Sort 80,412 files into folders based on name begins with

Question

Situation:

• a folder with 80,412 files
• all files named with CamelCase, some have a dash (-) separating the first word
• files can be grouped by beginning with the same word or letters(i.e. "ClientslistRebecca..." and "Clientslist-Adam...")
• no file name begins with a number
• very few files begin with a unique word

I need to take the first word, find all files that begin with that word, and create a folder named with that word. Repeat for all files.

The main complication is that sometimes a name begins with multiple capitals, followed by a word that also begins with a capital (i.e. "AWDRebecca..."). Not sure how to resolve that, but I'm fine with doing some manual cleanup afterwards if I get folders with "AWDRebecca", "AWDReports", "AWDNovember", but having a folder "A" will make it harder.

Answer 1

A shell snippet like

for i in *; do
  case "$i" in
    [A-Z][A-Z]*) # $i has several leading caps
        dir=$(sed -E -e 's|([A-Z]+)[A-Z].*|\1|' <<< "$i")
        echo mkdir -p "$dir"
        echo mv "$i" "$dir"/
        ;;
    [A-Z][a-z]*) # $i has one leading cap
        dir=$(sed -E -e 's|([A-Z][a-z]+).*|\1|' <<< "$i")
        echo mkdir -p "$dir"
        echo mv "$i" "$dir"/
        ;;
    *) echo "Huh: $i"
       ;;
  esac
done

should work. As written it will just print the commands on the screen. If they look sensible, remove the four echo before mkdir/mv.

Also, for 80'000 files, the for i in * part may fail. If this is the case, run it in batches with for i in [A-E]*, for i in [F-K]* etc.

PS: It will fail for files with a ALLCAPS name. If you have does, please comment so I can add another rule.