Set logical status of CheckBox in matplotlib without triggering callbacks?
Is there a way to set the logical status of CheckButtons in matplotlib withput triggering onclick()
callbacks?
Here is my scenario:
I am displaying a graph with a large number of traces (50+) and want to give the user the option of displaying a subset of these on the basis of some fixed queries, which the user selects via CheckButton
. Every time a button is clicked the the script gets the status of all present CheckButton
s via get_status()
, computes the indices of the traces that meet the selected criteria, and updates the display accordingly.
In order to make life easier for the user I want to also have a "Select all" and "Clear all" buttons. The best way I can come up with to implement these is to force the logical state of all CheckButtons to True or False without triggering an on_click()
callback, then set the visibility of the traces accordingly.
However, CheckButton()
does not have a set_status()
method
https://matplotlib.org/api/widgets_api.html#matplotlib.widgets.CheckButtons
There is a set_active()
method but according the documetnation it triggers a callback if eventson
is true.
Can someone give me a couple of pointers how to go about doing this? Do I need to set eventson=False
for the wdget, then set_active()=True
followed by eventson=True
? Seems rather clunky, but I can't figure a more elegant way, even after spending way too many hours.
Thanks!
python python-2.7 matplotlib matplotlib-widget
add a comment |
Is there a way to set the logical status of CheckButtons in matplotlib withput triggering onclick()
callbacks?
Here is my scenario:
I am displaying a graph with a large number of traces (50+) and want to give the user the option of displaying a subset of these on the basis of some fixed queries, which the user selects via CheckButton
. Every time a button is clicked the the script gets the status of all present CheckButton
s via get_status()
, computes the indices of the traces that meet the selected criteria, and updates the display accordingly.
In order to make life easier for the user I want to also have a "Select all" and "Clear all" buttons. The best way I can come up with to implement these is to force the logical state of all CheckButtons to True or False without triggering an on_click()
callback, then set the visibility of the traces accordingly.
However, CheckButton()
does not have a set_status()
method
https://matplotlib.org/api/widgets_api.html#matplotlib.widgets.CheckButtons
There is a set_active()
method but according the documetnation it triggers a callback if eventson
is true.
Can someone give me a couple of pointers how to go about doing this? Do I need to set eventson=False
for the wdget, then set_active()=True
followed by eventson=True
? Seems rather clunky, but I can't figure a more elegant way, even after spending way too many hours.
Thanks!
python python-2.7 matplotlib matplotlib-widget
add a comment |
Is there a way to set the logical status of CheckButtons in matplotlib withput triggering onclick()
callbacks?
Here is my scenario:
I am displaying a graph with a large number of traces (50+) and want to give the user the option of displaying a subset of these on the basis of some fixed queries, which the user selects via CheckButton
. Every time a button is clicked the the script gets the status of all present CheckButton
s via get_status()
, computes the indices of the traces that meet the selected criteria, and updates the display accordingly.
In order to make life easier for the user I want to also have a "Select all" and "Clear all" buttons. The best way I can come up with to implement these is to force the logical state of all CheckButtons to True or False without triggering an on_click()
callback, then set the visibility of the traces accordingly.
However, CheckButton()
does not have a set_status()
method
https://matplotlib.org/api/widgets_api.html#matplotlib.widgets.CheckButtons
There is a set_active()
method but according the documetnation it triggers a callback if eventson
is true.
Can someone give me a couple of pointers how to go about doing this? Do I need to set eventson=False
for the wdget, then set_active()=True
followed by eventson=True
? Seems rather clunky, but I can't figure a more elegant way, even after spending way too many hours.
Thanks!
python python-2.7 matplotlib matplotlib-widget
Is there a way to set the logical status of CheckButtons in matplotlib withput triggering onclick()
callbacks?
Here is my scenario:
I am displaying a graph with a large number of traces (50+) and want to give the user the option of displaying a subset of these on the basis of some fixed queries, which the user selects via CheckButton
. Every time a button is clicked the the script gets the status of all present CheckButton
s via get_status()
, computes the indices of the traces that meet the selected criteria, and updates the display accordingly.
In order to make life easier for the user I want to also have a "Select all" and "Clear all" buttons. The best way I can come up with to implement these is to force the logical state of all CheckButtons to True or False without triggering an on_click()
callback, then set the visibility of the traces accordingly.
However, CheckButton()
does not have a set_status()
method
https://matplotlib.org/api/widgets_api.html#matplotlib.widgets.CheckButtons
There is a set_active()
method but according the documetnation it triggers a callback if eventson
is true.
Can someone give me a couple of pointers how to go about doing this? Do I need to set eventson=False
for the wdget, then set_active()=True
followed by eventson=True
? Seems rather clunky, but I can't figure a more elegant way, even after spending way too many hours.
Thanks!
python python-2.7 matplotlib matplotlib-widget
python python-2.7 matplotlib matplotlib-widget
asked Nov 13 '18 at 7:51
GroovyGeekGroovyGeek
11
11
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
I knew it, as soon as I post a question I figure out the answer. Here is what I came up with for anyone else that wants to do something like this
check.eventson = False
status = check.get_status()
for i,stat in enumerate(status):
if ((enable and not stat) or (not enable and stat)):
check.set_active(i)
check.eventson = True
here check is a reference to the CheckBox()
object and enable
is a variable set to True
if you want to enable all checkboxes, or False
otherwise.
add a comment |
Your Answer
StackExchange.ifUsing("editor", function ()
StackExchange.using("externalEditor", function ()
StackExchange.using("snippets", function ()
StackExchange.snippets.init();
);
);
, "code-snippets");
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "1"
;
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: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
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%2fstackoverflow.com%2fquestions%2f53276214%2fset-logical-status-of-checkbox-in-matplotlib-without-triggering-callbacks%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
I knew it, as soon as I post a question I figure out the answer. Here is what I came up with for anyone else that wants to do something like this
check.eventson = False
status = check.get_status()
for i,stat in enumerate(status):
if ((enable and not stat) or (not enable and stat)):
check.set_active(i)
check.eventson = True
here check is a reference to the CheckBox()
object and enable
is a variable set to True
if you want to enable all checkboxes, or False
otherwise.
add a comment |
I knew it, as soon as I post a question I figure out the answer. Here is what I came up with for anyone else that wants to do something like this
check.eventson = False
status = check.get_status()
for i,stat in enumerate(status):
if ((enable and not stat) or (not enable and stat)):
check.set_active(i)
check.eventson = True
here check is a reference to the CheckBox()
object and enable
is a variable set to True
if you want to enable all checkboxes, or False
otherwise.
add a comment |
I knew it, as soon as I post a question I figure out the answer. Here is what I came up with for anyone else that wants to do something like this
check.eventson = False
status = check.get_status()
for i,stat in enumerate(status):
if ((enable and not stat) or (not enable and stat)):
check.set_active(i)
check.eventson = True
here check is a reference to the CheckBox()
object and enable
is a variable set to True
if you want to enable all checkboxes, or False
otherwise.
I knew it, as soon as I post a question I figure out the answer. Here is what I came up with for anyone else that wants to do something like this
check.eventson = False
status = check.get_status()
for i,stat in enumerate(status):
if ((enable and not stat) or (not enable and stat)):
check.set_active(i)
check.eventson = True
here check is a reference to the CheckBox()
object and enable
is a variable set to True
if you want to enable all checkboxes, or False
otherwise.
answered Nov 13 '18 at 8:16
GroovyGeekGroovyGeek
11
11
add a comment |
add a comment |
Thanks for contributing an answer to Stack Overflow!
- 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.
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%2fstackoverflow.com%2fquestions%2f53276214%2fset-logical-status-of-checkbox-in-matplotlib-without-triggering-callbacks%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