Re: efficient way to use matched string in variable substitution

[Greg Wooledge]

On Mon, Aug 23, 2021 at 11:36:52AM -0700, L A Walsh wrote:
> Starting with a number N, is there
> an easy way to print its digits into an array?

"Easy"?  Or "efficient"?  Your subject header says one, but your body
says the other.

> arr=(${N//[0-9]/\1 })

> arr=(${N//[0-9]/$1 })

Obviously those don't work.


> >  for x in 0 1 2 3 4 5 6 7 8 9;do n=${n//$x/$x }; done
> >  arr=($n)

That doesn't look particularly elegant to me.  But we can throw it on the
pile.

unicorn:~$ f1() { local x n="$1" arr; for x in 0 1 2 3 4 5 6 7 8 9; do 
n=${n//$x/$x }; done; arr=($n); }
unicorn:~$ time for ((i=1; i<=10000; i++)); do f1 682390; done
real 0.444  user 0.444  sys 0.000


> >  for ((d=0; d<${#n};d+=1)); do arr+=(${n:$d:1}); done

That's the approach I'd start with, with a few minor changes.

unicorn:~$ f2() { local i n="${#1}" arr; for ((i=0; i<n; i++)); do 
arr+=("${1:i:1}"); done; }
unicorn:~$ time for ((i=1; i<=10000; i++)); do f2 682390; done
real 0.413  user 0.403  sys 0.000

Slightly more efficient, it seems.


Just for fun, let's actually treat your "number N" like a number, rather
than a string.

unicorn:~$ f3() { local n="$1" tmp arr i; while ((n > 0)); do 
tmp+=("$((n%10))"); ((n /= 10)); done; for ((i=${#tmp[@]}-1; i >= 0; i--)); do 
arr+=("${tmp[i]}"); done; }
unicorn:~$ time for ((i=1; i<=10000; i++)); do f3 682390; done
real 0.695  user 0.691  sys 0.000

This one is less efficient, no doubt because of the extra time needed
to reverse the temporary array.


If we take steps to avoid that reversal, let's see what happens:

unicorn:~$ f4() { local n="$1" i=${#1} arr; while ((n > 0)); do 
arr[--i]=$((n%10)); ((n /= 10)); done; }
unicorn:~$ time for ((i=1; i<=10000; i++)); do f4 682390; done
real 0.356  user 0.356  sys 0.000

Looks like we have a winner.