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Sort with assumptions
variable 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 52 mins ago
leastactionleastaction
23229
$endgroup$
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 52 mins ago
leastactionleastaction
23229
$endgroup$
1
$begingroup$
An interesting idea, but a symbolic list where you have an ordering for all the elements is rare
$endgroup$
– mikado
30 mins ago
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 52 mins ago
leastactionleastaction
23229
$endgroup$
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 52 mins ago
leastactionleastaction
23229
asked 52 mins ago
leastactionleastaction
23229
asked 52 mins ago
leastactionleastaction
23229
asked 52 mins ago
leastactionleastaction
23229
asked 52 mins ago
leastactionleastaction
23229
23229
1
$begingroup$
An interesting idea, but a symbolic list where you have an ordering for all the elements is rare
$endgroup$
– mikado
30 mins ago
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
30 mins ago
1
1
$begingroup$
An interesting idea, but a symbolic list where you have an ordering for all the elements is rare
$endgroup$
– mikado
30 mins ago
$begingroup$
An interesting idea, but a symbolic list where you have an ordering for all the elements is rare
$endgroup$
– mikado
30 mins ago
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 27 mins ago
Carl WollCarl Woll
70.8k394184
$endgroup$
$begingroup$
Thank you, Carl!
$endgroup$
– leastaction
24 mins ago
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 34 mins ago
Kuba♦Kuba
106k12209530
$endgroup$
$begingroup$
Thanks! Seems to work like a charm, but can you shed some light on why?
$endgroup$
– leastaction
30 mins ago
$begingroup$
@leastaction just take a look atlist /. _Subscript -> 1and atOrdering[list /. _Subscript -> 1].
$endgroup$
– Kuba♦
30 mins ago
$begingroup$
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
26 mins ago
1
$begingroup$
@leastaction sure, which means there isn't one correct answer so every valid within constraints is correct?
$endgroup$
– Kuba♦
23 mins ago
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 26 mins ago
MarcoBMarcoB
37.6k556113
$endgroup$
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 25 mins ago
mikadomikado
6,7071929
$endgroup$
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 27 mins ago
Carl WollCarl Woll
70.8k394184
$endgroup$
$begingroup$
Thank you, Carl!
$endgroup$
– leastaction
24 mins ago
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 27 mins ago
Carl WollCarl Woll
70.8k394184
$endgroup$
$begingroup$
Thank you, Carl!
$endgroup$
– leastaction
24 mins ago
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 27 mins ago
Carl WollCarl Woll
70.8k394184
$endgroup$
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 27 mins ago
Carl WollCarl Woll
70.8k394184
answered 27 mins ago
Carl WollCarl Woll
70.8k394184
answered 27 mins ago
Carl WollCarl Woll
70.8k394184
answered 27 mins ago
Carl WollCarl Woll
70.8k394184
70.8k394184
$begingroup$
Thank you, Carl!
$endgroup$
– leastaction
24 mins ago
add a comment |
$begingroup$
Thank you, Carl!
$endgroup$
– leastaction
24 mins ago
$begingroup$
Thank you, Carl!
$endgroup$
– leastaction
24 mins ago
$begingroup$
Thank you, Carl!
$endgroup$
– leastaction
24 mins ago
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 34 mins ago
Kuba♦Kuba
106k12209530
$endgroup$
$begingroup$
Thanks! Seems to work like a charm, but can you shed some light on why?
$endgroup$
– leastaction
30 mins ago
$begingroup$
@leastaction just take a look atlist /. _Subscript -> 1and atOrdering[list /. _Subscript -> 1].
$endgroup$
– Kuba♦
30 mins ago
$begingroup$
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
26 mins ago
1
$begingroup$
@leastaction sure, which means there isn't one correct answer so every valid within constraints is correct?
$endgroup$
– Kuba♦
23 mins ago
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 34 mins ago
Kuba♦Kuba
106k12209530
$endgroup$
$begingroup$
Thanks! Seems to work like a charm, but can you shed some light on why?
$endgroup$
– leastaction
30 mins ago
$begingroup$
@leastaction just take a look atlist /. _Subscript -> 1and atOrdering[list /. _Subscript -> 1].
$endgroup$
– Kuba♦
30 mins ago
$begingroup$
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
26 mins ago
1
$begingroup$
@leastaction sure, which means there isn't one correct answer so every valid within constraints is correct?
$endgroup$
– Kuba♦
23 mins ago
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 34 mins ago
Kuba♦Kuba
106k12209530
$endgroup$
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 34 mins ago
Kuba♦Kuba
106k12209530
answered 34 mins ago
Kuba♦Kuba
106k12209530
answered 34 mins ago
Kuba♦Kuba
106k12209530
answered 34 mins ago
Kuba♦Kuba
106k12209530
106k12209530
$begingroup$
Thanks! Seems to work like a charm, but can you shed some light on why?
$endgroup$
– leastaction
30 mins ago
$begingroup$
@leastaction just take a look atlist /. _Subscript -> 1and atOrdering[list /. _Subscript -> 1].
$endgroup$
– Kuba♦
30 mins ago
$begingroup$
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
26 mins ago
1
$begingroup$
@leastaction sure, which means there isn't one correct answer so every valid within constraints is correct?
$endgroup$
– Kuba♦
23 mins ago
add a comment |
$begingroup$
Thanks! Seems to work like a charm, but can you shed some light on why?
$endgroup$
– leastaction
30 mins ago
$begingroup$
@leastaction just take a look atlist /. _Subscript -> 1and atOrdering[list /. _Subscript -> 1].
$endgroup$
– Kuba♦
30 mins ago
$begingroup$
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
26 mins ago
1
$begingroup$
@leastaction sure, which means there isn't one correct answer so every valid within constraints is correct?
$endgroup$
– Kuba♦
23 mins ago
$begingroup$
Thanks! Seems to work like a charm, but can you shed some light on why?
$endgroup$
– leastaction
30 mins ago
$begingroup$
Thanks! Seems to work like a charm, but can you shed some light on why?
$endgroup$
– leastaction
30 mins ago
$begingroup$
@leastaction just take a look at
list /. _Subscript -> 1 and at Ordering[list /. _Subscript -> 1].$endgroup$
– Kuba♦
30 mins ago
$begingroup$
@leastaction just take a look at
list /. _Subscript -> 1 and at Ordering[list /. _Subscript -> 1].$endgroup$
– Kuba♦
30 mins ago
$begingroup$
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
26 mins ago
$begingroup$
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
26 mins ago
1
1
$begingroup$
@leastaction sure, which means there isn't one correct answer so every valid within constraints is correct?
$endgroup$
– Kuba♦
23 mins ago
$begingroup$
@leastaction sure, which means there isn't one correct answer so every valid within constraints is correct?
$endgroup$
– Kuba♦
23 mins ago
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 26 mins ago
MarcoBMarcoB
37.6k556113
$endgroup$
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 26 mins ago
MarcoBMarcoB
37.6k556113
$endgroup$
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 26 mins ago
MarcoBMarcoB
37.6k556113
$endgroup$
Sort[list, TrueQ@Simplify[#1 < #2, _Subscript > 0] &]
(* Out: -Subscript[x, 3] - Subscript[x, 9], -Subscript[x, 9], 0, Subscript[x, 7] *)
answered 26 mins ago
MarcoBMarcoB
37.6k556113
answered 26 mins ago
MarcoBMarcoB
37.6k556113
answered 26 mins ago
MarcoBMarcoB
37.6k556113
answered 26 mins ago
MarcoBMarcoB
37.6k556113
37.6k556113
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 25 mins ago
mikadomikado
6,7071929
$endgroup$
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 25 mins ago
mikadomikado
6,7071929
$endgroup$
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 25 mins ago
mikadomikado
6,7071929
$endgroup$
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 25 mins ago
mikadomikado
6,7071929
edited 18 mins ago
answered 25 mins ago
mikadomikado
6,7071929
answered 25 mins ago
mikadomikado
6,7071929
answered 25 mins ago
mikadomikado
6,7071929
6,7071929
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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
30 mins ago