Terse Method to Swap Lowest for Highest?Efficient method for Inserting arrays into arraysSwap elements in list without copyBetter method to swap the values of two 2-D arraysHow to get this list with a terse methodBuilt-in (or Terse) Method to Combine and Transpose DatasetsAre there more readable and terse method can get this listefficiently method for generating a sequenceSimple method to sort versionsFunction for SortBySwap Elements of a continuous List, possible?
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Terse Method to Swap Lowest for Highest?
Efficient method for Inserting arrays into arraysSwap elements in list without copyBetter method to swap the values of two 2-D arraysHow to get this list with a terse methodBuilt-in (or Terse) Method to Combine and Transpose DatasetsAre there more readable and terse method can get this listefficiently method for generating a sequenceSimple method to sort versionsFunction for SortBySwap Elements of a continuous List, possible?
$begingroup$
I have built a solution to swap the lowest values with the highest values in a list.
With
SeedRandom[987]
test = RandomSample@*Join @@ Range @@@ 6, 10, 56, 60, 1, 5, -5, -1
-1, 2, 7, 8, 60, 57, 58, 10, 9, 4, -5, -3, 3, 59, 1, 5, -4, 6, -2, 56
Then
swapPositions =
PermutationReplace[
Ordering@Ordering@test,
With[len = Length@test,
Cycles@
Transpose@Range @@ 1, Floor[len/2], Reverse@*Range @@ Ceiling[len/2] + 1, len
]
];
Sort[test][[swapPositions]]
56, 9, 4, 3, -5, -2, -3, 1, 2, 7, 60, 58, 8, -4, 10, 6, 59, 5, 57, -1
The largest half of the numbers have had their positions swapped with lowest half of the numbers.
However, it feels too verbose and I think Sort might be expensive in this case. Is there a built-in function or more terse method to achieve this. Of course with no loss in speed. The actual case is for list of length 100000 and more.
list-manipulation performance-tuning sorting permutation
edited Mar 23 at 2:15
J. M. is away♦
99k10311470
asked Mar 22 at 20:57
EdmundEdmund
26.9k330103
$endgroup$
add a comment |
$begingroup$
I have built a solution to swap the lowest values with the highest values in a list.
With
SeedRandom[987]
test = RandomSample@*Join @@ Range @@@ 6, 10, 56, 60, 1, 5, -5, -1
-1, 2, 7, 8, 60, 57, 58, 10, 9, 4, -5, -3, 3, 59, 1, 5, -4, 6, -2, 56
Then
swapPositions =
PermutationReplace[
Ordering@Ordering@test,
With[len = Length@test,
Cycles@
Transpose@Range @@ 1, Floor[len/2], Reverse@*Range @@ Ceiling[len/2] + 1, len
]
];
Sort[test][[swapPositions]]
56, 9, 4, 3, -5, -2, -3, 1, 2, 7, 60, 58, 8, -4, 10, 6, 59, 5, 57, -1
The largest half of the numbers have had their positions swapped with lowest half of the numbers.
However, it feels too verbose and I think Sort might be expensive in this case. Is there a built-in function or more terse method to achieve this. Of course with no loss in speed. The actual case is for list of length 100000 and more.
list-manipulation performance-tuning sorting permutation
edited Mar 23 at 2:15
J. M. is away♦
99k10311470
asked Mar 22 at 20:57
EdmundEdmund
26.9k330103
$endgroup$
add a comment |
$begingroup$
I have built a solution to swap the lowest values with the highest values in a list.
With
SeedRandom[987]
test = RandomSample@*Join @@ Range @@@ 6, 10, 56, 60, 1, 5, -5, -1
-1, 2, 7, 8, 60, 57, 58, 10, 9, 4, -5, -3, 3, 59, 1, 5, -4, 6, -2, 56
Then
swapPositions =
PermutationReplace[
Ordering@Ordering@test,
With[len = Length@test,
Cycles@
Transpose@Range @@ 1, Floor[len/2], Reverse@*Range @@ Ceiling[len/2] + 1, len
]
];
Sort[test][[swapPositions]]
56, 9, 4, 3, -5, -2, -3, 1, 2, 7, 60, 58, 8, -4, 10, 6, 59, 5, 57, -1
The largest half of the numbers have had their positions swapped with lowest half of the numbers.
However, it feels too verbose and I think Sort might be expensive in this case. Is there a built-in function or more terse method to achieve this. Of course with no loss in speed. The actual case is for list of length 100000 and more.
list-manipulation performance-tuning sorting permutation
edited Mar 23 at 2:15
J. M. is away♦
99k10311470
asked Mar 22 at 20:57
EdmundEdmund
26.9k330103
$endgroup$
I have built a solution to swap the lowest values with the highest values in a list.
With
SeedRandom[987]
test = RandomSample@*Join @@ Range @@@ 6, 10, 56, 60, 1, 5, -5, -1
-1, 2, 7, 8, 60, 57, 58, 10, 9, 4, -5, -3, 3, 59, 1, 5, -4, 6, -2, 56
Then
swapPositions =
PermutationReplace[
Ordering@Ordering@test,
With[len = Length@test,
Cycles@
Transpose@Range @@ 1, Floor[len/2], Reverse@*Range @@ Ceiling[len/2] + 1, len
]
];
Sort[test][[swapPositions]]
56, 9, 4, 3, -5, -2, -3, 1, 2, 7, 60, 58, 8, -4, 10, 6, 59, 5, 57, -1
The largest half of the numbers have had their positions swapped with lowest half of the numbers.
However, it feels too verbose and I think Sort might be expensive in this case. Is there a built-in function or more terse method to achieve this. Of course with no loss in speed. The actual case is for list of length 100000 and more.
list-manipulation performance-tuning sorting permutation
list-manipulation performance-tuning sorting permutation
edited Mar 23 at 2:15
J. M. is away♦
99k10311470
asked Mar 22 at 20:57
EdmundEdmund
26.9k330103
edited Mar 23 at 2:15
J. M. is away♦
99k10311470
asked Mar 22 at 20:57
EdmundEdmund
26.9k330103
edited Mar 23 at 2:15
J. M. is away♦
99k10311470
edited Mar 23 at 2:15
J. M. is away♦
99k10311470
edited Mar 23 at 2:15
J. M. is away♦
99k10311470
99k10311470
asked Mar 22 at 20:57
EdmundEdmund
26.9k330103
asked Mar 22 at 20:57
EdmundEdmund
26.9k330103
asked Mar 22 at 20:57
EdmundEdmund
26.9k330103
26.9k330103
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
How about:
Module[tmp = test,
With[ord=Ordering[tmp],
tmp[[ord]] = Reverse @ tmp[[ord]]];
tmp
]
56, 9, 4, 3, -5, -2, -3, 1, 2, 7, 60, 58, 8, -4, 10, 6, 59, 5, 57, -1
answered Mar 22 at 21:11
Carl WollCarl Woll
76.8k3101201
$endgroup$
1
$begingroup$
That is so obvious I want to cry. Thanks (+1).
$endgroup$
– Edmund
Mar 22 at 21:15
add a comment |
$begingroup$
This is equivalent to Carl's procedure, except that it uses one less scratch list:
With[ord = Ordering[test],
test[[PermutationProduct[Reverse[ord], InversePermutation[ord]]]]]
56, 9, 4, 3, -5, -2, -3, 1, 2, 7, 60, 58, 8, -4, 10, 6, 59, 5, 57, -1
Recall that list[[perm]] = list is equivalent to list = list[[InversePermutation[perm]]], where perm is a permutation list. (The situation is equivalent to list.pmat being the same as Transpose[pmat].list if pmat is a permutation matrix.) You can then use PermutationProduct[] to compose successive permutations.
(This was supposed to be a comment, but it got too long.)
answered Mar 23 at 2:14
J. M. is away♦J. M. is away
99k10311470
$endgroup$
$begingroup$
This solution doesn't copy the list so may be faster than Carl's. (+1).
$endgroup$
– Edmund
Mar 23 at 3:38
$begingroup$
FWIW, I consistently get 56, -2, 6, -4, 5, 1, 59, 3, -3, -5, 4, 9, 10, 58, 57, 60, 8, 7, 2, -1 from this.
$endgroup$
– Christopher Lamb
Mar 23 at 16:01
$begingroup$
@Rabbit, what version number of Mathematica is giving that result?
$endgroup$
– J. M. is away♦
Mar 23 at 16:10
$begingroup$
11.3.0.0 (5944644, 2018030701) Win 10. I did a trace, which might have had the needed info but I didn't catch it. Started w/ fresh kernel, & repeated, w/ same result. Baffled.
$endgroup$
– Christopher Lamb
Mar 23 at 16:16
1
$begingroup$
At least on my machine, the questioner's original (very verbose) proposed solution is faster than any other proposal. In order of presentation: 0.0000114366 v. 0.0000357762 v. 0.0000164219 (AbsoluteTiming in seconds).
$endgroup$
– CElliott
Apr 3 at 14:10
|
show 3 more comments
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
How about:
Module[tmp = test,
With[ord=Ordering[tmp],
tmp[[ord]] = Reverse @ tmp[[ord]]];
tmp
]
56, 9, 4, 3, -5, -2, -3, 1, 2, 7, 60, 58, 8, -4, 10, 6, 59, 5, 57, -1
answered Mar 22 at 21:11
Carl WollCarl Woll
76.8k3101201
$endgroup$
1
$begingroup$
That is so obvious I want to cry. Thanks (+1).
$endgroup$
– Edmund
Mar 22 at 21:15
add a comment |
$begingroup$
How about:
Module[tmp = test,
With[ord=Ordering[tmp],
tmp[[ord]] = Reverse @ tmp[[ord]]];
tmp
]
56, 9, 4, 3, -5, -2, -3, 1, 2, 7, 60, 58, 8, -4, 10, 6, 59, 5, 57, -1
answered Mar 22 at 21:11
Carl WollCarl Woll
76.8k3101201
$endgroup$
1
$begingroup$
That is so obvious I want to cry. Thanks (+1).
$endgroup$
– Edmund
Mar 22 at 21:15
add a comment |
$begingroup$
How about:
Module[tmp = test,
With[ord=Ordering[tmp],
tmp[[ord]] = Reverse @ tmp[[ord]]];
tmp
]
56, 9, 4, 3, -5, -2, -3, 1, 2, 7, 60, 58, 8, -4, 10, 6, 59, 5, 57, -1
answered Mar 22 at 21:11
Carl WollCarl Woll
76.8k3101201
$endgroup$
How about:
Module[tmp = test,
With[ord=Ordering[tmp],
tmp[[ord]] = Reverse @ tmp[[ord]]];
tmp
]
56, 9, 4, 3, -5, -2, -3, 1, 2, 7, 60, 58, 8, -4, 10, 6, 59, 5, 57, -1
answered Mar 22 at 21:11
Carl WollCarl Woll
76.8k3101201
answered Mar 22 at 21:11
Carl WollCarl Woll
76.8k3101201
answered Mar 22 at 21:11
Carl WollCarl Woll
76.8k3101201
answered Mar 22 at 21:11
Carl WollCarl Woll
76.8k3101201
76.8k3101201
1
$begingroup$
That is so obvious I want to cry. Thanks (+1).
$endgroup$
– Edmund
Mar 22 at 21:15
add a comment |
1
$begingroup$
That is so obvious I want to cry. Thanks (+1).
$endgroup$
– Edmund
Mar 22 at 21:15
1
1
$begingroup$
That is so obvious I want to cry. Thanks (+1).
$endgroup$
– Edmund
Mar 22 at 21:15
$begingroup$
That is so obvious I want to cry. Thanks (+1).
$endgroup$
– Edmund
Mar 22 at 21:15
add a comment |
$begingroup$
This is equivalent to Carl's procedure, except that it uses one less scratch list:
With[ord = Ordering[test],
test[[PermutationProduct[Reverse[ord], InversePermutation[ord]]]]]
56, 9, 4, 3, -5, -2, -3, 1, 2, 7, 60, 58, 8, -4, 10, 6, 59, 5, 57, -1
Recall that list[[perm]] = list is equivalent to list = list[[InversePermutation[perm]]], where perm is a permutation list. (The situation is equivalent to list.pmat being the same as Transpose[pmat].list if pmat is a permutation matrix.) You can then use PermutationProduct[] to compose successive permutations.
(This was supposed to be a comment, but it got too long.)
answered Mar 23 at 2:14
J. M. is away♦J. M. is away
99k10311470
$endgroup$
$begingroup$
This solution doesn't copy the list so may be faster than Carl's. (+1).
$endgroup$
– Edmund
Mar 23 at 3:38
$begingroup$
FWIW, I consistently get 56, -2, 6, -4, 5, 1, 59, 3, -3, -5, 4, 9, 10, 58, 57, 60, 8, 7, 2, -1 from this.
$endgroup$
– Christopher Lamb
Mar 23 at 16:01
$begingroup$
@Rabbit, what version number of Mathematica is giving that result?
$endgroup$
– J. M. is away♦
Mar 23 at 16:10
$begingroup$
11.3.0.0 (5944644, 2018030701) Win 10. I did a trace, which might have had the needed info but I didn't catch it. Started w/ fresh kernel, & repeated, w/ same result. Baffled.
$endgroup$
– Christopher Lamb
Mar 23 at 16:16
1
$begingroup$
At least on my machine, the questioner's original (very verbose) proposed solution is faster than any other proposal. In order of presentation: 0.0000114366 v. 0.0000357762 v. 0.0000164219 (AbsoluteTiming in seconds).
$endgroup$
– CElliott
Apr 3 at 14:10
|
show 3 more comments
$begingroup$
This is equivalent to Carl's procedure, except that it uses one less scratch list:
With[ord = Ordering[test],
test[[PermutationProduct[Reverse[ord], InversePermutation[ord]]]]]
56, 9, 4, 3, -5, -2, -3, 1, 2, 7, 60, 58, 8, -4, 10, 6, 59, 5, 57, -1
Recall that list[[perm]] = list is equivalent to list = list[[InversePermutation[perm]]], where perm is a permutation list. (The situation is equivalent to list.pmat being the same as Transpose[pmat].list if pmat is a permutation matrix.) You can then use PermutationProduct[] to compose successive permutations.
(This was supposed to be a comment, but it got too long.)
answered Mar 23 at 2:14
J. M. is away♦J. M. is away
99k10311470
$endgroup$
$begingroup$
This solution doesn't copy the list so may be faster than Carl's. (+1).
$endgroup$
– Edmund
Mar 23 at 3:38
$begingroup$
FWIW, I consistently get 56, -2, 6, -4, 5, 1, 59, 3, -3, -5, 4, 9, 10, 58, 57, 60, 8, 7, 2, -1 from this.
$endgroup$
– Christopher Lamb
Mar 23 at 16:01
$begingroup$
@Rabbit, what version number of Mathematica is giving that result?
$endgroup$
– J. M. is away♦
Mar 23 at 16:10
$begingroup$
11.3.0.0 (5944644, 2018030701) Win 10. I did a trace, which might have had the needed info but I didn't catch it. Started w/ fresh kernel, & repeated, w/ same result. Baffled.
$endgroup$
– Christopher Lamb
Mar 23 at 16:16
1
$begingroup$
At least on my machine, the questioner's original (very verbose) proposed solution is faster than any other proposal. In order of presentation: 0.0000114366 v. 0.0000357762 v. 0.0000164219 (AbsoluteTiming in seconds).
$endgroup$
– CElliott
Apr 3 at 14:10
|
show 3 more comments
$begingroup$
This is equivalent to Carl's procedure, except that it uses one less scratch list:
With[ord = Ordering[test],
test[[PermutationProduct[Reverse[ord], InversePermutation[ord]]]]]
56, 9, 4, 3, -5, -2, -3, 1, 2, 7, 60, 58, 8, -4, 10, 6, 59, 5, 57, -1
Recall that list[[perm]] = list is equivalent to list = list[[InversePermutation[perm]]], where perm is a permutation list. (The situation is equivalent to list.pmat being the same as Transpose[pmat].list if pmat is a permutation matrix.) You can then use PermutationProduct[] to compose successive permutations.
(This was supposed to be a comment, but it got too long.)
answered Mar 23 at 2:14
J. M. is away♦J. M. is away
99k10311470
$endgroup$
This is equivalent to Carl's procedure, except that it uses one less scratch list:
With[ord = Ordering[test],
test[[PermutationProduct[Reverse[ord], InversePermutation[ord]]]]]
56, 9, 4, 3, -5, -2, -3, 1, 2, 7, 60, 58, 8, -4, 10, 6, 59, 5, 57, -1
Recall that list[[perm]] = list is equivalent to list = list[[InversePermutation[perm]]], where perm is a permutation list. (The situation is equivalent to list.pmat being the same as Transpose[pmat].list if pmat is a permutation matrix.) You can then use PermutationProduct[] to compose successive permutations.
(This was supposed to be a comment, but it got too long.)
answered Mar 23 at 2:14
J. M. is away♦J. M. is away
99k10311470
edited Mar 23 at 2:27
answered Mar 23 at 2:14
J. M. is away♦J. M. is away
99k10311470
answered Mar 23 at 2:14
J. M. is away♦J. M. is away
99k10311470
answered Mar 23 at 2:14
J. M. is away♦J. M. is away
99k10311470
99k10311470
$begingroup$
This solution doesn't copy the list so may be faster than Carl's. (+1).
$endgroup$
– Edmund
Mar 23 at 3:38
$begingroup$
FWIW, I consistently get 56, -2, 6, -4, 5, 1, 59, 3, -3, -5, 4, 9, 10, 58, 57, 60, 8, 7, 2, -1 from this.
$endgroup$
– Christopher Lamb
Mar 23 at 16:01
$begingroup$
@Rabbit, what version number of Mathematica is giving that result?
$endgroup$
– J. M. is away♦
Mar 23 at 16:10
$begingroup$
11.3.0.0 (5944644, 2018030701) Win 10. I did a trace, which might have had the needed info but I didn't catch it. Started w/ fresh kernel, & repeated, w/ same result. Baffled.
$endgroup$
– Christopher Lamb
Mar 23 at 16:16
1
$begingroup$
At least on my machine, the questioner's original (very verbose) proposed solution is faster than any other proposal. In order of presentation: 0.0000114366 v. 0.0000357762 v. 0.0000164219 (AbsoluteTiming in seconds).
$endgroup$
– CElliott
Apr 3 at 14:10
|
show 3 more comments
$begingroup$
This solution doesn't copy the list so may be faster than Carl's. (+1).
$endgroup$
– Edmund
Mar 23 at 3:38
$begingroup$
FWIW, I consistently get 56, -2, 6, -4, 5, 1, 59, 3, -3, -5, 4, 9, 10, 58, 57, 60, 8, 7, 2, -1 from this.
$endgroup$
– Christopher Lamb
Mar 23 at 16:01
$begingroup$
@Rabbit, what version number of Mathematica is giving that result?
$endgroup$
– J. M. is away♦
Mar 23 at 16:10
$begingroup$
11.3.0.0 (5944644, 2018030701) Win 10. I did a trace, which might have had the needed info but I didn't catch it. Started w/ fresh kernel, & repeated, w/ same result. Baffled.
$endgroup$
– Christopher Lamb
Mar 23 at 16:16
1
$begingroup$
At least on my machine, the questioner's original (very verbose) proposed solution is faster than any other proposal. In order of presentation: 0.0000114366 v. 0.0000357762 v. 0.0000164219 (AbsoluteTiming in seconds).
$endgroup$
– CElliott
Apr 3 at 14:10
$begingroup$
This solution doesn't copy the list so may be faster than Carl's. (+1).
$endgroup$
– Edmund
Mar 23 at 3:38
$begingroup$
This solution doesn't copy the list so may be faster than Carl's. (+1).
$endgroup$
– Edmund
Mar 23 at 3:38
$begingroup$
FWIW, I consistently get 56, -2, 6, -4, 5, 1, 59, 3, -3, -5, 4, 9, 10, 58, 57, 60, 8, 7, 2, -1 from this.
$endgroup$
– Christopher Lamb
Mar 23 at 16:01
$begingroup$
FWIW, I consistently get 56, -2, 6, -4, 5, 1, 59, 3, -3, -5, 4, 9, 10, 58, 57, 60, 8, 7, 2, -1 from this.
$endgroup$
– Christopher Lamb
Mar 23 at 16:01
$begingroup$
@Rabbit, what version number of Mathematica is giving that result?
$endgroup$
– J. M. is away♦
Mar 23 at 16:10
$begingroup$
@Rabbit, what version number of Mathematica is giving that result?
$endgroup$
– J. M. is away♦
Mar 23 at 16:10
$begingroup$
11.3.0.0 (5944644, 2018030701) Win 10. I did a trace, which might have had the needed info but I didn't catch it. Started w/ fresh kernel, & repeated, w/ same result. Baffled.
$endgroup$
– Christopher Lamb
Mar 23 at 16:16
$begingroup$
11.3.0.0 (5944644, 2018030701) Win 10. I did a trace, which might have had the needed info but I didn't catch it. Started w/ fresh kernel, & repeated, w/ same result. Baffled.
$endgroup$
– Christopher Lamb
Mar 23 at 16:16
1
1
$begingroup$
At least on my machine, the questioner's original (very verbose) proposed solution is faster than any other proposal. In order of presentation: 0.0000114366 v. 0.0000357762 v. 0.0000164219 (AbsoluteTiming in seconds).
$endgroup$
– CElliott
Apr 3 at 14:10
$begingroup$
At least on my machine, the questioner's original (very verbose) proposed solution is faster than any other proposal. In order of presentation: 0.0000114366 v. 0.0000357762 v. 0.0000164219 (AbsoluteTiming in seconds).
$endgroup$
– CElliott
Apr 3 at 14:10
|
show 3 more comments
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Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
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');
);
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
