Using sed to replace numbers with '@' inserts '@' between every character

Question

I want to replaced all number with '@' symbol. I am using the below sed command , but not getting the desired result.

command -

echo "abc 434 pankaj 444" | sed 's/[0-9]*/@/g'

Result -

@a@b@c@ @ @p@a@n@k@a@j@ @

Answer 1

Well, quite simply, [0-9]* matches strings that consist entirely of zero or more digits, include empty strings. Anything that matches an empty string, matches between any two characters, so the replacement @ is added between all letters in the input. Strings of multiple digits are replaced with one @ since the expression matches all consecutive digits at once.

So in the input string ab43 the matches to [0-9]+ are (with some whitespace added for clarity):

    a   b   434
  ^   ^     ^^^- here, a string of some digits
  ^   ^- here, a zero-length string
  ^- here, a zero-length string

Use [0-9] to match exactly one digit, or [0-9][0-9]* to match one or more (or [0-9]+ in extended regular expressions).

Answer 2

Replace each digit with @

echo "abc 434 pankaj 444" | sed 's/[0-9]/\@/g'
#Output: abc @@@ pankaj @@@

or replace each number with @

echo "abc 434 pankaj 444" | sed 's/[0-9]\+/\@/g'
#Output: abc @ pankaj @

depending on the desired output you need

Answer 3

echo "abc 434 pankaj 444"| tr "[0-9]" "@"

abc @@@ pankaj @@@

Answer 4

Replace * with any special character can be achieved by using tr command.

#!/bin/bash
mystring="This is my number: 12345"
echo $mystring | sed 's/[0-9]/*/g'

Answer 5

Depending on whether you want to replace digits or whole numbers (runs of consecutive digits) with @, you may use the tr command without or with its -s option:

$ echo "abc 434 pankaj 444" | tr '[:digit:]' '[@*]'
abc @@@ pankaj @@@

$ echo "abc 434 pankaj 444" | tr -s '[:digit:]' '[@*]'
abc @ pankaj @

The -s option of tr makes the utility "squeeze" any consecutively occurring character from the second argument found in the data. In this case, using -s makes tr combine runs of @ characters into single @ characters.

The two arguments [:digit:] and [@*] tell tr to replace digits with @ characters. Most (all?) implementations of tr allow you to write @ in place of [@*] as the second argument. The [@*] literally means "as many @ as there are characters in the set described by the first argument".

Note that tr will squeeze any pre-existing runs of consecutive @ characters in the input if -s is in effect.

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Your approach with sed does not work as expected because you insert a @ character wherever there are zero or more matches of a digit. There are zero digits between each character in the input, so the expression matches between each character.

A modified substitution that would work correctly and replace runs of digits with a single @ would be

sed 's/[[:digit:]]\{1,\}/@/g'

By using \{1,\} in place of * (which in itself is the same as \{0,\}) we force at least one match of the preceding expression. You will see \{1,\} written as + in extended regular expressions.

By removing \{1,\} from the expression, leaving the substitution as s/[[:digit:]]/@/g, you replace each individual digit with a @.

Answer 6

You can achieve this by awk command too

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echo "abc 434 pankaj 444" |awk '{gsub("[0-9]","@",$0);print $0}'

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