text =~ ^([^[:space:]]+).*[[:space:]]([^[:space:]]) ]]; then
 declare -p BASH_REMATCH
 echo "$BASH_REMATCH[1]$BASH_REMATCH[2]"
fi

declare -ar BASH_REMATCH='([0]="hello world u" [1]="hello" [2]="u")'
hellou

share|improve this answer

answered Apr 14 at 12:05

glenn jackmanglenn jackman

53.9k674115

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  just curious to understand to know, why is declare -p needed here? The array is already populated right?
  
  – Inian
  Apr 15 at 5:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  It's not needed at all. Merely to show the contents of the array, to see the results of the regex match with capturing parentheses.
  
  – glenn jackman
  Apr 15 at 10:42
  
  

  
  
  
  

add a comment | 

Using bash:

text="hello world unix"
if [[ $text =~ ^([^[:space:]]+).*[[:space:]]([^[:space:]]) ]]; then
 declare -p BASH_REMATCH
 echo "$BASH_REMATCH[1]$BASH_REMATCH[2]"
fi

declare -ar BASH_REMATCH='([0]="hello world u" [1]="hello" [2]="u")'
hellou

share|improve this answer

answered Apr 14 at 12:05

glenn jackmanglenn jackman

53.9k674115

text="hello world unix"
if [[ $text =~ ^([^[:space:]]+).*[[:space:]]([^[:space:]]) ]]; then
 declare -p BASH_REMATCH
 echo "$BASH_REMATCH[1]$BASH_REMATCH[2]"
fi

declare -ar BASH_REMATCH='([0]="hello world u" [1]="hello" [2]="u")'
hellou

share|improve this answer

answered Apr 14 at 12:05

glenn jackmanglenn jackman

53.9k674115

answered Apr 14 at 12:05

glenn jackmanglenn jackman

53.9k674115

answered Apr 14 at 12:05

glenn jackmanglenn jackman

53.9k674115