Sort with assumptionsvariable 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
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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
$endgroup$
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
$endgroup$
1
$begingroup$
An interesting idea, but a symbolic list where you have an ordering for all the elements is rare
$endgroup$
– mikado
33 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
$endgroup$
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 55 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
33 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
33 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
33 mins ago
$begingroup$
An interesting idea, but a symbolic list where you have an ordering for all the elements is rare
$endgroup$
– mikado
33 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
$left-x_3-x_9,-x_9,0,x_7,x_9right$
$endgroup$
$begingroup$
Thank you, Carl!
$endgroup$
– leastaction
27 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.
$endgroup$
$begingroup$
Thanks! Seems to work like a charm, but can you shed some light on why?
$endgroup$
– leastaction
33 mins ago
$begingroup$
@leastaction just take a look atlist /. _Subscript -> 1and atOrdering[list /. _Subscript -> 1].
$endgroup$
– Kuba♦
32 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
29 mins ago
1
$begingroup$
@leastaction sure, which means there isn't one correct answer so every valid within constraints is correct?
$endgroup$
– Kuba♦
26 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] *)
$endgroup$
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.
$endgroup$
add a comment |
Your Answer
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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
$left-x_3-x_9,-x_9,0,x_7,x_9right$
$endgroup$
$begingroup$
Thank you, Carl!
$endgroup$
– leastaction
27 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
$left-x_3-x_9,-x_9,0,x_7,x_9right$
$endgroup$
$begingroup$
Thank you, Carl!
$endgroup$
– leastaction
27 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
$left-x_3-x_9,-x_9,0,x_7,x_9right$
$endgroup$
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 29 mins ago
Carl WollCarl Woll
70.8k394184
70.8k394184
$begingroup$
Thank you, Carl!
$endgroup$
– leastaction
27 mins ago
add a comment |
$begingroup$
Thank you, Carl!
$endgroup$
– leastaction
27 mins ago
$begingroup$
Thank you, Carl!
$endgroup$
– leastaction
27 mins ago
$begingroup$
Thank you, Carl!
$endgroup$
– leastaction
27 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.
$endgroup$
$begingroup$
Thanks! Seems to work like a charm, but can you shed some light on why?
$endgroup$
– leastaction
33 mins ago
$begingroup$
@leastaction just take a look atlist /. _Subscript -> 1and atOrdering[list /. _Subscript -> 1].
$endgroup$
– Kuba♦
32 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
29 mins ago
1
$begingroup$
@leastaction sure, which means there isn't one correct answer so every valid within constraints is correct?
$endgroup$
– Kuba♦
26 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.
$endgroup$
$begingroup$
Thanks! Seems to work like a charm, but can you shed some light on why?
$endgroup$
– leastaction
33 mins ago
$begingroup$
@leastaction just take a look atlist /. _Subscript -> 1and atOrdering[list /. _Subscript -> 1].
$endgroup$
– Kuba♦
32 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
29 mins ago
1
$begingroup$
@leastaction sure, which means there isn't one correct answer so every valid within constraints is correct?
$endgroup$
– Kuba♦
26 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.
$endgroup$
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 37 mins ago
Kuba♦Kuba
106k12209530
106k12209530
$begingroup$
Thanks! Seems to work like a charm, but can you shed some light on why?
$endgroup$
– leastaction
33 mins ago
$begingroup$
@leastaction just take a look atlist /. _Subscript -> 1and atOrdering[list /. _Subscript -> 1].
$endgroup$
– Kuba♦
32 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
29 mins ago
1
$begingroup$
@leastaction sure, which means there isn't one correct answer so every valid within constraints is correct?
$endgroup$
– Kuba♦
26 mins ago
add a comment |
$begingroup$
Thanks! Seems to work like a charm, but can you shed some light on why?
$endgroup$
– leastaction
33 mins ago
$begingroup$
@leastaction just take a look atlist /. _Subscript -> 1and atOrdering[list /. _Subscript -> 1].
$endgroup$
– Kuba♦
32 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
29 mins ago
1
$begingroup$
@leastaction sure, which means there isn't one correct answer so every valid within constraints is correct?
$endgroup$
– Kuba♦
26 mins ago
$begingroup$
Thanks! Seems to work like a charm, but can you shed some light on why?
$endgroup$
– leastaction
33 mins ago
$begingroup$
Thanks! Seems to work like a charm, but can you shed some light on why?
$endgroup$
– leastaction
33 mins ago
$begingroup$
@leastaction just take a look at
list /. _Subscript -> 1 and at Ordering[list /. _Subscript -> 1].$endgroup$
– Kuba♦
32 mins ago
$begingroup$
@leastaction just take a look at
list /. _Subscript -> 1 and at Ordering[list /. _Subscript -> 1].$endgroup$
– Kuba♦
32 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
29 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
29 mins ago
1
1
$begingroup$
@leastaction sure, which means there isn't one correct answer so every valid within constraints is correct?
$endgroup$
– Kuba♦
26 mins ago
$begingroup$
@leastaction sure, which means there isn't one correct answer so every valid within constraints is correct?
$endgroup$
– Kuba♦
26 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] *)
$endgroup$
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] *)
$endgroup$
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] *)
$endgroup$
Sort[list, TrueQ@Simplify[#1 < #2, _Subscript > 0] &]
(* Out: -Subscript[x, 3] - Subscript[x, 9], -Subscript[x, 9], 0, Subscript[x, 7] *)
answered 29 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.
$endgroup$
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.
$endgroup$
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.
$endgroup$
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.
edited 21 mins ago
answered 28 mins ago
mikadomikado
6,7071929
6,7071929
add a comment |
add a comment |
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StackExchange.helpers.onClickDraftSave('#login-link');
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onScroll = function(e)
var $elem = $('.new-login-left'),
docViewTop = $window.scrollTop(),
docViewBottom = docViewTop + $window.height(),
elemTop = $elem.offset().top,
elemBottom = elemTop + $elem.height();
if ((docViewTop elemBottom))
StackExchange.using('gps', function() StackExchange.gps.track('embedded_signup_form.view', location: 'question_page' ); );
$window.unbind('scroll', onScroll);
;
$window.on('scroll', onScroll);
);
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StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
var $window = $(window),
onScroll = function(e)
var $elem = $('.new-login-left'),
docViewTop = $window.scrollTop(),
docViewBottom = docViewTop + $window.height(),
elemTop = $elem.offset().top,
elemBottom = elemTop + $elem.height();
if ((docViewTop elemBottom))
StackExchange.using('gps', function() StackExchange.gps.track('embedded_signup_form.view', location: 'question_page' ); );
$window.unbind('scroll', onScroll);
;
$window.on('scroll', onScroll);
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
var $window = $(window),
onScroll = function(e)
var $elem = $('.new-login-left'),
docViewTop = $window.scrollTop(),
docViewBottom = docViewTop + $window.height(),
elemTop = $elem.offset().top,
elemBottom = elemTop + $elem.height();
if ((docViewTop elemBottom))
StackExchange.using('gps', function() StackExchange.gps.track('embedded_signup_form.view', location: 'question_page' ); );
$window.unbind('scroll', onScroll);
;
$window.on('scroll', onScroll);
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
$begingroup$
An interesting idea, but a symbolic list where you have an ordering for all the elements is rare
$endgroup$
– mikado
33 mins ago