CODEWITHSUNDEEP
arraysbashelement
harnen
harnen

Reputation: 403

Deleting number only elements from an array/string in bash

I have problem with deleting digit only elements from my string/array (conversion is easy) in bash. The trick is, in the array I have elements containing both digits and other character and I want to keep them.

So for

VAR="a2b a22 12b  417 900 600 86400 3600"

The output should be

"a2b a22 12b"

The furthest I could go is:

echo ${VAR}# |  sed 's/ [0-9][0-9]*[$ ]/ /g'

but it still doesn't solve the problem. I tried to do it in an array, but without "$" and "^" I'm not able to prevent deletion of some parts of "good elements".

Can anyone help me with that?

Upvotes: 1

Views: 24

Answers (1)

John1024
John1024

Reputation: 113924

There are two issues with your code. One is that [$ ] will match a literal dollar sign but does, as one might hope, not match the end-of-the-line. The other is that, while g indicates global matching, the matches are not allowed to overlap which would be needed for it to work as you want.

If you have GNU sed, then a simple solution is to avoid matching the spaces and instead use \< and \> to mark word boundaries:

$ echo ${VAR} |  sed -E 's/\<[0-9][0-9]*\>/ /g'
a2b a22 12b

Alternatively, without the GNU extensions, you can use looping:

$ echo ${VAR} |  sed 's/^[0-9][0-9]* / /; :a; s/ [0-9][0-9]* / /g; t a; s/ [0-9][0-9]*$/ /'
a2b a22 12b

The code :a indicates a label. The code t a indicates a test. If the previous substitute command did make a substitution, then the sed jumps to label a.

(The above was tested under GNU sed. It should work with BSD/OSX sed with only minor adjustments.)

Upvotes: 3

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