[SOLVED] bash curly braces and normal parenthesis filter words inclusive space to array

blumenwesen · 07-30-2022, 09:46 AM

Hello, try using curly braces and normal parenthesis to remove numbers and first space and read words up to the comma into an array.

Code:

str="a1 word word, a2 word sign, a1b sign word, a2b sign sign"
te2=${str// /.}; arr=(${te2//,./ })
for z in ${arr[@]}; do z=${z#*.}; echo ${z//./ }; done # result

How to separate only from the comma with normal parenthesis?
Is it possible within curly braces common strings continue defining?

Code:

a="a1 word word"
echo ${a#[a-z][0-9]* } # correct
a="a1b2 sign word"
echo ${a#([a-z][0-9]+)* } # wrong=a1b2 sign word / correct=word word

teckk · 07-30-2022, 02:12 PM

Solvable in a number of ways.

Real basic way:

Code:

str="a1 word word, a2 word sign, a1b sign word, a2b sign sign"

arr=()

IFS=',' read -ra A <<< "${str[@]}"
for i in "${A[@]}"; do
    if [[ "$i" = \ *  ]]; then 
        arr+=$(cut -d " " -f 2-4 <<< "${i#?}")
        arr+=" "
    else
        arr+=$(cut -d " " -f 2-4 <<< "$i")
        arr+=" "
    fi
done

echo "${arr[@]}"

MadeInGermany · 07-31-2022, 10:02 AM

Code:

a="a1 word word"
a="a1b2 sign word"

Both words end with a number, and if the following space should be stripped of, too, it becomes

Code:

echo "${a#*[0-9] }"

I don't understand the whole exercise yet. Maybe the followi g makes sense?

Code:

str="a1 word word, a2 word sign, a1b sign word, a2b sign sign"
IFS="," read -ra arr <<< "$str"
arr=( "${arr[@]#*[0-9] }" )

teckk · 07-31-2022, 10:39 AM

Quote:

str="a1 word word, a2 word sign, a1b sign word, a2b sign sign"
IFS="," read -ra arr <<< "$str"
arr=( "${arr[@]#*[0-9] }" )

echo "${arr[@]}"
word word word sign a1b sign word a2b sign sign

This is what I get from that.

I thought that I could tell what the OP wanted from post. Maybe not.

blumenwesen · 07-31-2022, 12:37 PM

thank you

NevemTeve · 07-31-2022, 02:18 PM

Is it solved now? If not, please provide some <input,expected output> pairs.

MadeInGermany · 08-01-2022, 12:38 AM

Still guessing what output is required.

The following variant removes the first word from each array member. (Delete from the left, up to the first non-space and space.)

Code:

str="a1 word word, a2 word sign, a1b sign word, a2b sign sign"
arr=( "${arr[@]#*[! ] }" )

The modifier works on each array member. (Saving a loop.)

Print the array members newline-separated:

Code:

printf "%s\n" "${arr[@]}"

The echo prints its arguments space-separated

Code:

echo "${arr[@]}t"

blumenwesen · 08-01-2022, 09:58 AM

yes solved, are two direct ways.

Code:

IFS="," read -ra arr <<< "a1 word word, a2 word sign, a1b sign word, a2b sign sign"
arr=( "${arr[@]#*[! ] }" ) # or arr=( "${arr[@]#*[a-z0-9] }" )
echo ${arr[0]}
echo ${arr[1]}
echo ${arr[2]}
echo ${arr[3]}

MadeInGermany · 08-08-2022, 02:12 AM

Note that

Code:

arr=( "${arr[@]#*[! ] }" )

is an operation on all the (quoted=protected!) array members, without a loop.
The operation here is the # modifier: strip off characters from the left until a not-space followed by a space.
Likewise you can print all the (quoted!) array members:

Code:

orintf "%s\n" "${arr[@]}"

A loop would need another array:

Code:

arr2=()
for a in "${arr[@]}"
do
  arr2+=( "${a#*[! ] }" )
done

Three methods to loop over an array:

Code:

# 1. Loop over the member values
for a in "${arr2[@]}"
do
  echo "$a"
done
echo
# 2a. Loop over the indexes and get the values
for i in ${!arr2[@]}
do
  echo "$[arr2[$i]}"
done
echo
# 2b. Loop over the indexes and get the values
for (( i=0; i < ${#arr2[@]}; i++ ))
do
  echo "$[arr2[$i]}"
done