Hello,
Does someone know how to programmatically know which editable shape is currently activated ?
I would like to allow the user to delete the selected shape by clicking a button. The only way found, was by forcing the user to first move the shape (the fillcolor is changed to red using the “on_change” callback). Then, I can detect the Red shape and delete it.
My approach is not elegant and apparently, the callback “on_change” is lost after the first click on the button (probably because the way I use to delete the shape but I haven’t found any proposal in the documentation)
Any help is more than welcome !
Patrick
import plotly.graph_objects as go
from ipywidgets import Button, VBox
fig = go.FigureWidget()
# Set axes properties
fig.update_xaxes(range=[0, 7], showgrid=False)
fig.update_yaxes(range=[0, 3.5])
shp1 = go.layout.Shape(
type="rect",
x0=1,
y0=1,
x1=2,
y1=3,
fillcolor="Blue",
editable=True)
shp2 = go.layout.Shape(
type="rect",
x0=3,
y0=1,
x1=6,
y1=2,
fillcolor="Blue",
editable=True)
# Add shapes
fig.add_shape(shp1)
fig.add_shape(shp2)
fig.update_shapes(dict(xref='x', yref='y'))
def change_color(shp,*args,**kwargs):
shp.update(dict(fillcolor='Red'))
def delete_shapes(*args,**kwargs):
with output:
try:
list_shapes=[shp for shp in fig.layout.shapes if shp.fillcolor!='Red']
self.fig.update_layout(shapes=list_shapes)
except Exception as e:
raise e
for shp in fig.layout.shapes:
shp.on_change(change_color,'x0','y0','x1','y1')
button.on_click(delete_shapes)
VBox([button,fig])
