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Sort a list of lists by increasing order of elements

Implementing a function which generalizes the merging step in merge sortSort lists according to the order of anotherSorting a list with secondary criterionSorting lists element by elementHow to sort my list?Sorting Lists of lists of arbitrary lengthArrangement Order and Multiple Columns Sorting Order of Associations with Missing KeysSort using indicesInserting an integer into a sorted listSort a list in a descending order

3

$begingroup$

What is the quickest way to sort a list of lists of integers, such that the first sorting criterion is the first element of the list, the next is the second element of the list and so on?

E.g. The list

list = 1,3,4,5, 1,2,4,3, 1,1,2,8, 1,3,5,6, 1,2,3,4

would under these rules be sorted to

sortedlist = 1,1,2,8, 1,2,3,4, 1,2,4,3, 1,3,4,5, 1,3,5,6

list-manipulation sorting

share|improve this question

asked 9 hours ago

nonreligiousnonreligious

6266 bronze badges

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Simply Sort[list] , No ?
  $endgroup$
  – andre314
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, you're right - I was trying to do something more complicated than this simple example which was a list of lists of lists, but I could use this to do Table[Sort[list[[i]] , i, 1, Length[list]]. Thanks
  $endgroup$
  – nonreligious
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @nonreligious - Rather than use Table, map the Sort onto the list, i.e., Sort /@ list
  $endgroup$
  – Bob Hanlon
  8 hours ago
  

  
  
  
  

add a comment | 

3

$begingroup$

What is the quickest way to sort a list of lists of integers, such that the first sorting criterion is the first element of the list, the next is the second element of the list and so on?

E.g. The list

list = 1,3,4,5, 1,2,4,3, 1,1,2,8, 1,3,5,6, 1,2,3,4

would under these rules be sorted to

sortedlist = 1,1,2,8, 1,2,3,4, 1,2,4,3, 1,3,4,5, 1,3,5,6

list-manipulation sorting

share|improve this question

asked 9 hours ago

nonreligiousnonreligious

6266 bronze badges

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Simply Sort[list] , No ?
  $endgroup$
  – andre314
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, you're right - I was trying to do something more complicated than this simple example which was a list of lists of lists, but I could use this to do Table[Sort[list[[i]] , i, 1, Length[list]]. Thanks
  $endgroup$
  – nonreligious
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @nonreligious - Rather than use Table, map the Sort onto the list, i.e., Sort /@ list
  $endgroup$
  – Bob Hanlon
  8 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

What is the quickest way to sort a list of lists of integers, such that the first sorting criterion is the first element of the list, the next is the second element of the list and so on?

E.g. The list

list = 1,3,4,5, 1,2,4,3, 1,1,2,8, 1,3,5,6, 1,2,3,4

would under these rules be sorted to

sortedlist = 1,1,2,8, 1,2,3,4, 1,2,4,3, 1,3,4,5, 1,3,5,6

list-manipulation sorting

share|improve this question

asked 9 hours ago

nonreligiousnonreligious

6266 bronze badges

$endgroup$

share|improve this question

asked 9 hours ago

nonreligiousnonreligious

6266 bronze badges

share|improve this question

asked 9 hours ago

nonreligiousnonreligious

6266 bronze badges

asked 9 hours ago

nonreligiousnonreligious

6266 bronze badges

asked 9 hours ago

nonreligiousnonreligious

6266 bronze badges

2 Answers
2

active

oldest

votes

5

$begingroup$

If the sublists have equal lengths,

list[[Ordering[list]]]

1, 1, 2, 8, 1, 2, 3, 4, 1, 2, 4, 3, 1, 3, 4, 5, 1, 3, 5, 6

With possibly unequal lengths:

list[[Ordering[PadRight @ list]]] 

1, 1, 2, 8, 1, 2, 3, 4, 1, 2, 4, 3, 1, 3, 4, 5, 1, 3, 5, 6

share|improve this answer

edited 8 hours ago

answered 9 hours ago

kglrkglr

201k1010 gold badges230230 silver badges459459 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I see, I had thought I would need to use SortBy[list, #[[1]] &, #[[2]] &, #[[3]] &] but this works.
  $endgroup$
  – nonreligious
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  If the lengths are unequal the later parts that are being explicitly referenced may not exist. This would result in a Part::partw error message.
  $endgroup$
  – Bob Hanlon
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @BobHanlon, right. Ordering with PadRight works without that issue.
  $endgroup$
  – kglr
  8 hours ago
  

  
  
  
  

add a comment | 

1

$begingroup$

With equal lengths, the simplest answer, as andre314 says in the comment above, is simply Sort:

Sort[list]

1, 1, 2, 8, 1, 2, 3, 4, 1, 2, 4, 3, 1, 3, 4, 5, 1, 3, 5, 6

With unequal lengths, kglr's Ordering + PadRight approach is probably optimal.

share|improve this answer

answered 7 hours ago

Carl WollCarl Woll

86.2k33 gold badges110110 silver badges220220 bronze badges

$endgroup$

add a comment | 

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5

$begingroup$

If the sublists have equal lengths,

list[[Ordering[list]]]

1, 1, 2, 8, 1, 2, 3, 4, 1, 2, 4, 3, 1, 3, 4, 5, 1, 3, 5, 6

With possibly unequal lengths:

list[[Ordering[PadRight @ list]]] 

1, 1, 2, 8, 1, 2, 3, 4, 1, 2, 4, 3, 1, 3, 4, 5, 1, 3, 5, 6

share|improve this answer

edited 8 hours ago

answered 9 hours ago

kglrkglr

201k1010 gold badges230230 silver badges459459 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I see, I had thought I would need to use SortBy[list, #[[1]] &, #[[2]] &, #[[3]] &] but this works.
  $endgroup$
  – nonreligious
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  If the lengths are unequal the later parts that are being explicitly referenced may not exist. This would result in a Part::partw error message.
  $endgroup$
  – Bob Hanlon
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @BobHanlon, right. Ordering with PadRight works without that issue.
  $endgroup$
  – kglr
  8 hours ago
  

  
  
  
  

add a comment | 

5

$begingroup$

If the sublists have equal lengths,

list[[Ordering[list]]]

1, 1, 2, 8, 1, 2, 3, 4, 1, 2, 4, 3, 1, 3, 4, 5, 1, 3, 5, 6

With possibly unequal lengths:

list[[Ordering[PadRight @ list]]] 

1, 1, 2, 8, 1, 2, 3, 4, 1, 2, 4, 3, 1, 3, 4, 5, 1, 3, 5, 6

share|improve this answer

edited 8 hours ago

answered 9 hours ago

kglrkglr

201k1010 gold badges230230 silver badges459459 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I see, I had thought I would need to use SortBy[list, #[[1]] &, #[[2]] &, #[[3]] &] but this works.
  $endgroup$
  – nonreligious
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  If the lengths are unequal the later parts that are being explicitly referenced may not exist. This would result in a Part::partw error message.
  $endgroup$
  – Bob Hanlon
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @BobHanlon, right. Ordering with PadRight works without that issue.
  $endgroup$
  – kglr
  8 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

If the sublists have equal lengths,

list[[Ordering[list]]]

1, 1, 2, 8, 1, 2, 3, 4, 1, 2, 4, 3, 1, 3, 4, 5, 1, 3, 5, 6

With possibly unequal lengths:

list[[Ordering[PadRight @ list]]] 

1, 1, 2, 8, 1, 2, 3, 4, 1, 2, 4, 3, 1, 3, 4, 5, 1, 3, 5, 6

share|improve this answer

edited 8 hours ago

answered 9 hours ago

kglrkglr

201k1010 gold badges230230 silver badges459459 bronze badges

$endgroup$

list[[Ordering[list]]]

list[[Ordering[PadRight @ list]]] 

share|improve this answer

edited 8 hours ago

answered 9 hours ago

kglrkglr

201k1010 gold badges230230 silver badges459459 bronze badges

answered 9 hours ago

kglrkglr

201k1010 gold badges230230 silver badges459459 bronze badges

answered 9 hours ago

kglrkglr

201k1010 gold badges230230 silver badges459459 bronze badges

1

$begingroup$

With equal lengths, the simplest answer, as andre314 says in the comment above, is simply Sort:

Sort[list]

1, 1, 2, 8, 1, 2, 3, 4, 1, 2, 4, 3, 1, 3, 4, 5, 1, 3, 5, 6

With unequal lengths, kglr's Ordering + PadRight approach is probably optimal.

share|improve this answer

answered 7 hours ago

Carl WollCarl Woll

86.2k33 gold badges110110 silver badges220220 bronze badges

$endgroup$

add a comment | 

1

$begingroup$

With equal lengths, the simplest answer, as andre314 says in the comment above, is simply Sort:

Sort[list]

1, 1, 2, 8, 1, 2, 3, 4, 1, 2, 4, 3, 1, 3, 4, 5, 1, 3, 5, 6

With unequal lengths, kglr's Ordering + PadRight approach is probably optimal.

share|improve this answer

answered 7 hours ago

Carl WollCarl Woll

86.2k33 gold badges110110 silver badges220220 bronze badges

$endgroup$

add a comment | 

$begingroup$

With equal lengths, the simplest answer, as andre314 says in the comment above, is simply Sort:

Sort[list]

1, 1, 2, 8, 1, 2, 3, 4, 1, 2, 4, 3, 1, 3, 4, 5, 1, 3, 5, 6

With unequal lengths, kglr's Ordering + PadRight approach is probably optimal.

share|improve this answer

answered 7 hours ago

Carl WollCarl Woll

86.2k33 gold badges110110 silver badges220220 bronze badges

$endgroup$

Sort[list]

share|improve this answer

answered 7 hours ago

Carl WollCarl Woll

86.2k33 gold badges110110 silver badges220220 bronze badges

answered 7 hours ago

Carl WollCarl Woll

86.2k33 gold badges110110 silver badges220220 bronze badges

answered 7 hours ago

Carl WollCarl Woll

86.2k33 gold badges110110 silver badges220220 bronze badges