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Sort with assumptions
The Next CEO of Stack Overflowvariable sized lists and using lists as variablesRetaining and reusing a one-to-one mapping from a sortSorting a matrix alphanumericallyRearranging a ListHow can I check if one expression implies another?Generating an Array of VectorsImporting, sorting and exporting listsDeleting Lonely Numbers From a ListApplying multiple functions to a single column in a tableFind positions in which list elements are equal
$begingroup$
I have a list which looks like this
list = 0, Subscript[x,7], -Subscript[x,3]-Subscript[x,9], -Subscript[x,9];
and all the $x_i$'s are positive semidefinite (i.e. nonnegative) real numbers. I would like to be able to sort this into
sortedlist = -Subscript[x,3]-Subscript[x,9], -Subscript[x,9], 0, Subscript[x,7]
How do I achieve this? I tried
Assuming[Subscript[x,3] > 0 && Subscript[x,7] > 0 && Subscript[x,9] > 0, Sort[list]]
But this obviously does not work. In general, I'd like to be able to impose more constraints on the $x_i's$ when they're being sorted.
list-manipulation symbolic array sorting
asked Mar 19 at 21:09
leastactionleastaction
244210
$endgroup$
add a comment |
$begingroup$
I have a list which looks like this
list = 0, Subscript[x,7], -Subscript[x,3]-Subscript[x,9], -Subscript[x,9];
and all the $x_i$'s are positive semidefinite (i.e. nonnegative) real numbers. I would like to be able to sort this into
sortedlist = -Subscript[x,3]-Subscript[x,9], -Subscript[x,9], 0, Subscript[x,7]
How do I achieve this? I tried
Assuming[Subscript[x,3] > 0 && Subscript[x,7] > 0 && Subscript[x,9] > 0, Sort[list]]
But this obviously does not work. In general, I'd like to be able to impose more constraints on the $x_i's$ when they're being sorted.
list-manipulation symbolic array sorting
asked Mar 19 at 21:09
leastactionleastaction
244210
$endgroup$
1
$begingroup$
An interesting idea, but a symbolic list where you have an ordering for all the elements is rare
$endgroup$
– mikado
Mar 19 at 21:31
add a comment |
$begingroup$
I have a list which looks like this
list = 0, Subscript[x,7], -Subscript[x,3]-Subscript[x,9], -Subscript[x,9];
and all the $x_i$'s are positive semidefinite (i.e. nonnegative) real numbers. I would like to be able to sort this into
sortedlist = -Subscript[x,3]-Subscript[x,9], -Subscript[x,9], 0, Subscript[x,7]
How do I achieve this? I tried
Assuming[Subscript[x,3] > 0 && Subscript[x,7] > 0 && Subscript[x,9] > 0, Sort[list]]
But this obviously does not work. In general, I'd like to be able to impose more constraints on the $x_i's$ when they're being sorted.
list-manipulation symbolic array sorting
asked Mar 19 at 21:09
leastactionleastaction
244210
$endgroup$
I have a list which looks like this
list = 0, Subscript[x,7], -Subscript[x,3]-Subscript[x,9], -Subscript[x,9];
and all the $x_i$'s are positive semidefinite (i.e. nonnegative) real numbers. I would like to be able to sort this into
sortedlist = -Subscript[x,3]-Subscript[x,9], -Subscript[x,9], 0, Subscript[x,7]
How do I achieve this? I tried
Assuming[Subscript[x,3] > 0 && Subscript[x,7] > 0 && Subscript[x,9] > 0, Sort[list]]
But this obviously does not work. In general, I'd like to be able to impose more constraints on the $x_i's$ when they're being sorted.
list-manipulation symbolic array sorting
list-manipulation symbolic array sorting
asked Mar 19 at 21:09
leastactionleastaction
244210
asked Mar 19 at 21:09
leastactionleastaction
244210
asked Mar 19 at 21:09
leastactionleastaction
244210
asked Mar 19 at 21:09
leastactionleastaction
244210
asked Mar 19 at 21:09
leastactionleastaction
244210
244210
1
$begingroup$
An interesting idea, but a symbolic list where you have an ordering for all the elements is rare
$endgroup$
– mikado
Mar 19 at 21:31
add a comment |
1
$begingroup$
An interesting idea, but a symbolic list where you have an ordering for all the elements is rare
$endgroup$
– mikado
Mar 19 at 21:31
1
1
$begingroup$
An interesting idea, but a symbolic list where you have an ordering for all the elements is rare
$endgroup$
– mikado
Mar 19 at 21:31
$begingroup$
An interesting idea, but a symbolic list where you have an ordering for all the elements is rare
$endgroup$
– mikado
Mar 19 at 21:31
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Here is a possibility:
sortWithAssumptions[list_, assum_] := Module[order,
order[a_, b_] := Simplify[a < b, assum];
Sort[list, order]
]
For your example:
sortWithAssumptions[
0,Subscript[x,7],-Subscript[x,3]-Subscript[x,9],-Subscript[x,9],
Subscript[x,3]>0&&Subscript[x,7]>0&&Subscript[x,9]>0
] //TeXForm
$left-x_3-x_9,-x_9,0,x_7right$
Another example:
sortWithAssumptions[
0,Subscript[x,7],-Subscript[x,3]-Subscript[x,9],-Subscript[x,9], Subscript[x,9],
Subscript[x,3]>0&&Subscript[x,7]>0&&Subscript[x,9]>0&&Subscript[x,7]<Subscript[x,9]
] //TeXForm
$left-x_3-x_9,-x_9,0,x_7,x_9right$
answered Mar 19 at 21:35
Carl WollCarl Woll
72.1k395186
$endgroup$
$begingroup$
Thank you, Carl!
$endgroup$
– leastaction
Mar 19 at 21:38
add a comment |
$begingroup$
How about:
list[[Ordering[list /. _Subscript -> 1]]]
-Subscript[x, 3] - Subscript[x, 9], -Subscript[x, 9], 0, Subscript[x, 7]
So basically we sort it the way it would be sorted with all subscripts == 1.
answered Mar 19 at 21:28
Kuba♦Kuba
107k12210533
$endgroup$
$begingroup$
Thanks! Seems to work like a charm, but can you shed some light on why?
$endgroup$
– leastaction
Mar 19 at 21:31
$begingroup$
@leastaction just take a look atlist /. _Subscript -> 1and atOrdering[list /. _Subscript -> 1].
$endgroup$
– Kuba♦
Mar 19 at 21:32
$begingroup$
I see, so you basically assigned the value $x_i = 1$ to each $x_i$. It works in this case, but generically, $x_i$'s may be different.
$endgroup$
– leastaction
Mar 19 at 21:36
1
$begingroup$
@leastaction sure, which means there isn't only one correct answer and every valid within constraints is correct.
$endgroup$
– Kuba♦
Mar 19 at 21:38
add a comment |
$begingroup$
Sort[list, TrueQ@Simplify[#1 < #2, _Subscript > 0] &]
(* Out: -Subscript[x, 3] - Subscript[x, 9], -Subscript[x, 9], 0, Subscript[x, 7] *)
answered Mar 19 at 21:35
MarcoBMarcoB
38.2k556114
$endgroup$
add a comment |
$begingroup$
In this case, we can use RankedMin and FullSimplify to get the answer you seek
Assuming[
Subscript[x, 3] > 0 && Subscript[x, 7] > 0 && Subscript[x, 9] > 0,
FullSimplify[Table[RankedMin[list, i], i, 1, Length[list]]]]
(* -Subscript[x, 3] - Subscript[x, 9], -Subscript[x, 9], 0, Subscript[x, 7] *)
This has the advantage of not returning a (potentially) wrong answer if the sort order is uncertain.
answered Mar 19 at 21:36
mikadomikado
6,8321929
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here is a possibility:
sortWithAssumptions[list_, assum_] := Module[order,
order[a_, b_] := Simplify[a < b, assum];
Sort[list, order]
]
For your example:
sortWithAssumptions[
0,Subscript[x,7],-Subscript[x,3]-Subscript[x,9],-Subscript[x,9],
Subscript[x,3]>0&&Subscript[x,7]>0&&Subscript[x,9]>0
] //TeXForm
$left-x_3-x_9,-x_9,0,x_7right$
Another example:
sortWithAssumptions[
0,Subscript[x,7],-Subscript[x,3]-Subscript[x,9],-Subscript[x,9], Subscript[x,9],
Subscript[x,3]>0&&Subscript[x,7]>0&&Subscript[x,9]>0&&Subscript[x,7]<Subscript[x,9]
] //TeXForm
$left-x_3-x_9,-x_9,0,x_7,x_9right$
answered Mar 19 at 21:35
Carl WollCarl Woll
72.1k395186
$endgroup$
$begingroup$
Thank you, Carl!
$endgroup$
– leastaction
Mar 19 at 21:38
add a comment |
$begingroup$
Here is a possibility:
sortWithAssumptions[list_, assum_] := Module[order,
order[a_, b_] := Simplify[a < b, assum];
Sort[list, order]
]
For your example:
sortWithAssumptions[
0,Subscript[x,7],-Subscript[x,3]-Subscript[x,9],-Subscript[x,9],
Subscript[x,3]>0&&Subscript[x,7]>0&&Subscript[x,9]>0
] //TeXForm
$left-x_3-x_9,-x_9,0,x_7right$
Another example:
sortWithAssumptions[
0,Subscript[x,7],-Subscript[x,3]-Subscript[x,9],-Subscript[x,9], Subscript[x,9],
Subscript[x,3]>0&&Subscript[x,7]>0&&Subscript[x,9]>0&&Subscript[x,7]<Subscript[x,9]
] //TeXForm
$left-x_3-x_9,-x_9,0,x_7,x_9right$
answered Mar 19 at 21:35
Carl WollCarl Woll
72.1k395186
$endgroup$
$begingroup$
Thank you, Carl!
$endgroup$
– leastaction
Mar 19 at 21:38
add a comment |
$begingroup$
Here is a possibility:
sortWithAssumptions[list_, assum_] := Module[order,
order[a_, b_] := Simplify[a < b, assum];
Sort[list, order]
]
For your example:
sortWithAssumptions[
0,Subscript[x,7],-Subscript[x,3]-Subscript[x,9],-Subscript[x,9],
Subscript[x,3]>0&&Subscript[x,7]>0&&Subscript[x,9]>0
] //TeXForm
$left-x_3-x_9,-x_9,0,x_7right$
Another example:
sortWithAssumptions[
0,Subscript[x,7],-Subscript[x,3]-Subscript[x,9],-Subscript[x,9], Subscript[x,9],
Subscript[x,3]>0&&Subscript[x,7]>0&&Subscript[x,9]>0&&Subscript[x,7]<Subscript[x,9]
] //TeXForm
$left-x_3-x_9,-x_9,0,x_7,x_9right$
answered Mar 19 at 21:35
Carl WollCarl Woll
72.1k395186
$endgroup$
Here is a possibility:
sortWithAssumptions[list_, assum_] := Module[order,
order[a_, b_] := Simplify[a < b, assum];
Sort[list, order]
]
For your example:
sortWithAssumptions[
0,Subscript[x,7],-Subscript[x,3]-Subscript[x,9],-Subscript[x,9],
Subscript[x,3]>0&&Subscript[x,7]>0&&Subscript[x,9]>0
] //TeXForm
$left-x_3-x_9,-x_9,0,x_7right$
Another example:
sortWithAssumptions[
0,Subscript[x,7],-Subscript[x,3]-Subscript[x,9],-Subscript[x,9], Subscript[x,9],
Subscript[x,3]>0&&Subscript[x,7]>0&&Subscript[x,9]>0&&Subscript[x,7]<Subscript[x,9]
] //TeXForm
$left-x_3-x_9,-x_9,0,x_7,x_9right$
answered Mar 19 at 21:35
Carl WollCarl Woll
72.1k395186
answered Mar 19 at 21:35
Carl WollCarl Woll
72.1k395186
answered Mar 19 at 21:35
Carl WollCarl Woll
72.1k395186
answered Mar 19 at 21:35
Carl WollCarl Woll
72.1k395186
72.1k395186
$begingroup$
Thank you, Carl!
$endgroup$
– leastaction
Mar 19 at 21:38
add a comment |
$begingroup$
Thank you, Carl!
$endgroup$
– leastaction
Mar 19 at 21:38
$begingroup$
Thank you, Carl!
$endgroup$
– leastaction
Mar 19 at 21:38
$begingroup$
Thank you, Carl!
$endgroup$
– leastaction
Mar 19 at 21:38
add a comment |
$begingroup$
How about:
list[[Ordering[list /. _Subscript -> 1]]]
-Subscript[x, 3] - Subscript[x, 9], -Subscript[x, 9], 0, Subscript[x, 7]
So basically we sort it the way it would be sorted with all subscripts == 1.
answered Mar 19 at 21:28
Kuba♦Kuba
107k12210533
$endgroup$
$begingroup$
Thanks! Seems to work like a charm, but can you shed some light on why?
$endgroup$
– leastaction
Mar 19 at 21:31
$begingroup$
@leastaction just take a look atlist /. _Subscript -> 1and atOrdering[list /. _Subscript -> 1].
$endgroup$
– Kuba♦
Mar 19 at 21:32
$begingroup$
I see, so you basically assigned the value $x_i = 1$ to each $x_i$. It works in this case, but generically, $x_i$'s may be different.
$endgroup$
– leastaction
Mar 19 at 21:36
1
$begingroup$
@leastaction sure, which means there isn't only one correct answer and every valid within constraints is correct.
$endgroup$
– Kuba♦
Mar 19 at 21:38
add a comment |
$begingroup$
How about:
list[[Ordering[list /. _Subscript -> 1]]]
-Subscript[x, 3] - Subscript[x, 9], -Subscript[x, 9], 0, Subscript[x, 7]
So basically we sort it the way it would be sorted with all subscripts == 1.
answered Mar 19 at 21:28
Kuba♦Kuba
107k12210533
$endgroup$
$begingroup$
Thanks! Seems to work like a charm, but can you shed some light on why?
$endgroup$
– leastaction
Mar 19 at 21:31
$begingroup$
@leastaction just take a look atlist /. _Subscript -> 1and atOrdering[list /. _Subscript -> 1].
$endgroup$
– Kuba♦
Mar 19 at 21:32
$begingroup$
I see, so you basically assigned the value $x_i = 1$ to each $x_i$. It works in this case, but generically, $x_i$'s may be different.
$endgroup$
– leastaction
Mar 19 at 21:36
1
$begingroup$
@leastaction sure, which means there isn't only one correct answer and every valid within constraints is correct.
$endgroup$
– Kuba♦
Mar 19 at 21:38
add a comment |
$begingroup$
How about:
list[[Ordering[list /. _Subscript -> 1]]]
-Subscript[x, 3] - Subscript[x, 9], -Subscript[x, 9], 0, Subscript[x, 7]
So basically we sort it the way it would be sorted with all subscripts == 1.
answered Mar 19 at 21:28
Kuba♦Kuba
107k12210533
$endgroup$
How about:
list[[Ordering[list /. _Subscript -> 1]]]
-Subscript[x, 3] - Subscript[x, 9], -Subscript[x, 9], 0, Subscript[x, 7]
So basically we sort it the way it would be sorted with all subscripts == 1.
answered Mar 19 at 21:28
Kuba♦Kuba
107k12210533
answered Mar 19 at 21:28
Kuba♦Kuba
107k12210533
answered Mar 19 at 21:28
Kuba♦Kuba
107k12210533
answered Mar 19 at 21:28
Kuba♦Kuba
107k12210533
107k12210533
$begingroup$
Thanks! Seems to work like a charm, but can you shed some light on why?
$endgroup$
– leastaction
Mar 19 at 21:31
$begingroup$
@leastaction just take a look atlist /. _Subscript -> 1and atOrdering[list /. _Subscript -> 1].
$endgroup$
– Kuba♦
Mar 19 at 21:32
$begingroup$
I see, so you basically assigned the value $x_i = 1$ to each $x_i$. It works in this case, but generically, $x_i$'s may be different.
$endgroup$
– leastaction
Mar 19 at 21:36
1
$begingroup$
@leastaction sure, which means there isn't only one correct answer and every valid within constraints is correct.
$endgroup$
– Kuba♦
Mar 19 at 21:38
add a comment |
$begingroup$
Thanks! Seems to work like a charm, but can you shed some light on why?
$endgroup$
– leastaction
Mar 19 at 21:31
$begingroup$
@leastaction just take a look atlist /. _Subscript -> 1and atOrdering[list /. _Subscript -> 1].
$endgroup$
– Kuba♦
Mar 19 at 21:32
$begingroup$
I see, so you basically assigned the value $x_i = 1$ to each $x_i$. It works in this case, but generically, $x_i$'s may be different.
$endgroup$
– leastaction
Mar 19 at 21:36
1
$begingroup$
@leastaction sure, which means there isn't only one correct answer and every valid within constraints is correct.
$endgroup$
– Kuba♦
Mar 19 at 21:38
$begingroup$
Thanks! Seems to work like a charm, but can you shed some light on why?
$endgroup$
– leastaction
Mar 19 at 21:31
$begingroup$
Thanks! Seems to work like a charm, but can you shed some light on why?
$endgroup$
– leastaction
Mar 19 at 21:31
$begingroup$
@leastaction just take a look at
list /. _Subscript -> 1 and at Ordering[list /. _Subscript -> 1].$endgroup$
– Kuba♦
Mar 19 at 21:32
$begingroup$
@leastaction just take a look at
list /. _Subscript -> 1 and at Ordering[list /. _Subscript -> 1].$endgroup$
– Kuba♦
Mar 19 at 21:32
$begingroup$
I see, so you basically assigned the value $x_i = 1$ to each $x_i$. It works in this case, but generically, $x_i$'s may be different.
$endgroup$
– leastaction
Mar 19 at 21:36
$begingroup$
I see, so you basically assigned the value $x_i = 1$ to each $x_i$. It works in this case, but generically, $x_i$'s may be different.
$endgroup$
– leastaction
Mar 19 at 21:36
1
1
$begingroup$
@leastaction sure, which means there isn't only one correct answer and every valid within constraints is correct.
$endgroup$
– Kuba♦
Mar 19 at 21:38
$begingroup$
@leastaction sure, which means there isn't only one correct answer and every valid within constraints is correct.
$endgroup$
– Kuba♦
Mar 19 at 21:38
add a comment |
$begingroup$
Sort[list, TrueQ@Simplify[#1 < #2, _Subscript > 0] &]
(* Out: -Subscript[x, 3] - Subscript[x, 9], -Subscript[x, 9], 0, Subscript[x, 7] *)
answered Mar 19 at 21:35
MarcoBMarcoB
38.2k556114
$endgroup$
add a comment |
$begingroup$
Sort[list, TrueQ@Simplify[#1 < #2, _Subscript > 0] &]
(* Out: -Subscript[x, 3] - Subscript[x, 9], -Subscript[x, 9], 0, Subscript[x, 7] *)
answered Mar 19 at 21:35
MarcoBMarcoB
38.2k556114
$endgroup$
add a comment |
$begingroup$
Sort[list, TrueQ@Simplify[#1 < #2, _Subscript > 0] &]
(* Out: -Subscript[x, 3] - Subscript[x, 9], -Subscript[x, 9], 0, Subscript[x, 7] *)
answered Mar 19 at 21:35
MarcoBMarcoB
38.2k556114
$endgroup$
Sort[list, TrueQ@Simplify[#1 < #2, _Subscript > 0] &]
(* Out: -Subscript[x, 3] - Subscript[x, 9], -Subscript[x, 9], 0, Subscript[x, 7] *)
answered Mar 19 at 21:35
MarcoBMarcoB
38.2k556114
answered Mar 19 at 21:35
MarcoBMarcoB
38.2k556114
answered Mar 19 at 21:35
MarcoBMarcoB
38.2k556114
answered Mar 19 at 21:35
MarcoBMarcoB
38.2k556114
38.2k556114
add a comment |
add a comment |
$begingroup$
In this case, we can use RankedMin and FullSimplify to get the answer you seek
Assuming[
Subscript[x, 3] > 0 && Subscript[x, 7] > 0 && Subscript[x, 9] > 0,
FullSimplify[Table[RankedMin[list, i], i, 1, Length[list]]]]
(* -Subscript[x, 3] - Subscript[x, 9], -Subscript[x, 9], 0, Subscript[x, 7] *)
This has the advantage of not returning a (potentially) wrong answer if the sort order is uncertain.
answered Mar 19 at 21:36
mikadomikado
6,8321929
$endgroup$
add a comment |
$begingroup$
In this case, we can use RankedMin and FullSimplify to get the answer you seek
Assuming[
Subscript[x, 3] > 0 && Subscript[x, 7] > 0 && Subscript[x, 9] > 0,
FullSimplify[Table[RankedMin[list, i], i, 1, Length[list]]]]
(* -Subscript[x, 3] - Subscript[x, 9], -Subscript[x, 9], 0, Subscript[x, 7] *)
This has the advantage of not returning a (potentially) wrong answer if the sort order is uncertain.
answered Mar 19 at 21:36
mikadomikado
6,8321929
$endgroup$
add a comment |
$begingroup$
In this case, we can use RankedMin and FullSimplify to get the answer you seek
Assuming[
Subscript[x, 3] > 0 && Subscript[x, 7] > 0 && Subscript[x, 9] > 0,
FullSimplify[Table[RankedMin[list, i], i, 1, Length[list]]]]
(* -Subscript[x, 3] - Subscript[x, 9], -Subscript[x, 9], 0, Subscript[x, 7] *)
This has the advantage of not returning a (potentially) wrong answer if the sort order is uncertain.
answered Mar 19 at 21:36
mikadomikado
6,8321929
$endgroup$
In this case, we can use RankedMin and FullSimplify to get the answer you seek
Assuming[
Subscript[x, 3] > 0 && Subscript[x, 7] > 0 && Subscript[x, 9] > 0,
FullSimplify[Table[RankedMin[list, i], i, 1, Length[list]]]]
(* -Subscript[x, 3] - Subscript[x, 9], -Subscript[x, 9], 0, Subscript[x, 7] *)
This has the advantage of not returning a (potentially) wrong answer if the sort order is uncertain.
answered Mar 19 at 21:36
mikadomikado
6,8321929
edited Mar 19 at 21:43
answered Mar 19 at 21:36
mikadomikado
6,8321929
answered Mar 19 at 21:36
mikadomikado
6,8321929
answered Mar 19 at 21:36
mikadomikado
6,8321929
6,8321929
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$begingroup$
An interesting idea, but a symbolic list where you have an ordering for all the elements is rare
$endgroup$
– mikado
Mar 19 at 21:31