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Copy digits from end of string to another

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0

How could I copy digits from one end of a string to another end of a string? So example,

Input -

Example123:Hello

Exp12:Hey1

Exp:heylo

expected output -

Example123:Hello123

Exp12:Hey112

Exp:heylo

I'm open to using sed or awk, seperator must be accounted for, so row1 is the row to extract digits from and row 2 is the row to place digits

awk sed

share|improve this question

edited 1 hour ago

Sparhawk

10.7k848102

asked 2 hours ago

user357075user357075

1

New contributor

user357075 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  From your expected output, it seems you want to copy, not swap those digits from the end of the first field to the end of the line.
  
  – Stéphane Chazelas
  1 hour ago
  

  
  
  
  

add a comment | 

0

How could I copy digits from one end of a string to another end of a string? So example,

Input -

Example123:Hello

Exp12:Hey1

Exp:heylo

expected output -

Example123:Hello123

Exp12:Hey112

Exp:heylo

I'm open to using sed or awk, seperator must be accounted for, so row1 is the row to extract digits from and row 2 is the row to place digits

awk sed

share|improve this question

edited 1 hour ago

Sparhawk

10.7k848102

asked 2 hours ago

user357075user357075

1

New contributor

user357075 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  From your expected output, it seems you want to copy, not swap those digits from the end of the first field to the end of the line.
  
  – Stéphane Chazelas
  1 hour ago
  

  
  
  
  

add a comment | 

How could I copy digits from one end of a string to another end of a string? So example,

Input -

Example123:Hello

Exp12:Hey1

Exp:heylo

expected output -

Example123:Hello123

Exp12:Hey112

Exp:heylo

I'm open to using sed or awk, seperator must be accounted for, so row1 is the row to extract digits from and row 2 is the row to place digits

awk sed

share|improve this question

edited 1 hour ago

Sparhawk

10.7k848102

asked 2 hours ago

user357075user357075

1

New contributor

user357075 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

Example123:Hello

Exp12:Hey1

Exp:heylo

Example123:Hello123

Exp12:Hey112

Exp:heylo

share|improve this question

edited 1 hour ago

Sparhawk

10.7k848102

asked 2 hours ago

user357075user357075

1

New contributor

user357075 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this question

edited 1 hour ago

Sparhawk

10.7k848102

asked 2 hours ago

user357075user357075

1

New contributor

user357075 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited 1 hour ago

Sparhawk

10.7k848102

edited 1 hour ago

Sparhawk

10.7k848102

asked 2 hours ago

user357075user357075

1

New contributor

user357075 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 2 hours ago

user357075user357075

1

3 Answers
3

active

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3

Assuming there's one and only one occurrence of : in each line of the input, you could do something like:

sed 's/([[:digit:]]*):.*/&1/' < input

If there can be more than one : (and you want to append the digits to the end of the line, not the second field), that becomes more complicated, like:

sed 's/^([^:]*[^:[:digit:]]){0,1}([[:digit:]]*):.*/&2/' < input

share|improve this answer

edited 1 hour ago

answered 1 hour ago

Stéphane ChazelasStéphane Chazelas

321k57613983

add a comment | 

1

You can use awk here

awk -F':' '{ j=$1; gsub(/[^0-9]+/,"",j); printf "%s%sn",$0,j;}' 

Explanation

-F':': tells awk to use colon as the separator

j=$1; gsub(/[^0-9]+/,"",j) : assigns the first column to to a temporary variable j, and then removes anything that is not a digit from j

printf "%s%sn",$0,j;}' : finally prints the original string appended with j that carries our number

share|improve this answer

answered 1 hour ago

amisaxamisax

1,636615

add a comment | 

1

With the match() function of GNU awk to store the matching group in an array, you could do

awk -F: -v OFS=: 'match($1, /([0-9]+)$/ , arr) { $2 = $2 arr[1] } 1' file

On any POSIX compliant awk you could do

awk -F: -v OFS=: 'match($1, /[[:digit:]]+$/) { $2 = $2 substr($0, RSTART, RLENGTH) }; 1' file

See How to permanently change a file using awk? ("in-place" edits, as with "sed -i") to make the file change persistent.

share|improve this answer

edited 1 hour ago

Stéphane Chazelas

321k57613983

answered 1 hour ago

InianInian

6,4401734

add a comment | 

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3

Assuming there's one and only one occurrence of : in each line of the input, you could do something like:

sed 's/([[:digit:]]*):.*/&1/' < input

If there can be more than one : (and you want to append the digits to the end of the line, not the second field), that becomes more complicated, like:

sed 's/^([^:]*[^:[:digit:]]){0,1}([[:digit:]]*):.*/&2/' < input

share|improve this answer

edited 1 hour ago

answered 1 hour ago

Stéphane ChazelasStéphane Chazelas

321k57613983

add a comment | 

3

Assuming there's one and only one occurrence of : in each line of the input, you could do something like:

sed 's/([[:digit:]]*):.*/&1/' < input

If there can be more than one : (and you want to append the digits to the end of the line, not the second field), that becomes more complicated, like:

sed 's/^([^:]*[^:[:digit:]]){0,1}([[:digit:]]*):.*/&2/' < input

share|improve this answer

edited 1 hour ago

answered 1 hour ago

Stéphane ChazelasStéphane Chazelas

321k57613983

add a comment | 

Assuming there's one and only one occurrence of : in each line of the input, you could do something like:

sed 's/([[:digit:]]*):.*/&1/' < input

If there can be more than one : (and you want to append the digits to the end of the line, not the second field), that becomes more complicated, like:

sed 's/^([^:]*[^:[:digit:]]){0,1}([[:digit:]]*):.*/&2/' < input

share|improve this answer

edited 1 hour ago

answered 1 hour ago

Stéphane ChazelasStéphane Chazelas

321k57613983

sed 's/([[:digit:]]*):.*/&1/' < input

sed 's/^([^:]*[^:[:digit:]]){0,1}([[:digit:]]*):.*/&2/' < input

share|improve this answer

edited 1 hour ago

answered 1 hour ago

Stéphane ChazelasStéphane Chazelas

321k57613983

answered 1 hour ago

Stéphane ChazelasStéphane Chazelas

321k57613983

answered 1 hour ago

Stéphane ChazelasStéphane Chazelas

321k57613983

1

You can use awk here

awk -F':' '{ j=$1; gsub(/[^0-9]+/,"",j); printf "%s%sn",$0,j;}' 

Explanation

-F':': tells awk to use colon as the separator

j=$1; gsub(/[^0-9]+/,"",j) : assigns the first column to to a temporary variable j, and then removes anything that is not a digit from j

printf "%s%sn",$0,j;}' : finally prints the original string appended with j that carries our number

share|improve this answer

answered 1 hour ago

amisaxamisax

1,636615

add a comment | 

1

You can use awk here

awk -F':' '{ j=$1; gsub(/[^0-9]+/,"",j); printf "%s%sn",$0,j;}' 

Explanation

-F':': tells awk to use colon as the separator

j=$1; gsub(/[^0-9]+/,"",j) : assigns the first column to to a temporary variable j, and then removes anything that is not a digit from j

printf "%s%sn",$0,j;}' : finally prints the original string appended with j that carries our number

share|improve this answer

answered 1 hour ago

amisaxamisax

1,636615

add a comment | 

You can use awk here

awk -F':' '{ j=$1; gsub(/[^0-9]+/,"",j); printf "%s%sn",$0,j;}' 

Explanation

-F':': tells awk to use colon as the separator

j=$1; gsub(/[^0-9]+/,"",j) : assigns the first column to to a temporary variable j, and then removes anything that is not a digit from j

printf "%s%sn",$0,j;}' : finally prints the original string appended with j that carries our number

share|improve this answer

answered 1 hour ago

amisaxamisax

1,636615

awk -F':' '{ j=$1; gsub(/[^0-9]+/,"",j); printf "%s%sn",$0,j;}' 

share|improve this answer

answered 1 hour ago

amisaxamisax

1,636615

answered 1 hour ago

amisaxamisax

1,636615

answered 1 hour ago

amisaxamisax

1,636615

1

With the match() function of GNU awk to store the matching group in an array, you could do

awk -F: -v OFS=: 'match($1, /([0-9]+)$/ , arr) { $2 = $2 arr[1] } 1' file

On any POSIX compliant awk you could do

awk -F: -v OFS=: 'match($1, /[[:digit:]]+$/) { $2 = $2 substr($0, RSTART, RLENGTH) }; 1' file

See How to permanently change a file using awk? ("in-place" edits, as with "sed -i") to make the file change persistent.

share|improve this answer

edited 1 hour ago

Stéphane Chazelas

321k57613983

answered 1 hour ago

InianInian

6,4401734

add a comment | 

1

With the match() function of GNU awk to store the matching group in an array, you could do

awk -F: -v OFS=: 'match($1, /([0-9]+)$/ , arr) { $2 = $2 arr[1] } 1' file

On any POSIX compliant awk you could do

awk -F: -v OFS=: 'match($1, /[[:digit:]]+$/) { $2 = $2 substr($0, RSTART, RLENGTH) }; 1' file

See How to permanently change a file using awk? ("in-place" edits, as with "sed -i") to make the file change persistent.

share|improve this answer

edited 1 hour ago

Stéphane Chazelas

321k57613983

answered 1 hour ago

InianInian

6,4401734

add a comment | 

With the match() function of GNU awk to store the matching group in an array, you could do

awk -F: -v OFS=: 'match($1, /([0-9]+)$/ , arr) { $2 = $2 arr[1] } 1' file

On any POSIX compliant awk you could do

awk -F: -v OFS=: 'match($1, /[[:digit:]]+$/) { $2 = $2 substr($0, RSTART, RLENGTH) }; 1' file

See How to permanently change a file using awk? ("in-place" edits, as with "sed -i") to make the file change persistent.

share|improve this answer

edited 1 hour ago

Stéphane Chazelas

321k57613983

answered 1 hour ago

InianInian

6,4401734

awk -F: -v OFS=: 'match($1, /([0-9]+)$/ , arr) { $2 = $2 arr[1] } 1' file

awk -F: -v OFS=: 'match($1, /[[:digit:]]+$/) { $2 = $2 substr($0, RSTART, RLENGTH) }; 1' file

share|improve this answer

edited 1 hour ago

Stéphane Chazelas

321k57613983

answered 1 hour ago

InianInian

6,4401734

edited 1 hour ago

Stéphane Chazelas

321k57613983

edited 1 hour ago

Stéphane Chazelas

321k57613983

answered 1 hour ago

InianInian

6,4401734

answered 1 hour ago

InianInian

6,4401734