More efficient way of finding missing number

Question

I have a variable called DISK_INFO with the following contents:

diskid   HGST     HUSMH8010BSS204  serial             no  no  [0] Slot00
diskid   HGST     HUH728080AL4204  serial             no  no  [0] Slot02
diskid   HGST     HUH728080AL4204  serial             no  no  [0] Slot03
diskid   HGST     HUH728080AL4204  serial             no  no  [0] Slot04
diskid   HGST     HUH728080AL4204  serial             no  no  [0] Slot05
diskid   HGST     HUH728080AL4204  serial             no  no  [0] Slot06
diskid   HGST     HUH728080AL4204  serial             no  no  [0] Slot07
diskid   HGST     HUH728080AL4204  serial             no  no  [0] Slot08
diskid   HGST     HUH728080AL4204  serial             no  no  [0] Slot09
diskid   HGST     HUH728080AL4204  serial             no  no  [0] Slot10
diskid   HGST     HUH728080AL4204  serial             no  no  [0] Slot11
diskid   HGST     HUH728080AL4204  serial             no  no  [0] Slot12
diskid   HGST     HUH728080AL4204  serial             no  no  [0] Slot13
diskid   HGST     HUH728080AL4204  serial             no  no  [0] Slot14
diskid   HGST     HUH728080AL4204  serial             no  no  [0] Slot15
diskid   HGST     HUH728080AL4204  serial             no  no  [0] Slot16
diskid   HGST     HUH728080AL4204  serial             no  no  [0] Slot17
diskid   HGST     HUH728080AL4204  serial             no  no  [0] Slot18
diskid   HGST     HUH728080AL4204  serial             no  no  [0] Slot19
diskid   HGST     HUH728080AL4204  serial             no  no  [0] Slot20
diskid   HGST     HUH728080AL4204  serial             no  no  [0] Slot21
diskid   HGST     HUH728080AL4204  serial             no  no  [0] Slot22
diskid   HGST     HUH728080AL4204  serial             no  no  [0] Slot23
diskid   HGST     HUH728080AL4204  serial             no  no  [1] Slot00
diskid   HGST     HUH728080AL4204  serial             no  no  [1] Slot01
diskid   HGST     HUH728080AL4204  serial             no  no  [1] Slot02
diskid   HGST     HUH728080AL4204  serial             no  no  [1] Slot03
diskid   HGST     HUH728080AL4204  serial             no  no  [1] Slot04
diskid   HGST     HUH728080AL4204  serial             no  no  [1] Slot05
diskid   HGST     HUH728080AL4204  serial             no  no  [1] Slot06
diskid   HGST     HUH728080AL4204  serial             no  no  [1] Slot07
diskid   HGST     HUH728080AL4204  serial             no  no  [1] Slot08
diskid   HGST     HUH728080AL4204  serial             no  no  [1] Slot09
diskid   HGST     HUH728080AL4204  serial             no  no  [1] Slot10
diskid   HGST     HUH728080AL4204  serial             no  no  [1] Slot11
c2t0d0                  Kingston DataTraveler 2.0 -                    -   -   -

When a disk has failed it will be removed from this list, in this example the disk in enclosure 0 Slot 01 has failed.

Assuming enclosure 0 will always have 24 disks 00-23 and enclosure 1 will always have 12 disks 00-11, how can I efficiently and accurately determine the missing disk(s)?

I currently have the following but I'm sure this can be done in a single awk command:

enclosure0=($(awk '$7 ~ "[0]"{print $8}' <<<"$DISK_INFO" | sort -n))
enclosure1=($(awk '$7 ~ "[1]"{print $8}' <<<"$DISK_INFO" | sort -n))
for n in {00..23}; do
  grep -q "$n" <<<"${enclosure0[@]}" || missing+=("Enclosure 0 - Slot$n")
done
for n in {00..11}; do
  grep -q "$n" <<< "${enclosure1[@]}" || missing+=("Enclosure 1 - Slot$n")
done

xenoid · Accepted Answer · 2018-05-10 22:44:33Z

4

Without awk, for each enclosure:

{ printf '[0] Slot%s\n' {00..23} ; grep -Eo '\[0\] Slot..' disks ; } | sort | uniq -u

In slow-mo:

printf '[0] Slot%s\n' {00..23} generates the list of all possible disks
grep -Eo '\[0\] Slot..' disks extracts the existing disks
{..} concatenates the output of the two commands
sort | uniq -u extracts the lines that appear only once

You can replace the printf and grep steps by adequate functions, or the printf part by a similar grep on another file which is the expected list of disks.

answered May 10, 2018 at 22:44

xenoid

9,3384 gold badges29 silver badges52 bronze badges

Excellent approach, uses the GNU philosophy to enable extensibility and customization.

Spidey
– Spidey

2018-05-11 04:43:06 +00:00
Commented May 11, 2018 at 4:43

Add a comment |

Gilles 'SO- stop being evil' · Accepted Answer · 2018-05-10 21:58:48Z

Since you know in advance which items need to exist, create a list and tick them off as you see them.

awk '
    BEGIN {
        for (i = 0; i < 24; i++) missing[0][sprintf("%02d", i)] = 1;
        for (i = 0; i < 12; i++) missing[1][sprintf("%02d", i)] = 1;
    }
    $7 ~ /^\[[0-9]+\]$/ && $8 ~ /^Slot[0-9]+$/ {
        gsub(/[^0-9]/, "", $7);
        sub(/^[^0-9]+/, "", $8);
        delete missing[$7][$8];
    }
    END {
        for (enclosure in missing) {
            for (slot in missing[enclosure]) {
                printf "Missing enclosure %d Slot%s\n", enclosure, slot;
            }
        }
    }
'

Rakesh Sharma · Accepted Answer · 2018-05-12 17:21:14Z

perl -sle '
  my(@e, @AoA) = qw/ 24 12 /;
  $AoA[$1][$2]++ while /\[([01])]\h+(?:(?!\d)\S)+0*(\d+)$/mg;
  for my $enc ( 0 .. $#e ) {
     for my $m_slot ( grep { ! defined $AoA[$enc][$_] } 0 .. $e[$enc]-1 ) {
         print "in enclosure $enc - Slot$m_slot is missing.";
     }
  }
 ' -- -_="$DISK_INFO";

Explanation:

°  Initialize the array @e which holds the number of slots in the various enclosures. 

° The Disk info variable is passed into the command line as $_ initialized to $DISK_INFO. 

°  progressively scan and match the $_ variable using the while loop and look for the numbers in the '[..]' and the 'Slot...'  locations. Using these we update the array of array @AoA, it can be viewed as a matrix. 

°  Now once we have ingested all the data, its time to process it now in two for loops. 

° The outer for loops on the enclosures, in our case, they are two. 

° The inner for loop computes the indices of the current enclosure elements that are undefined, IOW, those slots that were never encountered during the data collection drive in the while loop.

Stack Exchange Network

More efficient way of finding missing number

3 Answers 3

You must log in to answer this question.

Hot Network Questions

More efficient way of finding missing number

3 Answers 3

You must log in to answer this question.

Related

Hot Network Questions