Find and replace in same string in bash

Question

I have a string with some words in them, example a=1 b=2 c=3 a=50. Now I want to parse this and create another string a=50 b=2 c=3 which is essentially the same as above except that if the same phrase before the = is encountered for the second time the first one is over written with the latest one, so in the end there are only unique phrases on the left of =. Here is what I got till now:

a="a=1 b=2 c=3 a=50"
o=()

for i in $a
do
    reg=${i%=*}
    if [[ ${o[*]} == *"$reg"* ]]
    then
        o=$(echo ${o[*]} | sed -e "s/\$reg=\S/\$i")
    else
        o+=( $i )
    fi
done

What am I doing wrong here?

Use an associative array whose keys are the words before =. — Barmar
– Barmar, Commented Dec 19, 2014 at 22:28

Charles Duffy · Accepted Answer · 2014-12-19 22:57:11Z

3

I'd take an entirely different approach, not based on regular expressions or string rewriting.

declare -A values=( )              # Initialize an associative array ("hash", "map")
while IFS= read -r -d' ' word; do  # iterate over input words, separated by spaces
  if [[ $word = *=* ]]; then       # ignore any word that doesn't have an "=" in it
    values[${word%%=*}]=${word#*=} # add everything before the "=" as a key...
  fi                               # ...with everything after the "=" as a value
done

for key in "${!values[@]}"; do     # Then iterate over keys we found
  value="${values[$key]}"          # ...extract the values for each...
  printf '%s=%s ' "$key" "$value"  # ...and print the pairs.
done
echo                               # When done iterating, print a newline.

Because the words are being processed first-to-last through the string, updates take effect before the print loop is reached.

edited Dec 19, 2014 at 22:57

answered Dec 19, 2014 at 22:33

Charles Duffy

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John1024 · Accepted Answer · 2014-12-19 23:30:19Z

3

Using `awk`

$ awk -F= -v RS=" |\n" '{a[$1]=$2} END{for (k in a) printf "%s=%s ",k,a[k]}' <<<"a=1 b=2 c=3 a=50"
a=50 b=2 c=3

How it works:

-F=

Set the field separator to be an equal sign.
-v RS=" |\n"

Set the record separator to be either a space or a newline.
a[$1]=$2

Update associative array a with the latest value.
END{for (k in a) printf "%s=%s ",k,a[k]}

In no particular order, print out the final values.

Using `bash`

Like Charles Duffy's approach, this uses read -d" " to parse the string. This approach, however, uses IFS="=" to separate names and values.

Two loops are required. The first gathers the values. The second reassembles the updated values in the original order:

a="a=1 b=2 c=3 a=50"
declare -A b
while IFS== read -d" " name value
do
    b["$name"]="$value"
done <<<"$a "

declare -A seen
while IFS== read -d" " name value
do
    [ "${seen[$name]}" ] || o="$o $name=${b["$name"]}"
    seen[$name]=1
done <<<"$a "
echo "$o"

edited Dec 19, 2014 at 23:30

answered Dec 19, 2014 at 22:51

John1024

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1 Comment

Charles Duffy Over a year ago

Nice. The only quibble I have here is that iterating through the string a second time, it reads to me like you could be printing keys twice. Perhaps use a second associative array to track what has and hasn't been printed to avoid that? Retaining order is a nice feature (and one I paid no attention to).

Charles Duffy · Accepted Answer · 2014-12-22 16:38:33Z

1

Easily done with perl:

echo "a=1 b=2 c=3 a=50" \
  | sed "s/ /\n/g" \
  | perl -e '
my %hash = ();
while(<>){
  $line = $_;
  if($line =~ m/(\S+)=(\S+)/) {
    $hash{$1} = $2;
  }
}
for $key (sort keys %hash) {
  print "$key=$hash{$key}\n";
}'

...or, all on one line:

echo "a=1 b=2 c=3 a=50" | sed "s/ /\n/g" | perl -e 'my %hash = (); while(<>){ $line = $_; if($line =~ m/(\S+)=(\S+)/) { $hash{$1} = $2; } } for $key (sort keys %hash) { print "$key=$hash{$key}\n"; }'

edited Dec 22, 2014 at 16:38

Charles Duffy

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answered Dec 19, 2014 at 23:03

Amit Anand

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