Function for SortBy












2














Let's say I have the following list



list = {{-1, 0, 0, 1}, {-1, 0, 1, 0}, {-1, 1, 0, 0}, {0, -1, 0, 1}, 
{0, -1, 1, 0}, {0, 0, -1, 1}, {0, 0, 1, -1}, {0, 1, -1, 0},
{0, 1, 0, -1}, {1, -1, 0, 0}, {1, 0, -1, 0}, {1, 0, 0, -1}}


What sort function (sfunc) used in SortBy [list, sfunc] can give me slist?



slist = {{0, 0, 1, -1}, {0, 0, -1, 1}, {0, -1, 0, 1}, {-1, 0, 0, 1}, 
{0, 1, 0, -1}, {0, 1, -1, 0}, {0, -1, 1, 0}, {-1, 0, 1, 0},
{1, 0, 0, -1}, {1, 0, -1, 0}, {1, -1, 0, 0}, {-1, 1, 0, 0}}


Few examples of sorted data



slist1 =  {{0, 0, 1, -2}, {0, 0, -2, 1}, {0, -2, 0, 1}, {-2, 0, 0, 1}, {0, 1, 0, -2}, {0, 1, -2, 0}, {0, -2, 1, 0}, {-2, 0, 1, 0}, {1, 0, 0, -2}, {1, 0, -2, 0}, {1, -2, 0, 0}, {-2, 1, 0, 0}, {0, 1, -1, -1}, {0, -1, 1, -1}, {0, -1, -1, 1}, {-1, 0, 1, -1}, {-1, 0, -1, 1},{-1, -1, 0, 1}, {1, 0, -1, -1}, {1, -1, 0, -1}, {1, -1, -1, 0}, {-1, 1, 0, -1}, {-1, 1, -1, 0}, {-1, -1, 1, 0}}


slist2 = {{0, 0, 2, -2}, {0, 0, -2, 2}, {0, -2, 0, 2}, {-2, 0, 0, 2}, {0, 1, 1, -2}, {0, 1, -2, 1}, {0, -2, 1, 1}, {-2, 0, 1, 1}, {0, 2, 0, -2}, {0, 2, -2, 0}, {0, -2, 2, 0}, {-2, 0, 2, 0}, {1, 0, 1, -2}, {1, 0, -2, 1}, {1, -2, 0, 1}, {-2, 1, 0, 1}, {1, 1, 0, -2}, {1, 1, -2, 0}, {1, -2, 1, 0}, {-2, 1, 1, 0}, {2, 0, 0, -2}, {2, 0, -2, 0}, {2, -2, 0, 0}, {-2, 2, 0, 0}, {0, 2, -1, -1}, {0, -1, 2, -1}, {0, -1, -1, 2}, {-1, 0, 2, -1}, {-1, 0, -1, 2}, {-1, -1, 0, 2}, {1, 1, -1, -1}, {1, -1, 1, -1}, {1, -1, -1, 1}, {-1, 1, 1, -1}, {-1, 1, -1, 1}, {-1, -1, 1, 1}, {2, 0, -1, -1}, {2, -1, 0, -1}, {2, -1, -1, 0}, {-1, 2, 0, -1}, {-1, 2, -1, 0}, {-1, -1, 2, 0}}









share|improve this question




















  • 4




    Explain the core idea behind your sorted list. It certainly isn't clear what you're seeking.
    – David G. Stork
    Jan 4 at 19:01






  • 2




    Can you give some examples with more complicated data?
    – MikeY
    Jan 4 at 19:33






  • 1




    @MikeY I have added 2 more sorted sets.
    – Hubble07
    Jan 4 at 20:10










  • So if you sort your data like this, it can probably be counted and therefore indexed in closed form.
    – MikeY
    2 days ago












  • @MikeY Yes I have decided to use this sorting. I have edited by bounty question. If you have time then you could answer that, then I can gladly give you the bounty.
    – Hubble07
    2 days ago
















2














Let's say I have the following list



list = {{-1, 0, 0, 1}, {-1, 0, 1, 0}, {-1, 1, 0, 0}, {0, -1, 0, 1}, 
{0, -1, 1, 0}, {0, 0, -1, 1}, {0, 0, 1, -1}, {0, 1, -1, 0},
{0, 1, 0, -1}, {1, -1, 0, 0}, {1, 0, -1, 0}, {1, 0, 0, -1}}


What sort function (sfunc) used in SortBy [list, sfunc] can give me slist?



slist = {{0, 0, 1, -1}, {0, 0, -1, 1}, {0, -1, 0, 1}, {-1, 0, 0, 1}, 
{0, 1, 0, -1}, {0, 1, -1, 0}, {0, -1, 1, 0}, {-1, 0, 1, 0},
{1, 0, 0, -1}, {1, 0, -1, 0}, {1, -1, 0, 0}, {-1, 1, 0, 0}}


Few examples of sorted data



slist1 =  {{0, 0, 1, -2}, {0, 0, -2, 1}, {0, -2, 0, 1}, {-2, 0, 0, 1}, {0, 1, 0, -2}, {0, 1, -2, 0}, {0, -2, 1, 0}, {-2, 0, 1, 0}, {1, 0, 0, -2}, {1, 0, -2, 0}, {1, -2, 0, 0}, {-2, 1, 0, 0}, {0, 1, -1, -1}, {0, -1, 1, -1}, {0, -1, -1, 1}, {-1, 0, 1, -1}, {-1, 0, -1, 1},{-1, -1, 0, 1}, {1, 0, -1, -1}, {1, -1, 0, -1}, {1, -1, -1, 0}, {-1, 1, 0, -1}, {-1, 1, -1, 0}, {-1, -1, 1, 0}}


slist2 = {{0, 0, 2, -2}, {0, 0, -2, 2}, {0, -2, 0, 2}, {-2, 0, 0, 2}, {0, 1, 1, -2}, {0, 1, -2, 1}, {0, -2, 1, 1}, {-2, 0, 1, 1}, {0, 2, 0, -2}, {0, 2, -2, 0}, {0, -2, 2, 0}, {-2, 0, 2, 0}, {1, 0, 1, -2}, {1, 0, -2, 1}, {1, -2, 0, 1}, {-2, 1, 0, 1}, {1, 1, 0, -2}, {1, 1, -2, 0}, {1, -2, 1, 0}, {-2, 1, 1, 0}, {2, 0, 0, -2}, {2, 0, -2, 0}, {2, -2, 0, 0}, {-2, 2, 0, 0}, {0, 2, -1, -1}, {0, -1, 2, -1}, {0, -1, -1, 2}, {-1, 0, 2, -1}, {-1, 0, -1, 2}, {-1, -1, 0, 2}, {1, 1, -1, -1}, {1, -1, 1, -1}, {1, -1, -1, 1}, {-1, 1, 1, -1}, {-1, 1, -1, 1}, {-1, -1, 1, 1}, {2, 0, -1, -1}, {2, -1, 0, -1}, {2, -1, -1, 0}, {-1, 2, 0, -1}, {-1, 2, -1, 0}, {-1, -1, 2, 0}}









share|improve this question




















  • 4




    Explain the core idea behind your sorted list. It certainly isn't clear what you're seeking.
    – David G. Stork
    Jan 4 at 19:01






  • 2




    Can you give some examples with more complicated data?
    – MikeY
    Jan 4 at 19:33






  • 1




    @MikeY I have added 2 more sorted sets.
    – Hubble07
    Jan 4 at 20:10










  • So if you sort your data like this, it can probably be counted and therefore indexed in closed form.
    – MikeY
    2 days ago












  • @MikeY Yes I have decided to use this sorting. I have edited by bounty question. If you have time then you could answer that, then I can gladly give you the bounty.
    – Hubble07
    2 days ago














2












2








2







Let's say I have the following list



list = {{-1, 0, 0, 1}, {-1, 0, 1, 0}, {-1, 1, 0, 0}, {0, -1, 0, 1}, 
{0, -1, 1, 0}, {0, 0, -1, 1}, {0, 0, 1, -1}, {0, 1, -1, 0},
{0, 1, 0, -1}, {1, -1, 0, 0}, {1, 0, -1, 0}, {1, 0, 0, -1}}


What sort function (sfunc) used in SortBy [list, sfunc] can give me slist?



slist = {{0, 0, 1, -1}, {0, 0, -1, 1}, {0, -1, 0, 1}, {-1, 0, 0, 1}, 
{0, 1, 0, -1}, {0, 1, -1, 0}, {0, -1, 1, 0}, {-1, 0, 1, 0},
{1, 0, 0, -1}, {1, 0, -1, 0}, {1, -1, 0, 0}, {-1, 1, 0, 0}}


Few examples of sorted data



slist1 =  {{0, 0, 1, -2}, {0, 0, -2, 1}, {0, -2, 0, 1}, {-2, 0, 0, 1}, {0, 1, 0, -2}, {0, 1, -2, 0}, {0, -2, 1, 0}, {-2, 0, 1, 0}, {1, 0, 0, -2}, {1, 0, -2, 0}, {1, -2, 0, 0}, {-2, 1, 0, 0}, {0, 1, -1, -1}, {0, -1, 1, -1}, {0, -1, -1, 1}, {-1, 0, 1, -1}, {-1, 0, -1, 1},{-1, -1, 0, 1}, {1, 0, -1, -1}, {1, -1, 0, -1}, {1, -1, -1, 0}, {-1, 1, 0, -1}, {-1, 1, -1, 0}, {-1, -1, 1, 0}}


slist2 = {{0, 0, 2, -2}, {0, 0, -2, 2}, {0, -2, 0, 2}, {-2, 0, 0, 2}, {0, 1, 1, -2}, {0, 1, -2, 1}, {0, -2, 1, 1}, {-2, 0, 1, 1}, {0, 2, 0, -2}, {0, 2, -2, 0}, {0, -2, 2, 0}, {-2, 0, 2, 0}, {1, 0, 1, -2}, {1, 0, -2, 1}, {1, -2, 0, 1}, {-2, 1, 0, 1}, {1, 1, 0, -2}, {1, 1, -2, 0}, {1, -2, 1, 0}, {-2, 1, 1, 0}, {2, 0, 0, -2}, {2, 0, -2, 0}, {2, -2, 0, 0}, {-2, 2, 0, 0}, {0, 2, -1, -1}, {0, -1, 2, -1}, {0, -1, -1, 2}, {-1, 0, 2, -1}, {-1, 0, -1, 2}, {-1, -1, 0, 2}, {1, 1, -1, -1}, {1, -1, 1, -1}, {1, -1, -1, 1}, {-1, 1, 1, -1}, {-1, 1, -1, 1}, {-1, -1, 1, 1}, {2, 0, -1, -1}, {2, -1, 0, -1}, {2, -1, -1, 0}, {-1, 2, 0, -1}, {-1, 2, -1, 0}, {-1, -1, 2, 0}}









share|improve this question















Let's say I have the following list



list = {{-1, 0, 0, 1}, {-1, 0, 1, 0}, {-1, 1, 0, 0}, {0, -1, 0, 1}, 
{0, -1, 1, 0}, {0, 0, -1, 1}, {0, 0, 1, -1}, {0, 1, -1, 0},
{0, 1, 0, -1}, {1, -1, 0, 0}, {1, 0, -1, 0}, {1, 0, 0, -1}}


What sort function (sfunc) used in SortBy [list, sfunc] can give me slist?



slist = {{0, 0, 1, -1}, {0, 0, -1, 1}, {0, -1, 0, 1}, {-1, 0, 0, 1}, 
{0, 1, 0, -1}, {0, 1, -1, 0}, {0, -1, 1, 0}, {-1, 0, 1, 0},
{1, 0, 0, -1}, {1, 0, -1, 0}, {1, -1, 0, 0}, {-1, 1, 0, 0}}


Few examples of sorted data



slist1 =  {{0, 0, 1, -2}, {0, 0, -2, 1}, {0, -2, 0, 1}, {-2, 0, 0, 1}, {0, 1, 0, -2}, {0, 1, -2, 0}, {0, -2, 1, 0}, {-2, 0, 1, 0}, {1, 0, 0, -2}, {1, 0, -2, 0}, {1, -2, 0, 0}, {-2, 1, 0, 0}, {0, 1, -1, -1}, {0, -1, 1, -1}, {0, -1, -1, 1}, {-1, 0, 1, -1}, {-1, 0, -1, 1},{-1, -1, 0, 1}, {1, 0, -1, -1}, {1, -1, 0, -1}, {1, -1, -1, 0}, {-1, 1, 0, -1}, {-1, 1, -1, 0}, {-1, -1, 1, 0}}


slist2 = {{0, 0, 2, -2}, {0, 0, -2, 2}, {0, -2, 0, 2}, {-2, 0, 0, 2}, {0, 1, 1, -2}, {0, 1, -2, 1}, {0, -2, 1, 1}, {-2, 0, 1, 1}, {0, 2, 0, -2}, {0, 2, -2, 0}, {0, -2, 2, 0}, {-2, 0, 2, 0}, {1, 0, 1, -2}, {1, 0, -2, 1}, {1, -2, 0, 1}, {-2, 1, 0, 1}, {1, 1, 0, -2}, {1, 1, -2, 0}, {1, -2, 1, 0}, {-2, 1, 1, 0}, {2, 0, 0, -2}, {2, 0, -2, 0}, {2, -2, 0, 0}, {-2, 2, 0, 0}, {0, 2, -1, -1}, {0, -1, 2, -1}, {0, -1, -1, 2}, {-1, 0, 2, -1}, {-1, 0, -1, 2}, {-1, -1, 0, 2}, {1, 1, -1, -1}, {1, -1, 1, -1}, {1, -1, -1, 1}, {-1, 1, 1, -1}, {-1, 1, -1, 1}, {-1, -1, 1, 1}, {2, 0, -1, -1}, {2, -1, 0, -1}, {2, -1, -1, 0}, {-1, 2, 0, -1}, {-1, 2, -1, 0}, {-1, -1, 2, 0}}






list-manipulation sorting






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Jan 4 at 20:06







Hubble07

















asked Jan 4 at 18:59









Hubble07Hubble07

2,916721




2,916721








  • 4




    Explain the core idea behind your sorted list. It certainly isn't clear what you're seeking.
    – David G. Stork
    Jan 4 at 19:01






  • 2




    Can you give some examples with more complicated data?
    – MikeY
    Jan 4 at 19:33






  • 1




    @MikeY I have added 2 more sorted sets.
    – Hubble07
    Jan 4 at 20:10










  • So if you sort your data like this, it can probably be counted and therefore indexed in closed form.
    – MikeY
    2 days ago












  • @MikeY Yes I have decided to use this sorting. I have edited by bounty question. If you have time then you could answer that, then I can gladly give you the bounty.
    – Hubble07
    2 days ago














  • 4




    Explain the core idea behind your sorted list. It certainly isn't clear what you're seeking.
    – David G. Stork
    Jan 4 at 19:01






  • 2




    Can you give some examples with more complicated data?
    – MikeY
    Jan 4 at 19:33






  • 1




    @MikeY I have added 2 more sorted sets.
    – Hubble07
    Jan 4 at 20:10










  • So if you sort your data like this, it can probably be counted and therefore indexed in closed form.
    – MikeY
    2 days ago












  • @MikeY Yes I have decided to use this sorting. I have edited by bounty question. If you have time then you could answer that, then I can gladly give you the bounty.
    – Hubble07
    2 days ago








4




4




Explain the core idea behind your sorted list. It certainly isn't clear what you're seeking.
– David G. Stork
Jan 4 at 19:01




Explain the core idea behind your sorted list. It certainly isn't clear what you're seeking.
– David G. Stork
Jan 4 at 19:01




2




2




Can you give some examples with more complicated data?
– MikeY
Jan 4 at 19:33




Can you give some examples with more complicated data?
– MikeY
Jan 4 at 19:33




1




1




@MikeY I have added 2 more sorted sets.
– Hubble07
Jan 4 at 20:10




@MikeY I have added 2 more sorted sets.
– Hubble07
Jan 4 at 20:10












So if you sort your data like this, it can probably be counted and therefore indexed in closed form.
– MikeY
2 days ago






So if you sort your data like this, it can probably be counted and therefore indexed in closed form.
– MikeY
2 days ago














@MikeY Yes I have decided to use this sorting. I have edited by bounty question. If you have time then you could answer that, then I can gladly give you the bounty.
– Hubble07
2 days ago




@MikeY Yes I have decided to use this sorting. I have edited by bounty question. If you have time then you could answer that, then I can gladly give you the bounty.
– Hubble07
2 days ago










1 Answer
1






active

oldest

votes


















8














EDITED TO ADD A SORT CRITERION

For the data sets, you are sorting on




  1. the number of negative numbers first, then

  2. the subset of just the nonnegative elements (using canonical ordering for lists), then

  3. the set gained when you replace negative terms with a '1' and nonnegative with '0'


  4. The subset of just the negative elements (using canonical ordering)



     funkySort[list_]:= SortBy[list,{
    Count[#, _?Negative] &,
    Select[#, NonNegative] &,
    Negative,
    Select[#, Negative] &
    }]


    slist == funkySort[slist[[RandomPermutation[Length[slist]]]]]
    slist1 == funkySort[slist1[[RandomPermutation[Length[slist1]]]]]
    slist2 == funkySort[slist2[[RandomPermutation[Length[slist2]]]]]




True



True



True







share|improve this answer























  • But slist != funkySort[list] . Are you sure this works. It doesn't seem to work on my system. Do you get really get slist1 and slist2 when your function is applied to any random ordering of those lists.
    – Hubble07
    2 days ago










  • Oops, copied over the wrong funkySort[ ] from my notebook. Fixed it...
    – MikeY
    2 days ago











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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









8














EDITED TO ADD A SORT CRITERION

For the data sets, you are sorting on




  1. the number of negative numbers first, then

  2. the subset of just the nonnegative elements (using canonical ordering for lists), then

  3. the set gained when you replace negative terms with a '1' and nonnegative with '0'


  4. The subset of just the negative elements (using canonical ordering)



     funkySort[list_]:= SortBy[list,{
    Count[#, _?Negative] &,
    Select[#, NonNegative] &,
    Negative,
    Select[#, Negative] &
    }]


    slist == funkySort[slist[[RandomPermutation[Length[slist]]]]]
    slist1 == funkySort[slist1[[RandomPermutation[Length[slist1]]]]]
    slist2 == funkySort[slist2[[RandomPermutation[Length[slist2]]]]]




True



True



True







share|improve this answer























  • But slist != funkySort[list] . Are you sure this works. It doesn't seem to work on my system. Do you get really get slist1 and slist2 when your function is applied to any random ordering of those lists.
    – Hubble07
    2 days ago










  • Oops, copied over the wrong funkySort[ ] from my notebook. Fixed it...
    – MikeY
    2 days ago
















8














EDITED TO ADD A SORT CRITERION

For the data sets, you are sorting on




  1. the number of negative numbers first, then

  2. the subset of just the nonnegative elements (using canonical ordering for lists), then

  3. the set gained when you replace negative terms with a '1' and nonnegative with '0'


  4. The subset of just the negative elements (using canonical ordering)



     funkySort[list_]:= SortBy[list,{
    Count[#, _?Negative] &,
    Select[#, NonNegative] &,
    Negative,
    Select[#, Negative] &
    }]


    slist == funkySort[slist[[RandomPermutation[Length[slist]]]]]
    slist1 == funkySort[slist1[[RandomPermutation[Length[slist1]]]]]
    slist2 == funkySort[slist2[[RandomPermutation[Length[slist2]]]]]




True



True



True







share|improve this answer























  • But slist != funkySort[list] . Are you sure this works. It doesn't seem to work on my system. Do you get really get slist1 and slist2 when your function is applied to any random ordering of those lists.
    – Hubble07
    2 days ago










  • Oops, copied over the wrong funkySort[ ] from my notebook. Fixed it...
    – MikeY
    2 days ago














8












8








8






EDITED TO ADD A SORT CRITERION

For the data sets, you are sorting on




  1. the number of negative numbers first, then

  2. the subset of just the nonnegative elements (using canonical ordering for lists), then

  3. the set gained when you replace negative terms with a '1' and nonnegative with '0'


  4. The subset of just the negative elements (using canonical ordering)



     funkySort[list_]:= SortBy[list,{
    Count[#, _?Negative] &,
    Select[#, NonNegative] &,
    Negative,
    Select[#, Negative] &
    }]


    slist == funkySort[slist[[RandomPermutation[Length[slist]]]]]
    slist1 == funkySort[slist1[[RandomPermutation[Length[slist1]]]]]
    slist2 == funkySort[slist2[[RandomPermutation[Length[slist2]]]]]




True



True



True







share|improve this answer














EDITED TO ADD A SORT CRITERION

For the data sets, you are sorting on




  1. the number of negative numbers first, then

  2. the subset of just the nonnegative elements (using canonical ordering for lists), then

  3. the set gained when you replace negative terms with a '1' and nonnegative with '0'


  4. The subset of just the negative elements (using canonical ordering)



     funkySort[list_]:= SortBy[list,{
    Count[#, _?Negative] &,
    Select[#, NonNegative] &,
    Negative,
    Select[#, Negative] &
    }]


    slist == funkySort[slist[[RandomPermutation[Length[slist]]]]]
    slist1 == funkySort[slist1[[RandomPermutation[Length[slist1]]]]]
    slist2 == funkySort[slist2[[RandomPermutation[Length[slist2]]]]]




True



True



True








share|improve this answer














share|improve this answer



share|improve this answer








edited yesterday









J. M. is computer-less

96.2k10300460




96.2k10300460










answered Jan 4 at 19:52









MikeYMikeY

2,317411




2,317411












  • But slist != funkySort[list] . Are you sure this works. It doesn't seem to work on my system. Do you get really get slist1 and slist2 when your function is applied to any random ordering of those lists.
    – Hubble07
    2 days ago










  • Oops, copied over the wrong funkySort[ ] from my notebook. Fixed it...
    – MikeY
    2 days ago


















  • But slist != funkySort[list] . Are you sure this works. It doesn't seem to work on my system. Do you get really get slist1 and slist2 when your function is applied to any random ordering of those lists.
    – Hubble07
    2 days ago










  • Oops, copied over the wrong funkySort[ ] from my notebook. Fixed it...
    – MikeY
    2 days ago
















But slist != funkySort[list] . Are you sure this works. It doesn't seem to work on my system. Do you get really get slist1 and slist2 when your function is applied to any random ordering of those lists.
– Hubble07
2 days ago




But slist != funkySort[list] . Are you sure this works. It doesn't seem to work on my system. Do you get really get slist1 and slist2 when your function is applied to any random ordering of those lists.
– Hubble07
2 days ago












Oops, copied over the wrong funkySort[ ] from my notebook. Fixed it...
– MikeY
2 days ago




Oops, copied over the wrong funkySort[ ] from my notebook. Fixed it...
– MikeY
2 days ago


















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