Manipulating a general length functionGeneral function taking general number of argumentsManipulating ListHow...
A flower in a hexagon
What's a good word to describe a public place that looks like it wouldn't be rough?
How should I handle players who ignore the session zero agreement?
Why did Jodrell Bank assist the Soviet Union to collect data from their spacecraft in the mid 1960's?
I am on the US no-fly list. What can I do in order to be allowed on flights which go through US airspace?
Program that converts a number to a letter of the alphabet
Inferring from (∃x)Fx to (∃x)(∃x)Fx using existential generalization?
What is better: yes / no radio, or simple checkbox?
How to acknowledge an embarrassing job interview, now that I work directly with the interviewer?
Rear brake cable temporary fix possible?
How to avoid being sexist when trying to employ someone to function in a very sexist environment?
How experienced do I need to be to go on a photography workshop?
If I delete my router's history can my ISP still provide it to my parents?
When does coming up with an idea constitute sufficient contribution for authorship?
Does Windows 10's telemetry include sending *.doc files if Word crashed?
Using loops to create tables
Strange Sign on Lab Door
Called into a meeting and told we are being made redundant (laid off) and "not to share outside". Can I tell my partner?
On what did Lego base the appearance of the new Hogwarts minifigs?
What's the most convenient time of year in the USA to end the world?
What is the wife of a henpecked husband called?
1 0 1 0 1 0 1 0 1 0 1
Are there any outlying considerations if I treat donning a shield as an object interaction during the first round of combat?
What is the best way to simulate grief?
Manipulating a general length function
General function taking general number of argumentsManipulating ListHow to delete mirror symmetric point pair efficientlyConvert Integer to Numeric with Replacement rulesManipulating arrayManipulating List?Coefficient of function with general argumentFinding negative sequences in a large list: optimizationDefining general odd functionLinear operator algebra: How to distinguish scalars and operators?Manipulating lists
$begingroup$
As a minimal example, define a function like s[1,2,3]; this is a function of three arguments in this case, but I'd like to apply this general idea to functions like s[3,4,5,6,7] which don't necessarily start at 1.
From an identity from theory, I need to re-express this as as sum of 2-argument functions that cycle over all of the possible combinations of the original function, but don't duplicate. Explicitly, this would be the same as writing
s[1,2,3]=f[1,2]+f[1,3]+f[2,3],
or in the longer more general case (and to illustrate my point)
s[3,4,5,6,7] = f[3,4]+f[3,5]+f[3,6]+f[3,7]+f[4,5]+f[4,6]+f[4,7]+f[5,6]+f[5,7]+f[6,7]
I was previously using the definition with simple replacement rules for each of the these chains, with general cases of s[i_,j_,k_] and such similar rules. However, this isn't optimal in larger problems, where I have many thousands of these results, and leads to remarkably long evaluation times.
To see my functional but slow solution, I have attached a partial screenshot here, but naturally, I think there must be a better method. I was considering using Partitions, as in this SE post or BlankSpaces, but I struggle to implement more complex uses of these. Any guidance or other ideas would be gratefully appreciated. Also, a title suggestion would be great as well!
list-manipulation performance-tuning pattern-matching
asked 16 hours ago
BradBrad
768
$endgroup$
add a comment |
$begingroup$
As a minimal example, define a function like s[1,2,3]; this is a function of three arguments in this case, but I'd like to apply this general idea to functions like s[3,4,5,6,7] which don't necessarily start at 1.
From an identity from theory, I need to re-express this as as sum of 2-argument functions that cycle over all of the possible combinations of the original function, but don't duplicate. Explicitly, this would be the same as writing
s[1,2,3]=f[1,2]+f[1,3]+f[2,3],
or in the longer more general case (and to illustrate my point)
s[3,4,5,6,7] = f[3,4]+f[3,5]+f[3,6]+f[3,7]+f[4,5]+f[4,6]+f[4,7]+f[5,6]+f[5,7]+f[6,7]
I was previously using the definition with simple replacement rules for each of the these chains, with general cases of s[i_,j_,k_] and such similar rules. However, this isn't optimal in larger problems, where I have many thousands of these results, and leads to remarkably long evaluation times.
To see my functional but slow solution, I have attached a partial screenshot here, but naturally, I think there must be a better method. I was considering using Partitions, as in this SE post or BlankSpaces, but I struggle to implement more complex uses of these. Any guidance or other ideas would be gratefully appreciated. Also, a title suggestion would be great as well!
list-manipulation performance-tuning pattern-matching
asked 16 hours ago
BradBrad
768
$endgroup$
$begingroup$
Look intoSubsets[listOfArguments, {2}]
$endgroup$
– MarcoB
15 hours ago
add a comment |
$begingroup$
As a minimal example, define a function like s[1,2,3]; this is a function of three arguments in this case, but I'd like to apply this general idea to functions like s[3,4,5,6,7] which don't necessarily start at 1.
From an identity from theory, I need to re-express this as as sum of 2-argument functions that cycle over all of the possible combinations of the original function, but don't duplicate. Explicitly, this would be the same as writing
s[1,2,3]=f[1,2]+f[1,3]+f[2,3],
or in the longer more general case (and to illustrate my point)
s[3,4,5,6,7] = f[3,4]+f[3,5]+f[3,6]+f[3,7]+f[4,5]+f[4,6]+f[4,7]+f[5,6]+f[5,7]+f[6,7]
I was previously using the definition with simple replacement rules for each of the these chains, with general cases of s[i_,j_,k_] and such similar rules. However, this isn't optimal in larger problems, where I have many thousands of these results, and leads to remarkably long evaluation times.
To see my functional but slow solution, I have attached a partial screenshot here, but naturally, I think there must be a better method. I was considering using Partitions, as in this SE post or BlankSpaces, but I struggle to implement more complex uses of these. Any guidance or other ideas would be gratefully appreciated. Also, a title suggestion would be great as well!
list-manipulation performance-tuning pattern-matching
asked 16 hours ago
BradBrad
768
$endgroup$
As a minimal example, define a function like s[1,2,3]; this is a function of three arguments in this case, but I'd like to apply this general idea to functions like s[3,4,5,6,7] which don't necessarily start at 1.
From an identity from theory, I need to re-express this as as sum of 2-argument functions that cycle over all of the possible combinations of the original function, but don't duplicate. Explicitly, this would be the same as writing
s[1,2,3]=f[1,2]+f[1,3]+f[2,3],
or in the longer more general case (and to illustrate my point)
s[3,4,5,6,7] = f[3,4]+f[3,5]+f[3,6]+f[3,7]+f[4,5]+f[4,6]+f[4,7]+f[5,6]+f[5,7]+f[6,7]
I was previously using the definition with simple replacement rules for each of the these chains, with general cases of s[i_,j_,k_] and such similar rules. However, this isn't optimal in larger problems, where I have many thousands of these results, and leads to remarkably long evaluation times.
To see my functional but slow solution, I have attached a partial screenshot here, but naturally, I think there must be a better method. I was considering using Partitions, as in this SE post or BlankSpaces, but I struggle to implement more complex uses of these. Any guidance or other ideas would be gratefully appreciated. Also, a title suggestion would be great as well!
list-manipulation performance-tuning pattern-matching
list-manipulation performance-tuning pattern-matching
asked 16 hours ago
BradBrad
768
asked 16 hours ago
BradBrad
768
asked 16 hours ago
BradBrad
768
asked 16 hours ago
BradBrad
768
asked 16 hours ago
BradBrad
768
768
$begingroup$
Look intoSubsets[listOfArguments, {2}]
$endgroup$
– MarcoB
15 hours ago
add a comment |
$begingroup$
Look intoSubsets[listOfArguments, {2}]
$endgroup$
– MarcoB
15 hours ago
$begingroup$
Look into
Subsets[listOfArguments, {2}]$endgroup$
– MarcoB
15 hours ago
$begingroup$
Look into
Subsets[listOfArguments, {2}]$endgroup$
– MarcoB
15 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Something like the following?
ClearAll[s]
s[seq__] := Total[f@@@Subsets[List[seq], {2}]]
answered 15 hours ago
MarcoBMarcoB
36.6k556112
$endgroup$
$begingroup$
Hi @MarcoB - Subsets was indeed the function I was looking for. I was thinking perhaps of more of a general replacement rule at the end, where in my final stage I remove all of the (previously defined) functions s, in exchange for these functions p. I will try and play around and see if I can figure out how to create a replacement rule based on this method. Thank you for your help:
$endgroup$
– Brad
15 hours ago
$begingroup$
After changing my definition of recursion to work with a symbolicsX, which actually makes evaluations much much faster overall, I can utilise this to get exactly what I need. Thank you all for your help, this is the fastest solution I found.
$endgroup$
– Brad
13 hours ago
add a comment |
$begingroup$
ClearAll[s]
s[f_] := Total @ Subsets[f @ ##, {2}] &;
s[f][a, b, c]
f[a, b] + f[a, c] + f[b, c]
s[f][3, 4, 5, 6, 7]
f[3, 4] + f[3, 5] + f[3, 6] + f[3, 7] + f[4, 5] + f[4, 6] + f[4, 7] +
f[5, 6] + f[5, 7] + f[6, 7]
answered 14 hours ago
kglrkglr
186k10203422
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "387"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
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
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f192481%2fmanipulating-a-general-length-function%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Something like the following?
ClearAll[s]
s[seq__] := Total[f@@@Subsets[List[seq], {2}]]
answered 15 hours ago
MarcoBMarcoB
36.6k556112
$endgroup$
$begingroup$
Hi @MarcoB - Subsets was indeed the function I was looking for. I was thinking perhaps of more of a general replacement rule at the end, where in my final stage I remove all of the (previously defined) functions s, in exchange for these functions p. I will try and play around and see if I can figure out how to create a replacement rule based on this method. Thank you for your help:
$endgroup$
– Brad
15 hours ago
$begingroup$
After changing my definition of recursion to work with a symbolicsX, which actually makes evaluations much much faster overall, I can utilise this to get exactly what I need. Thank you all for your help, this is the fastest solution I found.
$endgroup$
– Brad
13 hours ago
add a comment |
$begingroup$
Something like the following?
ClearAll[s]
s[seq__] := Total[f@@@Subsets[List[seq], {2}]]
answered 15 hours ago
MarcoBMarcoB
36.6k556112
$endgroup$
$begingroup$
Hi @MarcoB - Subsets was indeed the function I was looking for. I was thinking perhaps of more of a general replacement rule at the end, where in my final stage I remove all of the (previously defined) functions s, in exchange for these functions p. I will try and play around and see if I can figure out how to create a replacement rule based on this method. Thank you for your help:
$endgroup$
– Brad
15 hours ago
$begingroup$
After changing my definition of recursion to work with a symbolicsX, which actually makes evaluations much much faster overall, I can utilise this to get exactly what I need. Thank you all for your help, this is the fastest solution I found.
$endgroup$
– Brad
13 hours ago
add a comment |
$begingroup$
Something like the following?
ClearAll[s]
s[seq__] := Total[f@@@Subsets[List[seq], {2}]]
answered 15 hours ago
MarcoBMarcoB
36.6k556112
$endgroup$
Something like the following?
ClearAll[s]
s[seq__] := Total[f@@@Subsets[List[seq], {2}]]
answered 15 hours ago
MarcoBMarcoB
36.6k556112
answered 15 hours ago
MarcoBMarcoB
36.6k556112
answered 15 hours ago
MarcoBMarcoB
36.6k556112
answered 15 hours ago
MarcoBMarcoB
36.6k556112
36.6k556112
$begingroup$
Hi @MarcoB - Subsets was indeed the function I was looking for. I was thinking perhaps of more of a general replacement rule at the end, where in my final stage I remove all of the (previously defined) functions s, in exchange for these functions p. I will try and play around and see if I can figure out how to create a replacement rule based on this method. Thank you for your help:
$endgroup$
– Brad
15 hours ago
$begingroup$
After changing my definition of recursion to work with a symbolicsX, which actually makes evaluations much much faster overall, I can utilise this to get exactly what I need. Thank you all for your help, this is the fastest solution I found.
$endgroup$
– Brad
13 hours ago
add a comment |
$begingroup$
Hi @MarcoB - Subsets was indeed the function I was looking for. I was thinking perhaps of more of a general replacement rule at the end, where in my final stage I remove all of the (previously defined) functions s, in exchange for these functions p. I will try and play around and see if I can figure out how to create a replacement rule based on this method. Thank you for your help:
$endgroup$
– Brad
15 hours ago
$begingroup$
After changing my definition of recursion to work with a symbolicsX, which actually makes evaluations much much faster overall, I can utilise this to get exactly what I need. Thank you all for your help, this is the fastest solution I found.
$endgroup$
– Brad
13 hours ago
$begingroup$
Hi @MarcoB - Subsets was indeed the function I was looking for. I was thinking perhaps of more of a general replacement rule at the end, where in my final stage I remove all of the (previously defined) functions s, in exchange for these functions p. I will try and play around and see if I can figure out how to create a replacement rule based on this method. Thank you for your help:
$endgroup$
– Brad
15 hours ago
$begingroup$
Hi @MarcoB - Subsets was indeed the function I was looking for. I was thinking perhaps of more of a general replacement rule at the end, where in my final stage I remove all of the (previously defined) functions s, in exchange for these functions p. I will try and play around and see if I can figure out how to create a replacement rule based on this method. Thank you for your help:
$endgroup$
– Brad
15 hours ago
$begingroup$
After changing my definition of recursion to work with a symbolic
sX, which actually makes evaluations much much faster overall, I can utilise this to get exactly what I need. Thank you all for your help, this is the fastest solution I found.$endgroup$
– Brad
13 hours ago
$begingroup$
After changing my definition of recursion to work with a symbolic
sX, which actually makes evaluations much much faster overall, I can utilise this to get exactly what I need. Thank you all for your help, this is the fastest solution I found.$endgroup$
– Brad
13 hours ago
add a comment |
$begingroup$
ClearAll[s]
s[f_] := Total @ Subsets[f @ ##, {2}] &;
s[f][a, b, c]
f[a, b] + f[a, c] + f[b, c]
s[f][3, 4, 5, 6, 7]
f[3, 4] + f[3, 5] + f[3, 6] + f[3, 7] + f[4, 5] + f[4, 6] + f[4, 7] +
f[5, 6] + f[5, 7] + f[6, 7]
answered 14 hours ago
kglrkglr
186k10203422
$endgroup$
add a comment |
$begingroup$
ClearAll[s]
s[f_] := Total @ Subsets[f @ ##, {2}] &;
s[f][a, b, c]
f[a, b] + f[a, c] + f[b, c]
s[f][3, 4, 5, 6, 7]
f[3, 4] + f[3, 5] + f[3, 6] + f[3, 7] + f[4, 5] + f[4, 6] + f[4, 7] +
f[5, 6] + f[5, 7] + f[6, 7]
answered 14 hours ago
kglrkglr
186k10203422
$endgroup$
add a comment |
$begingroup$
ClearAll[s]
s[f_] := Total @ Subsets[f @ ##, {2}] &;
s[f][a, b, c]
f[a, b] + f[a, c] + f[b, c]
s[f][3, 4, 5, 6, 7]
f[3, 4] + f[3, 5] + f[3, 6] + f[3, 7] + f[4, 5] + f[4, 6] + f[4, 7] +
f[5, 6] + f[5, 7] + f[6, 7]
answered 14 hours ago
kglrkglr
186k10203422
$endgroup$
ClearAll[s]
s[f_] := Total @ Subsets[f @ ##, {2}] &;
s[f][a, b, c]
f[a, b] + f[a, c] + f[b, c]
s[f][3, 4, 5, 6, 7]
f[3, 4] + f[3, 5] + f[3, 6] + f[3, 7] + f[4, 5] + f[4, 6] + f[4, 7] +
f[5, 6] + f[5, 7] + f[6, 7]
answered 14 hours ago
kglrkglr
186k10203422
edited 14 hours ago
answered 14 hours ago
kglrkglr
186k10203422
answered 14 hours ago
kglrkglr
186k10203422
answered 14 hours ago
kglrkglr
186k10203422
186k10203422
add a comment |
add a comment |
Thanks for contributing an answer to Mathematica Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
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
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f192481%2fmanipulating-a-general-length-function%23new-answer', 'question_page');
}
);
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
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
$begingroup$
Look into
Subsets[listOfArguments, {2}]$endgroup$
– MarcoB
15 hours ago