Define an access by key ( structure[key] ) for a custom structure struct?
$begingroup$
I have some "complicated" data that I want to encapsulate into a "struct":
For instance (imagine that data have several fields):
data=<|"A"->6,"B"->2|>
var=myType[data]
The advantage is that it is easy to filter your arguments:
doSomething[myType[d_]]:=d
(* usage: *)
doSomething[var]
Now I want to allow this:
var["B"] <- must return 2
But I do not know how to do that, any idea?
What I have done so far:
myType /: Key[k_][myType[d_]:=d[k]
This works for syntax like:
Key["B"][var]
but not for
var["B"]
associations
$endgroup$
add a comment |
$begingroup$
I have some "complicated" data that I want to encapsulate into a "struct":
For instance (imagine that data have several fields):
data=<|"A"->6,"B"->2|>
var=myType[data]
The advantage is that it is easy to filter your arguments:
doSomething[myType[d_]]:=d
(* usage: *)
doSomething[var]
Now I want to allow this:
var["B"] <- must return 2
But I do not know how to do that, any idea?
What I have done so far:
myType /: Key[k_][myType[d_]:=d[k]
This works for syntax like:
Key["B"][var]
but not for
var["B"]
associations
$endgroup$
add a comment |
$begingroup$
I have some "complicated" data that I want to encapsulate into a "struct":
For instance (imagine that data have several fields):
data=<|"A"->6,"B"->2|>
var=myType[data]
The advantage is that it is easy to filter your arguments:
doSomething[myType[d_]]:=d
(* usage: *)
doSomething[var]
Now I want to allow this:
var["B"] <- must return 2
But I do not know how to do that, any idea?
What I have done so far:
myType /: Key[k_][myType[d_]:=d[k]
This works for syntax like:
Key["B"][var]
but not for
var["B"]
associations
$endgroup$
I have some "complicated" data that I want to encapsulate into a "struct":
For instance (imagine that data have several fields):
data=<|"A"->6,"B"->2|>
var=myType[data]
The advantage is that it is easy to filter your arguments:
doSomething[myType[d_]]:=d
(* usage: *)
doSomething[var]
Now I want to allow this:
var["B"] <- must return 2
But I do not know how to do that, any idea?
What I have done so far:
myType /: Key[k_][myType[d_]:=d[k]
This works for syntax like:
Key["B"][var]
but not for
var["B"]
associations
associations
edited Feb 1 at 14:19
Picaud Vincent
asked Feb 1 at 11:49


Picaud VincentPicaud Vincent
1,247518
1,247518
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
ClearAll[myType]
myType /: myType[data_Association][s_String] := data[[s]]
var = myType[<|"A" -> 6, "B" -> 2|>];
var["B"]
2
$endgroup$
$begingroup$
Thanks again Henrik, my questions are too easy :-)
$endgroup$
– Picaud Vincent
Feb 1 at 12:09
$begingroup$
You're welcome.
$endgroup$
– Henrik Schumacher
Feb 1 at 12:12
3
$begingroup$
This definitely works, though it's good to notice that this sets aSubValue
formyType
, not anUpValue
.
$endgroup$
– Sjoerd Smit
Feb 1 at 13:35
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%2f190653%2fdefine-an-access-by-key-structurekey-for-a-custom-structure-struct%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
ClearAll[myType]
myType /: myType[data_Association][s_String] := data[[s]]
var = myType[<|"A" -> 6, "B" -> 2|>];
var["B"]
2
$endgroup$
$begingroup$
Thanks again Henrik, my questions are too easy :-)
$endgroup$
– Picaud Vincent
Feb 1 at 12:09
$begingroup$
You're welcome.
$endgroup$
– Henrik Schumacher
Feb 1 at 12:12
3
$begingroup$
This definitely works, though it's good to notice that this sets aSubValue
formyType
, not anUpValue
.
$endgroup$
– Sjoerd Smit
Feb 1 at 13:35
add a comment |
$begingroup$
ClearAll[myType]
myType /: myType[data_Association][s_String] := data[[s]]
var = myType[<|"A" -> 6, "B" -> 2|>];
var["B"]
2
$endgroup$
$begingroup$
Thanks again Henrik, my questions are too easy :-)
$endgroup$
– Picaud Vincent
Feb 1 at 12:09
$begingroup$
You're welcome.
$endgroup$
– Henrik Schumacher
Feb 1 at 12:12
3
$begingroup$
This definitely works, though it's good to notice that this sets aSubValue
formyType
, not anUpValue
.
$endgroup$
– Sjoerd Smit
Feb 1 at 13:35
add a comment |
$begingroup$
ClearAll[myType]
myType /: myType[data_Association][s_String] := data[[s]]
var = myType[<|"A" -> 6, "B" -> 2|>];
var["B"]
2
$endgroup$
ClearAll[myType]
myType /: myType[data_Association][s_String] := data[[s]]
var = myType[<|"A" -> 6, "B" -> 2|>];
var["B"]
2
answered Feb 1 at 12:01


Henrik SchumacherHenrik Schumacher
59.8k582166
59.8k582166
$begingroup$
Thanks again Henrik, my questions are too easy :-)
$endgroup$
– Picaud Vincent
Feb 1 at 12:09
$begingroup$
You're welcome.
$endgroup$
– Henrik Schumacher
Feb 1 at 12:12
3
$begingroup$
This definitely works, though it's good to notice that this sets aSubValue
formyType
, not anUpValue
.
$endgroup$
– Sjoerd Smit
Feb 1 at 13:35
add a comment |
$begingroup$
Thanks again Henrik, my questions are too easy :-)
$endgroup$
– Picaud Vincent
Feb 1 at 12:09
$begingroup$
You're welcome.
$endgroup$
– Henrik Schumacher
Feb 1 at 12:12
3
$begingroup$
This definitely works, though it's good to notice that this sets aSubValue
formyType
, not anUpValue
.
$endgroup$
– Sjoerd Smit
Feb 1 at 13:35
$begingroup$
Thanks again Henrik, my questions are too easy :-)
$endgroup$
– Picaud Vincent
Feb 1 at 12:09
$begingroup$
Thanks again Henrik, my questions are too easy :-)
$endgroup$
– Picaud Vincent
Feb 1 at 12:09
$begingroup$
You're welcome.
$endgroup$
– Henrik Schumacher
Feb 1 at 12:12
$begingroup$
You're welcome.
$endgroup$
– Henrik Schumacher
Feb 1 at 12:12
3
3
$begingroup$
This definitely works, though it's good to notice that this sets a
SubValue
for myType
, not an UpValue
.$endgroup$
– Sjoerd Smit
Feb 1 at 13:35
$begingroup$
This definitely works, though it's good to notice that this sets a
SubValue
for myType
, not an UpValue
.$endgroup$
– Sjoerd Smit
Feb 1 at 13:35
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%2f190653%2fdefine-an-access-by-key-structurekey-for-a-custom-structure-struct%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