Group files into average size of the largest file












0















I have 6 files and would like to group them by 2 or 3 according to the average size.



file1.log 50G
file2.log 40G
file3.log 20G
file4.log 10G
file5.log 30G
file6.log 70G


File6 is 70G is the biggest file and I would like to group the rest of the files according to the biggest.



The output should look like this:




  1. Group by 1 should contain all the files - Parallel 1

  2. Group by 2 - Parallel 2


Output 1



file4.log 10G
file5.log 30G
file6.log 70G


Output 2



file1.log 50G
file2.log 40G
file3.log 20G


Notice average is both files are equals.



The third group parallel 3 should look like this:



output 1



file6.log 70G


output2



file1.log 50G
file3.log 20G


output3



file2.log 40G
file4.log 10G
file5.log 30G


It does not have to be the exact average, just divide the file the closest average possible.



Thanks!!










share|improve this question




















  • 1





    What have you tried?

    – Peschke
    Nov 27 '18 at 7:09











  • Here is what I have tried.

    – Kwa Arboncana
    Nov 27 '18 at 13:39











  • I don't see any code or research in your question. You may get better answers if you show what you have tried, rather than asking us to write code for you.

    – Peschke
    Nov 27 '18 at 16:22
















0















I have 6 files and would like to group them by 2 or 3 according to the average size.



file1.log 50G
file2.log 40G
file3.log 20G
file4.log 10G
file5.log 30G
file6.log 70G


File6 is 70G is the biggest file and I would like to group the rest of the files according to the biggest.



The output should look like this:




  1. Group by 1 should contain all the files - Parallel 1

  2. Group by 2 - Parallel 2


Output 1



file4.log 10G
file5.log 30G
file6.log 70G


Output 2



file1.log 50G
file2.log 40G
file3.log 20G


Notice average is both files are equals.



The third group parallel 3 should look like this:



output 1



file6.log 70G


output2



file1.log 50G
file3.log 20G


output3



file2.log 40G
file4.log 10G
file5.log 30G


It does not have to be the exact average, just divide the file the closest average possible.



Thanks!!










share|improve this question




















  • 1





    What have you tried?

    – Peschke
    Nov 27 '18 at 7:09











  • Here is what I have tried.

    – Kwa Arboncana
    Nov 27 '18 at 13:39











  • I don't see any code or research in your question. You may get better answers if you show what you have tried, rather than asking us to write code for you.

    – Peschke
    Nov 27 '18 at 16:22














0












0








0








I have 6 files and would like to group them by 2 or 3 according to the average size.



file1.log 50G
file2.log 40G
file3.log 20G
file4.log 10G
file5.log 30G
file6.log 70G


File6 is 70G is the biggest file and I would like to group the rest of the files according to the biggest.



The output should look like this:




  1. Group by 1 should contain all the files - Parallel 1

  2. Group by 2 - Parallel 2


Output 1



file4.log 10G
file5.log 30G
file6.log 70G


Output 2



file1.log 50G
file2.log 40G
file3.log 20G


Notice average is both files are equals.



The third group parallel 3 should look like this:



output 1



file6.log 70G


output2



file1.log 50G
file3.log 20G


output3



file2.log 40G
file4.log 10G
file5.log 30G


It does not have to be the exact average, just divide the file the closest average possible.



Thanks!!










share|improve this question
















I have 6 files and would like to group them by 2 or 3 according to the average size.



file1.log 50G
file2.log 40G
file3.log 20G
file4.log 10G
file5.log 30G
file6.log 70G


File6 is 70G is the biggest file and I would like to group the rest of the files according to the biggest.



The output should look like this:




  1. Group by 1 should contain all the files - Parallel 1

  2. Group by 2 - Parallel 2


Output 1



file4.log 10G
file5.log 30G
file6.log 70G


Output 2



file1.log 50G
file2.log 40G
file3.log 20G


Notice average is both files are equals.



The third group parallel 3 should look like this:



output 1



file6.log 70G


output2



file1.log 50G
file3.log 20G


output3



file2.log 40G
file4.log 10G
file5.log 30G


It does not have to be the exact average, just divide the file the closest average possible.



Thanks!!







shell-script awk sed






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 27 '18 at 6:12









Inian

4,015824




4,015824










asked Nov 27 '18 at 6:04









Kwa ArboncanaKwa Arboncana

53




53








  • 1





    What have you tried?

    – Peschke
    Nov 27 '18 at 7:09











  • Here is what I have tried.

    – Kwa Arboncana
    Nov 27 '18 at 13:39











  • I don't see any code or research in your question. You may get better answers if you show what you have tried, rather than asking us to write code for you.

    – Peschke
    Nov 27 '18 at 16:22














  • 1





    What have you tried?

    – Peschke
    Nov 27 '18 at 7:09











  • Here is what I have tried.

    – Kwa Arboncana
    Nov 27 '18 at 13:39











  • I don't see any code or research in your question. You may get better answers if you show what you have tried, rather than asking us to write code for you.

    – Peschke
    Nov 27 '18 at 16:22








1




1





What have you tried?

– Peschke
Nov 27 '18 at 7:09





What have you tried?

– Peschke
Nov 27 '18 at 7:09













Here is what I have tried.

– Kwa Arboncana
Nov 27 '18 at 13:39





Here is what I have tried.

– Kwa Arboncana
Nov 27 '18 at 13:39













I don't see any code or research in your question. You may get better answers if you show what you have tried, rather than asking us to write code for you.

– Peschke
Nov 27 '18 at 16:22





I don't see any code or research in your question. You may get better answers if you show what you have tried, rather than asking us to write code for you.

– Peschke
Nov 27 '18 at 16:22










2 Answers
2






active

oldest

votes


















0
















#!/usr/bin/env zsh

# To care about hidden filenames:
#setopt GLOB_DOTS

# Load the zstat builtin
zmodload -F zsh/stat b:zstat

# Get the regular files in the current directory,
# ordered by size (largest first)
files=( ./*(.OL) )

# Precalculate the filesizes
typeset -A filesizes
for file in "${files[@]}"; do
filesizes[$file]=$( zstat +size "$file" )
done

# The maximum size of a bin is the size of the largest file
maxsize=${filesizes[${files[1]}]}

binsizes=()
typeset -A filebins
for file in "${files[@]}"; do
filesize=${filesizes[$file]}
bin=1 # try fitting into first bin first
ok=0 # haven't yet found a bin for this file
for binsize in "${binsizes[@]}"; do
if (( filesize + binsize <= maxsize )); then
# File fits in this bin,
# update bin size and place file in bin
binsizes[$bin]=$(( filesize + binsize ))
filebins[$file]=$bin
ok=1 # now we're good
break
fi
# Try next bin
bin=$(( bin + 1 ))
done

if [ "$ok" -eq 0 ]; then
# Wasn't able to fit file in existing bin,
# create new bin
binsizes+=( "$filesize" )
filebins[$file]=${#binsizes[@]}
fi
done

# Do final output
printf 'Bin max size = %dn' "$maxsize"
for file in "${files[@]}"; do
printf '%d: %s (file size=%d / bin size=%d)n' "${filebins[$file]}" "$file"
"${filesizes[$file]}" "${binsizes[$filebins[$file]]}"
done | sort -n


The above zsh shell script does binning of all the files in the current directory with a maximum bin size based strictly on the size of the largest file. It implements a first-fit algorithm with the files ordered by decreasing size. This is what's called the "FFD" algorithm in the "Bin packing problem" Wikipedia article. The "MFFD" algorithm is non-trivial to implement in zsh in less than 200 or so lines of code, so I won't post it here.



Testing:



$ ls -l
total 450816
-rw-r--r-- 1 kk wheel 10485760 Jan 19 23:53 file-10.log
-rw-r--r-- 1 kk wheel 20971520 Jan 19 23:53 file-20.log
-rw-r--r-- 1 kk wheel 31457280 Jan 19 23:53 file-30.log
-rw-r--r-- 1 kk wheel 41943040 Jan 19 23:53 file-40.log
-rw-r--r-- 1 kk wheel 52428800 Jan 19 23:53 file-50.log
-rw-r--r-- 1 kk wheel 73400320 Jan 19 23:53 file-70.log




$ zsh ../script.sh
Bin max size = 73400320
1: ./file-70.log (file size=73400320 / bin size=73400320)
2: ./file-20.log (file size=20971520 / bin size=73400320)
2: ./file-50.log (file size=52428800 / bin size=73400320)
3: ./file-30.log (file size=31457280 / bin size=73400320)
3: ./file-40.log (file size=41943040 / bin size=73400320)
4: ./file-10.log (file size=10485760 / bin size=10485760)


The number at the start of each line above corresponds to the bin number assigned to the file.






share|improve this answer

































    0














    This seems to be pretty much equivalent to the Bin Packing problem.



    The Bin Packing problem is NP-hard, so there is no known shortcut to doing it, brute force (trying all the options in some sensible order that excludes silly attempts, like adding more files to an already oversized group) is the way to go.



    For six files, the brute force approach should be simple enough to do by hand; just list all the possible groupings, count how they split the file usage, and choose the one that gives you the smallest maximum group size.






    share|improve this answer























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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

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      active

      oldest

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      active

      oldest

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      0
















      #!/usr/bin/env zsh

      # To care about hidden filenames:
      #setopt GLOB_DOTS

      # Load the zstat builtin
      zmodload -F zsh/stat b:zstat

      # Get the regular files in the current directory,
      # ordered by size (largest first)
      files=( ./*(.OL) )

      # Precalculate the filesizes
      typeset -A filesizes
      for file in "${files[@]}"; do
      filesizes[$file]=$( zstat +size "$file" )
      done

      # The maximum size of a bin is the size of the largest file
      maxsize=${filesizes[${files[1]}]}

      binsizes=()
      typeset -A filebins
      for file in "${files[@]}"; do
      filesize=${filesizes[$file]}
      bin=1 # try fitting into first bin first
      ok=0 # haven't yet found a bin for this file
      for binsize in "${binsizes[@]}"; do
      if (( filesize + binsize <= maxsize )); then
      # File fits in this bin,
      # update bin size and place file in bin
      binsizes[$bin]=$(( filesize + binsize ))
      filebins[$file]=$bin
      ok=1 # now we're good
      break
      fi
      # Try next bin
      bin=$(( bin + 1 ))
      done

      if [ "$ok" -eq 0 ]; then
      # Wasn't able to fit file in existing bin,
      # create new bin
      binsizes+=( "$filesize" )
      filebins[$file]=${#binsizes[@]}
      fi
      done

      # Do final output
      printf 'Bin max size = %dn' "$maxsize"
      for file in "${files[@]}"; do
      printf '%d: %s (file size=%d / bin size=%d)n' "${filebins[$file]}" "$file"
      "${filesizes[$file]}" "${binsizes[$filebins[$file]]}"
      done | sort -n


      The above zsh shell script does binning of all the files in the current directory with a maximum bin size based strictly on the size of the largest file. It implements a first-fit algorithm with the files ordered by decreasing size. This is what's called the "FFD" algorithm in the "Bin packing problem" Wikipedia article. The "MFFD" algorithm is non-trivial to implement in zsh in less than 200 or so lines of code, so I won't post it here.



      Testing:



      $ ls -l
      total 450816
      -rw-r--r-- 1 kk wheel 10485760 Jan 19 23:53 file-10.log
      -rw-r--r-- 1 kk wheel 20971520 Jan 19 23:53 file-20.log
      -rw-r--r-- 1 kk wheel 31457280 Jan 19 23:53 file-30.log
      -rw-r--r-- 1 kk wheel 41943040 Jan 19 23:53 file-40.log
      -rw-r--r-- 1 kk wheel 52428800 Jan 19 23:53 file-50.log
      -rw-r--r-- 1 kk wheel 73400320 Jan 19 23:53 file-70.log




      $ zsh ../script.sh
      Bin max size = 73400320
      1: ./file-70.log (file size=73400320 / bin size=73400320)
      2: ./file-20.log (file size=20971520 / bin size=73400320)
      2: ./file-50.log (file size=52428800 / bin size=73400320)
      3: ./file-30.log (file size=31457280 / bin size=73400320)
      3: ./file-40.log (file size=41943040 / bin size=73400320)
      4: ./file-10.log (file size=10485760 / bin size=10485760)


      The number at the start of each line above corresponds to the bin number assigned to the file.






      share|improve this answer






























        0
















        #!/usr/bin/env zsh

        # To care about hidden filenames:
        #setopt GLOB_DOTS

        # Load the zstat builtin
        zmodload -F zsh/stat b:zstat

        # Get the regular files in the current directory,
        # ordered by size (largest first)
        files=( ./*(.OL) )

        # Precalculate the filesizes
        typeset -A filesizes
        for file in "${files[@]}"; do
        filesizes[$file]=$( zstat +size "$file" )
        done

        # The maximum size of a bin is the size of the largest file
        maxsize=${filesizes[${files[1]}]}

        binsizes=()
        typeset -A filebins
        for file in "${files[@]}"; do
        filesize=${filesizes[$file]}
        bin=1 # try fitting into first bin first
        ok=0 # haven't yet found a bin for this file
        for binsize in "${binsizes[@]}"; do
        if (( filesize + binsize <= maxsize )); then
        # File fits in this bin,
        # update bin size and place file in bin
        binsizes[$bin]=$(( filesize + binsize ))
        filebins[$file]=$bin
        ok=1 # now we're good
        break
        fi
        # Try next bin
        bin=$(( bin + 1 ))
        done

        if [ "$ok" -eq 0 ]; then
        # Wasn't able to fit file in existing bin,
        # create new bin
        binsizes+=( "$filesize" )
        filebins[$file]=${#binsizes[@]}
        fi
        done

        # Do final output
        printf 'Bin max size = %dn' "$maxsize"
        for file in "${files[@]}"; do
        printf '%d: %s (file size=%d / bin size=%d)n' "${filebins[$file]}" "$file"
        "${filesizes[$file]}" "${binsizes[$filebins[$file]]}"
        done | sort -n


        The above zsh shell script does binning of all the files in the current directory with a maximum bin size based strictly on the size of the largest file. It implements a first-fit algorithm with the files ordered by decreasing size. This is what's called the "FFD" algorithm in the "Bin packing problem" Wikipedia article. The "MFFD" algorithm is non-trivial to implement in zsh in less than 200 or so lines of code, so I won't post it here.



        Testing:



        $ ls -l
        total 450816
        -rw-r--r-- 1 kk wheel 10485760 Jan 19 23:53 file-10.log
        -rw-r--r-- 1 kk wheel 20971520 Jan 19 23:53 file-20.log
        -rw-r--r-- 1 kk wheel 31457280 Jan 19 23:53 file-30.log
        -rw-r--r-- 1 kk wheel 41943040 Jan 19 23:53 file-40.log
        -rw-r--r-- 1 kk wheel 52428800 Jan 19 23:53 file-50.log
        -rw-r--r-- 1 kk wheel 73400320 Jan 19 23:53 file-70.log




        $ zsh ../script.sh
        Bin max size = 73400320
        1: ./file-70.log (file size=73400320 / bin size=73400320)
        2: ./file-20.log (file size=20971520 / bin size=73400320)
        2: ./file-50.log (file size=52428800 / bin size=73400320)
        3: ./file-30.log (file size=31457280 / bin size=73400320)
        3: ./file-40.log (file size=41943040 / bin size=73400320)
        4: ./file-10.log (file size=10485760 / bin size=10485760)


        The number at the start of each line above corresponds to the bin number assigned to the file.






        share|improve this answer




























          0












          0








          0









          #!/usr/bin/env zsh

          # To care about hidden filenames:
          #setopt GLOB_DOTS

          # Load the zstat builtin
          zmodload -F zsh/stat b:zstat

          # Get the regular files in the current directory,
          # ordered by size (largest first)
          files=( ./*(.OL) )

          # Precalculate the filesizes
          typeset -A filesizes
          for file in "${files[@]}"; do
          filesizes[$file]=$( zstat +size "$file" )
          done

          # The maximum size of a bin is the size of the largest file
          maxsize=${filesizes[${files[1]}]}

          binsizes=()
          typeset -A filebins
          for file in "${files[@]}"; do
          filesize=${filesizes[$file]}
          bin=1 # try fitting into first bin first
          ok=0 # haven't yet found a bin for this file
          for binsize in "${binsizes[@]}"; do
          if (( filesize + binsize <= maxsize )); then
          # File fits in this bin,
          # update bin size and place file in bin
          binsizes[$bin]=$(( filesize + binsize ))
          filebins[$file]=$bin
          ok=1 # now we're good
          break
          fi
          # Try next bin
          bin=$(( bin + 1 ))
          done

          if [ "$ok" -eq 0 ]; then
          # Wasn't able to fit file in existing bin,
          # create new bin
          binsizes+=( "$filesize" )
          filebins[$file]=${#binsizes[@]}
          fi
          done

          # Do final output
          printf 'Bin max size = %dn' "$maxsize"
          for file in "${files[@]}"; do
          printf '%d: %s (file size=%d / bin size=%d)n' "${filebins[$file]}" "$file"
          "${filesizes[$file]}" "${binsizes[$filebins[$file]]}"
          done | sort -n


          The above zsh shell script does binning of all the files in the current directory with a maximum bin size based strictly on the size of the largest file. It implements a first-fit algorithm with the files ordered by decreasing size. This is what's called the "FFD" algorithm in the "Bin packing problem" Wikipedia article. The "MFFD" algorithm is non-trivial to implement in zsh in less than 200 or so lines of code, so I won't post it here.



          Testing:



          $ ls -l
          total 450816
          -rw-r--r-- 1 kk wheel 10485760 Jan 19 23:53 file-10.log
          -rw-r--r-- 1 kk wheel 20971520 Jan 19 23:53 file-20.log
          -rw-r--r-- 1 kk wheel 31457280 Jan 19 23:53 file-30.log
          -rw-r--r-- 1 kk wheel 41943040 Jan 19 23:53 file-40.log
          -rw-r--r-- 1 kk wheel 52428800 Jan 19 23:53 file-50.log
          -rw-r--r-- 1 kk wheel 73400320 Jan 19 23:53 file-70.log




          $ zsh ../script.sh
          Bin max size = 73400320
          1: ./file-70.log (file size=73400320 / bin size=73400320)
          2: ./file-20.log (file size=20971520 / bin size=73400320)
          2: ./file-50.log (file size=52428800 / bin size=73400320)
          3: ./file-30.log (file size=31457280 / bin size=73400320)
          3: ./file-40.log (file size=41943040 / bin size=73400320)
          4: ./file-10.log (file size=10485760 / bin size=10485760)


          The number at the start of each line above corresponds to the bin number assigned to the file.






          share|improve this answer

















          #!/usr/bin/env zsh

          # To care about hidden filenames:
          #setopt GLOB_DOTS

          # Load the zstat builtin
          zmodload -F zsh/stat b:zstat

          # Get the regular files in the current directory,
          # ordered by size (largest first)
          files=( ./*(.OL) )

          # Precalculate the filesizes
          typeset -A filesizes
          for file in "${files[@]}"; do
          filesizes[$file]=$( zstat +size "$file" )
          done

          # The maximum size of a bin is the size of the largest file
          maxsize=${filesizes[${files[1]}]}

          binsizes=()
          typeset -A filebins
          for file in "${files[@]}"; do
          filesize=${filesizes[$file]}
          bin=1 # try fitting into first bin first
          ok=0 # haven't yet found a bin for this file
          for binsize in "${binsizes[@]}"; do
          if (( filesize + binsize <= maxsize )); then
          # File fits in this bin,
          # update bin size and place file in bin
          binsizes[$bin]=$(( filesize + binsize ))
          filebins[$file]=$bin
          ok=1 # now we're good
          break
          fi
          # Try next bin
          bin=$(( bin + 1 ))
          done

          if [ "$ok" -eq 0 ]; then
          # Wasn't able to fit file in existing bin,
          # create new bin
          binsizes+=( "$filesize" )
          filebins[$file]=${#binsizes[@]}
          fi
          done

          # Do final output
          printf 'Bin max size = %dn' "$maxsize"
          for file in "${files[@]}"; do
          printf '%d: %s (file size=%d / bin size=%d)n' "${filebins[$file]}" "$file"
          "${filesizes[$file]}" "${binsizes[$filebins[$file]]}"
          done | sort -n


          The above zsh shell script does binning of all the files in the current directory with a maximum bin size based strictly on the size of the largest file. It implements a first-fit algorithm with the files ordered by decreasing size. This is what's called the "FFD" algorithm in the "Bin packing problem" Wikipedia article. The "MFFD" algorithm is non-trivial to implement in zsh in less than 200 or so lines of code, so I won't post it here.



          Testing:



          $ ls -l
          total 450816
          -rw-r--r-- 1 kk wheel 10485760 Jan 19 23:53 file-10.log
          -rw-r--r-- 1 kk wheel 20971520 Jan 19 23:53 file-20.log
          -rw-r--r-- 1 kk wheel 31457280 Jan 19 23:53 file-30.log
          -rw-r--r-- 1 kk wheel 41943040 Jan 19 23:53 file-40.log
          -rw-r--r-- 1 kk wheel 52428800 Jan 19 23:53 file-50.log
          -rw-r--r-- 1 kk wheel 73400320 Jan 19 23:53 file-70.log




          $ zsh ../script.sh
          Bin max size = 73400320
          1: ./file-70.log (file size=73400320 / bin size=73400320)
          2: ./file-20.log (file size=20971520 / bin size=73400320)
          2: ./file-50.log (file size=52428800 / bin size=73400320)
          3: ./file-30.log (file size=31457280 / bin size=73400320)
          3: ./file-40.log (file size=41943040 / bin size=73400320)
          4: ./file-10.log (file size=10485760 / bin size=10485760)


          The number at the start of each line above corresponds to the bin number assigned to the file.







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited yesterday

























          answered Jan 19 at 23:50









          KusalanandaKusalananda

          127k16239393




          127k16239393

























              0














              This seems to be pretty much equivalent to the Bin Packing problem.



              The Bin Packing problem is NP-hard, so there is no known shortcut to doing it, brute force (trying all the options in some sensible order that excludes silly attempts, like adding more files to an already oversized group) is the way to go.



              For six files, the brute force approach should be simple enough to do by hand; just list all the possible groupings, count how they split the file usage, and choose the one that gives you the smallest maximum group size.






              share|improve this answer




























                0














                This seems to be pretty much equivalent to the Bin Packing problem.



                The Bin Packing problem is NP-hard, so there is no known shortcut to doing it, brute force (trying all the options in some sensible order that excludes silly attempts, like adding more files to an already oversized group) is the way to go.



                For six files, the brute force approach should be simple enough to do by hand; just list all the possible groupings, count how they split the file usage, and choose the one that gives you the smallest maximum group size.






                share|improve this answer


























                  0












                  0








                  0







                  This seems to be pretty much equivalent to the Bin Packing problem.



                  The Bin Packing problem is NP-hard, so there is no known shortcut to doing it, brute force (trying all the options in some sensible order that excludes silly attempts, like adding more files to an already oversized group) is the way to go.



                  For six files, the brute force approach should be simple enough to do by hand; just list all the possible groupings, count how they split the file usage, and choose the one that gives you the smallest maximum group size.






                  share|improve this answer













                  This seems to be pretty much equivalent to the Bin Packing problem.



                  The Bin Packing problem is NP-hard, so there is no known shortcut to doing it, brute force (trying all the options in some sensible order that excludes silly attempts, like adding more files to an already oversized group) is the way to go.



                  For six files, the brute force approach should be simple enough to do by hand; just list all the possible groupings, count how they split the file usage, and choose the one that gives you the smallest maximum group size.







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered Nov 27 '18 at 8:24









                  BassBass

                  22113




                  22113






























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