Reputation: 1063
I am trying to convert all negative numbers to positive numbers and have so far come up with this
echo "-32 45 -45 -72" | sed -re 's/\-([0-9])([0-9])\ /\1\2/p'
but it is not working as it outputs:
3245 -45 -72
I thought by using \1\2 I would have got the positive number back ?
Where am I going wrong ?
Upvotes: 2
Views: 10829
Reputation: 21
You are dealing with numbers as with a string of characters. More appropriate would be to store numbers in an array and use built in Shell Parameter Expansion to remove the minus sign:
[~] $ # Creating and array with an arbitrary name:
[~] $ array17=(-32 45 -45 -72)
[~] $ # Calling all elements of the array and removing the first minus sign:
[~] $ echo ${array17[*]/-}
32 45 45 72
[~] $
Upvotes: 2
Reputation: 58381
This might work for you:
echo "-32 45 -45 -72" | sed 's/-\([0-9]\+\)/\1/g'
Reason why your regex is failing is
Your only doing a single substitution (no g
)
Your replacement has no space at the end.
The last number has no space following so it will always fail.
This would work too but less elegantly (and only for 2 digit numbers):
echo "-32 45 -45 -72" | sed -rn 's/-([0-9])([0-9])(\s?)/\1\2\3/gp'
Of course for this example only:
echo "-32 45 -45 -72" | tr -d '-'
Upvotes: 3
Reputation: 14004
My first thought is not using sed, if you don't have to. awk can understand that they're numbers and convert them thusly:
echo "-32 45 -45 -72" | awk -vRS=" " -vORS=" " '{ print ($1 < 0) ? ($1 * -1) : $1 }'
-vRS sets the "record separator" to a space, and -vORS sets the "output record separator" to a space. Then it simply checks each value, sees if it's less than 0, and multiplies it by -1 if it is, and if it's not, just prints the number.
In my opinion, if you don't have to use sed, this is more "correct," since it treats numbers like numbers.
Upvotes: 4
Reputation: 27618
Why not just remove the -'s?
[root@vm ~]# echo "-32 45 -45 -72" | sed 's/-//g'
32 45 45 72
Upvotes: 12