parse and expand interval

Question

In my script I need to expand an interval, e.g.:

input: 1,5-7

to get something like the following:

output: 1,5,6,7

I've found other solutions here, but they involve python and I can't use it in my script.

Can there be multiple ranges in the input like `2,8-10,12-14` or just a single one ? — Tuxdude, Mar 09 '13 at 17:49
Surely it would be better, but also single it's good enough! — Tom, Mar 09 '13 at 17:51

score 2 · Accepted Answer · answered Mar 09 '13 at 18:08

Solution with Just Bash 4 Builtins

You can use Bash range expansions. For example, assuming you've already parsed your input you can perform a series of successive operations to transform your range into a comma-separated series. For example:

value1=1
value2='5-7'
value2=${value2/-/..}
value2=`eval echo {$value2}`
echo "input: $value1,${value2// /,}"

All the usual caveats about the dangers of eval apply, and you'd definitely be better off solving this problem in Perl, Ruby, Python, or AWK. If you can't or won't, then you should at least consider including some pipeline tools like tr or sed in your conversions to avoid the need for eval.

Ansgar Wiechers · Answer 2 · 2015-04-30T16:07:13.647

2

Try something like this:

#!/bin/bash

for f in ${1//,/ }; do
  if [[ $f =~ - ]]; then
    a+=( $(seq ${f%-*} 1 ${f#*-}) )
  else
    a+=( $f )
  fi  
done

a=${a[*]}
a=${a// /,}

echo $a

Edit: As @Maxim_united mentioned in the comments, appending might be preferable to re-creating the array over and over again.

edited Apr 30 '15 at 16:07

answered Mar 09 '13 at 18:12

Ansgar Wiechers

193,178
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1

I think using `a+=( )` would be more efficient, no? – Maxim_united Apr 30 '15 at 14:55

score 1 · Answer 3 · answered Mar 09 '13 at 18:24

This should work with multiple ranges too.

#! /bin/bash
input="1,5-7,13-18,22"
result_str=""
for num in $(tr ',' ' ' <<< "$input"); do
    if [[ "$num" == *-* ]]; then
        res=$(seq -s ',' $(sed -n 's#\([0-9]\+\)-\([0-9]\+\).*#\1 \2#p' <<< "$num"))
    else
        res="$num"
    fi
    result_str="$result_str,$res"
done
echo ${result_str:1}

Will produce the following output:

1,5,6,7,13,14,15,16,17,18,22

score 0 · Answer 4 · answered Mar 09 '13 at 18:02

0

expand_commas()
{
    local arg
    local st en i

    set -- ${1//,/ }
    for arg
    do
        case $arg in
        [0-9]*-[0-9]*)
            st=${arg%-*}
            en=${arg#*-}
            for ((i = st; i <= en; i++))
            do
                echo $i
            done
        ;;
        *)
            echo $arg
        ;;
        esac
    done
}

Usage:

result=$(expand_commas arg)

eg:

result=$(expand_commas 1,5-7,9-12,3)
echo $result

You'll have to turn the separated words back into commas, of course.

It's a bit fragile with bad inputs but it's entirely in bash.

answered Mar 09 '13 at 18:02

torek

448,244
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It works! I've piped to this function | uniq | sort -n to remove duplicates and sort the output. Thank you! – Tom Mar 09 '13 at 18:08
If you have bash v4, combine with @CodeGnome's trick to eliminate the loop (the output will have range values on one line though). – torek Mar 09 '13 at 18:16

score 0 · Answer 5 · answered Mar 09 '13 at 18:09

Here's my stab at it:

input=1,5-7,10,17-20
IFS=, read -a chunks <<< "$input"

output=()
for chunk in "${chunks[@]}"
do
    IFS=- read -a args <<< "$chunk"
    if (( ${#args[@]} == 1 )) # single number                                   
    then
        output+=(${args[*]})
    else # range                                                                
        output+=($(seq "${args[@]}"))
    fi
done
joined=$(sed -e 's/ /,/g' <<< "${output[*]}")
echo $joined

Basically split on commas, then interpret each piece. Then join back together with commas at the end.

score 0 · Answer 6 · edited Jun 20 '20 at 09:12

A generic bash solution using the sequence expression `{x..y}'

#!/bin/bash
function doIt() {
  local inp="${@/,/ }"
  declare -a args=( $(echo ${inp/-/..}) )
  local item
  local sep
  for item in "${args[@]}"
  do
    case ${item} in
        *..*)  eval "for i in {${item}} ; do echo -n \${sep}\${i}; sep=, ; done";;
        *) echo -n ${sep}${item};;
    esac
    sep=,
  done
}
doIt "1,5-7"

Should work with any input following the sample in the question. Also with multiple occurrences of x-y
Use only bash builtins

score 0 · Answer 7 · answered Apr 30 '15 at 15:27

Using ideas from both @Ansgar Wiechers and @CodeGnome:

input="1,5-7,13-18,22"
for s in ${input//,/ }
do
    if [[ $f =~ - ]]
    then
        a+=( $(eval echo {${s//-/..}}) )
    else
        a+=( $s )
    fi  
done
oldIFS=$IFS; IFS=$','; echo "${a[*]}"; IFS=$oldIFS

Works in Bash 3

score 0 · Answer 8 · answered Aug 23 '16 at 12:29

Considering all the other answers, I came up with this solution, which does not use any sub-shells (but one call to eval for brace expansion) or separate processes:

# range list is assumed to be in $1 (e.g. 1-3,5,9-13)

# convert $1 to an array of ranges ("1-3" "5" "9-13")
IFS=,
local range=($1)
unset IFS

list=() # initialize result list
local r
for r in "${range[@]}"; do
    if [[ $r == *-* ]]; then
        # if the range is of the form "x-y",
        # * convert to a brace expression "{x..y}",
        # * using eval, this gets expanded to "x" "x+1" … "y" and
        # * append this to the list array
        eval list+=( {${r/-/..}} )
    else
        # otherwise, it is a simple number and can be appended to the array
        list+=($r)
    fi
done

# test output
echo ${list[@]}

parse and expand interval

8 Answers8

Solution with Just Bash 4 Builtins

A generic bash solution using the sequence expression `{x..y}'