Mdthbs

Ordering a word list

During the Space Shuttle Columbia Disaster of 2003, Why Did The Flight Director Say, "Lock the doors."?

Unique combinations of a list of tuples

Look mom! I made my own (Base 10) numeral system!

Dropdowns & Chevrons for Right to Left languages

How can you evade tax by getting employment income just in equity, then using this equity as collateral to take out loan?

Double blind peer review when paper cites author's GitHub repo for code

Ex-contractor published company source code and secrets online

Blocking people from taking pictures of me with smartphone

Are there any financial disadvantages to living significantly "below your means"?

Can a spacecraft use an accelerometer to determine its orientation?

In reversi, can you overwrite two chips in one move?

Accidentals - some in brackets, some not

Does a code snippet compile? Or does it get compiled?

Is refreshing multiple times a test case for web applications?

As a 16 year old, how can I keep my money safe from my mother?

Can I call myself an assistant professor without a PhD?

What are the uses and limitations of Persuasion, Insight, and Deception against other PCs?

Why isn’t SHA-3 in wider use?

How to mark beverage cans in a cooler for a blind person?

How quickly could a country build a tall concrete wall around a city?

Could one become a successful researcher by writing some really good papers while being outside academia?

Are there any differences in causality between linear and logistic regression?

How can I tell if a flight itinerary is fake?

changing number of arguments to a function in secondary evaluation

using a Mathematica function to define a new functionConstruct a function whose definition depends on the values of its arguments and evaluates a SectionHow to get the parameter number of inbuilt functionHow to hold evaluation of a value that is passed to a functionFormatting Display PrecisionUsing DeleteCases with a defined function with two arguments as a patternFunctions that remember some arguments while not remembering other arguments

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0;
}

5

$begingroup$

So I have a function that produces answers involving uninstantiated(?) function names, e.g.

 Out[1]= f[a,b,c,d].

I then want to evalute this leftover result by, say, replacing this f[a,b,c,d] with a defined function g[a,b,c,d,e].

I think something maybe to do with slots, but my poor attempts have still failed, eg.

 Out[1] /. f[##] &@g[##,e]



 Out[1] /. f[___] -> g



 Out[1] /. f[#1, #2, #3, #4] -> g[#1, #2, #3, #4, e]

functions argument-patterns

share|improve this question

asked 9 hours ago

natenate

27511 silver badge77 bronze badges

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  You're mixing the concepts of pure functions (# and &) with replacement (/., ->, :>). Use slots with the first one, use patterns (Pattern in the documentation, x_ or x__ for example in usage) with the second. E.g. f[a, b, c, d] /. f[x__] :> g[x, e].
  $endgroup$
  – eyorble
  9 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Aargh. This comment is an answer! I'll accept it if you want
  $endgroup$
  – nate
  9 hours ago
  

  
  
  
  

add a comment | 

5

$begingroup$

So I have a function that produces answers involving uninstantiated(?) function names, e.g.

 Out[1]= f[a,b,c,d].

I then want to evalute this leftover result by, say, replacing this f[a,b,c,d] with a defined function g[a,b,c,d,e].

I think something maybe to do with slots, but my poor attempts have still failed, eg.

 Out[1] /. f[##] &@g[##,e]



 Out[1] /. f[___] -> g



 Out[1] /. f[#1, #2, #3, #4] -> g[#1, #2, #3, #4, e]

functions argument-patterns

share|improve this question

asked 9 hours ago

natenate

27511 silver badge77 bronze badges

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  You're mixing the concepts of pure functions (# and &) with replacement (/., ->, :>). Use slots with the first one, use patterns (Pattern in the documentation, x_ or x__ for example in usage) with the second. E.g. f[a, b, c, d] /. f[x__] :> g[x, e].
  $endgroup$
  – eyorble
  9 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Aargh. This comment is an answer! I'll accept it if you want
  $endgroup$
  – nate
  9 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

So I have a function that produces answers involving uninstantiated(?) function names, e.g.

 Out[1]= f[a,b,c,d].

I then want to evalute this leftover result by, say, replacing this f[a,b,c,d] with a defined function g[a,b,c,d,e].

I think something maybe to do with slots, but my poor attempts have still failed, eg.

 Out[1] /. f[##] &@g[##,e]



 Out[1] /. f[___] -> g



 Out[1] /. f[#1, #2, #3, #4] -> g[#1, #2, #3, #4, e]

functions argument-patterns

share|improve this question

asked 9 hours ago

natenate

27511 silver badge77 bronze badges

$endgroup$

 Out[1]= f[a,b,c,d].

 Out[1] /. f[##] &@g[##,e]



 Out[1] /. f[___] -> g



 Out[1] /. f[#1, #2, #3, #4] -> g[#1, #2, #3, #4, e]

share|improve this question

asked 9 hours ago

natenate

27511 silver badge77 bronze badges

share|improve this question

asked 9 hours ago

natenate

27511 silver badge77 bronze badges

asked 9 hours ago

natenate

27511 silver badge77 bronze badges

asked 9 hours ago

natenate

27511 silver badge77 bronze badges

2 Answers
2

active

oldest

votes

6

$begingroup$

From my own comment:

You're mixing the concepts of pure functions (# and &) with replacement (/., ->, :>). Use slots with the first one, use patterns (Pattern in the documentation, x_ or x__ for example in usage) with the second.

For this problem:

f[a, b, c, d] /. f[x__] :> g[x, e]

g[a, b, c, d, e]

Note the use of __, which is two underscores: This is a pattern which grabs 1 or more elements (see also ___ which grabs 0 or more elements), and inserts them as a Sequence when used in replacement. Thus, it doesn't get inserted as a List.

If you wanted to do this with pure functions, it becomes a bit more complicated:

g[Sequence @@ #, e] &[f[a, b, c, d]]

g[a, b, c, d, e]

But note that this isn't dependent on its argument being in the form of f[...], it will replace any functional head. To avoid that requires conditionals of some variety, e.g:

If[Head[#] === f, g[Sequence @@ #, e], Undefined] &[f[a, b, c, d]]

share|improve this answer

edited 9 hours ago

answered 9 hours ago

eyorbleeyorble

6,49311 gold badge1111 silver badges3030 bronze badges

$endgroup$

add a comment | 

5

$begingroup$

You can also use Apply:

g[##, e] & @@ f[a, b, c, d]

g[a, b, c, d, e]

or ReplaceAll with replacement rule f -> (g[##, e] &):

f[a, b, c, d] /.  f -> (g[##, e] &)

g[a, b, c, d, e]

share|improve this answer

answered 8 hours ago

kglrkglr

210k1010 gold badges241241 silver badges480480 bronze badges

$endgroup$

add a comment | 

Your Answer

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
});


}
});

draft saved

draft discarded

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

Name

Email

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f203569%2fchanging-number-of-arguments-to-a-function-in-secondary-evaluation%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

6

$begingroup$

From my own comment:

You're mixing the concepts of pure functions (# and &) with replacement (/., ->, :>). Use slots with the first one, use patterns (Pattern in the documentation, x_ or x__ for example in usage) with the second.

For this problem:

f[a, b, c, d] /. f[x__] :> g[x, e]

g[a, b, c, d, e]

Note the use of __, which is two underscores: This is a pattern which grabs 1 or more elements (see also ___ which grabs 0 or more elements), and inserts them as a Sequence when used in replacement. Thus, it doesn't get inserted as a List.

If you wanted to do this with pure functions, it becomes a bit more complicated:

g[Sequence @@ #, e] &[f[a, b, c, d]]

g[a, b, c, d, e]

But note that this isn't dependent on its argument being in the form of f[...], it will replace any functional head. To avoid that requires conditionals of some variety, e.g:

If[Head[#] === f, g[Sequence @@ #, e], Undefined] &[f[a, b, c, d]]

share|improve this answer

edited 9 hours ago

answered 9 hours ago

eyorbleeyorble

6,49311 gold badge1111 silver badges3030 bronze badges

$endgroup$

add a comment | 

6

$begingroup$

From my own comment:

You're mixing the concepts of pure functions (# and &) with replacement (/., ->, :>). Use slots with the first one, use patterns (Pattern in the documentation, x_ or x__ for example in usage) with the second.

For this problem:

f[a, b, c, d] /. f[x__] :> g[x, e]

g[a, b, c, d, e]

Note the use of __, which is two underscores: This is a pattern which grabs 1 or more elements (see also ___ which grabs 0 or more elements), and inserts them as a Sequence when used in replacement. Thus, it doesn't get inserted as a List.

If you wanted to do this with pure functions, it becomes a bit more complicated:

g[Sequence @@ #, e] &[f[a, b, c, d]]

g[a, b, c, d, e]

But note that this isn't dependent on its argument being in the form of f[...], it will replace any functional head. To avoid that requires conditionals of some variety, e.g:

If[Head[#] === f, g[Sequence @@ #, e], Undefined] &[f[a, b, c, d]]

share|improve this answer

edited 9 hours ago

answered 9 hours ago

eyorbleeyorble

6,49311 gold badge1111 silver badges3030 bronze badges

$endgroup$

add a comment | 

$begingroup$

From my own comment:

You're mixing the concepts of pure functions (# and &) with replacement (/., ->, :>). Use slots with the first one, use patterns (Pattern in the documentation, x_ or x__ for example in usage) with the second.

For this problem:

f[a, b, c, d] /. f[x__] :> g[x, e]

g[a, b, c, d, e]

Note the use of __, which is two underscores: This is a pattern which grabs 1 or more elements (see also ___ which grabs 0 or more elements), and inserts them as a Sequence when used in replacement. Thus, it doesn't get inserted as a List.

If you wanted to do this with pure functions, it becomes a bit more complicated:

g[Sequence @@ #, e] &[f[a, b, c, d]]

g[a, b, c, d, e]

But note that this isn't dependent on its argument being in the form of f[...], it will replace any functional head. To avoid that requires conditionals of some variety, e.g:

If[Head[#] === f, g[Sequence @@ #, e], Undefined] &[f[a, b, c, d]]

share|improve this answer

edited 9 hours ago

answered 9 hours ago

eyorbleeyorble

6,49311 gold badge1111 silver badges3030 bronze badges

$endgroup$

f[a, b, c, d] /. f[x__] :> g[x, e]

g[Sequence @@ #, e] &[f[a, b, c, d]]

If[Head[#] === f, g[Sequence @@ #, e], Undefined] &[f[a, b, c, d]]

share|improve this answer

edited 9 hours ago

answered 9 hours ago

eyorbleeyorble

6,49311 gold badge1111 silver badges3030 bronze badges

answered 9 hours ago

eyorbleeyorble

6,49311 gold badge1111 silver badges3030 bronze badges

answered 9 hours ago

eyorbleeyorble

6,49311 gold badge1111 silver badges3030 bronze badges

5

$begingroup$

You can also use Apply:

g[##, e] & @@ f[a, b, c, d]

g[a, b, c, d, e]

or ReplaceAll with replacement rule f -> (g[##, e] &):

f[a, b, c, d] /.  f -> (g[##, e] &)

g[a, b, c, d, e]

share|improve this answer

answered 8 hours ago

kglrkglr

210k1010 gold badges241241 silver badges480480 bronze badges

$endgroup$

add a comment | 

5

$begingroup$

You can also use Apply:

g[##, e] & @@ f[a, b, c, d]

g[a, b, c, d, e]

or ReplaceAll with replacement rule f -> (g[##, e] &):

f[a, b, c, d] /.  f -> (g[##, e] &)

g[a, b, c, d, e]

share|improve this answer

answered 8 hours ago

kglrkglr

210k1010 gold badges241241 silver badges480480 bronze badges

$endgroup$

add a comment | 

$begingroup$

You can also use Apply:

g[##, e] & @@ f[a, b, c, d]

g[a, b, c, d, e]

or ReplaceAll with replacement rule f -> (g[##, e] &):

f[a, b, c, d] /.  f -> (g[##, e] &)

g[a, b, c, d, e]

share|improve this answer

answered 8 hours ago

kglrkglr

210k1010 gold badges241241 silver badges480480 bronze badges

$endgroup$

g[##, e] & @@ f[a, b, c, d]

f[a, b, c, d] /.  f -> (g[##, e] &)

share|improve this answer

answered 8 hours ago

kglrkglr

210k1010 gold badges241241 silver badges480480 bronze badges

answered 8 hours ago

kglrkglr

210k1010 gold badges241241 silver badges480480 bronze badges

answered 8 hours ago

kglrkglr

210k1010 gold badges241241 silver badges480480 bronze badges