Arc Shape with Path

[eekk]

Hey, I’ve been trying to plot some shapes in plotly, but I can’t seem to get an arc shape to work.

I.e., using the SVG

M 10 10 A 10 10 0 0 1 20 10

for example, like

trace=dict(x=[5,10], y=[5,10],
               mode='markers', 
               marker=dict(size=9, color='orange'))

axis=dict(showline=True, zeroline=False, showgrid=False)

layout=dict(width=450, height=450, autosize=False,
            xaxis=dict(axis, **dict(range=[1,20])),  
            yaxis=dict(axis,**dict(range=[1, 20])),
            shapes=[dict(type='path',
                    layer='below',
                    path="M 10 10 A 10 10 0 0 1 20 10",
                    fillcolor='red',
                    line=dict(color= 'red')
          
    )]
            )
fig=dict(data=[trace], layout=layout)
iplot(fig)

doesn’t plot the shape at all. Is there a workaround to make the arc command work ?

[empet]

@eek Plotly does not provide the command A for elliptical arc.

Here https://github.com/plotly/plotly.js/blob/master/src/components/shapes/constants.js are listed the available commands for paths.

[extraymond]

Thx for your reply! No wonder A is always not working in the path command.
Do you happen to know how can I draw a arc with custom degree(lets say 90 degree circle)?

[openafox]

Any word on how to do this or support for arc paths?

[empet]

@extraymond, @openafox

You can draw both arcs of circle or ellipse, and filled circle or ellipse calling a function that defines the corresponding svg paths:

import numpy as np
import plotly.graph_objects as go

def ellipse_arc(x_center=0, y_center=0, a=1, b =1, start_angle=0, end_angle=2*np.pi, N=100, closed= False):
    t = np.linspace(start_angle, end_angle, N)
    x = x_center + a*np.cos(t)
    y = y_center + b*np.sin(t)
    path = f'M {x[0]}, {y[0]}'
    for k in range(1, len(t)):
        path += f'L{x[k]}, {y[k]}'
    if closed:
        path += ' Z'
    return path    

fig = go.Figure()

# Create a minimal trace 
fig.add_trace(go.Scatter(
    x=[0],
    y=[0.2],
    marker_size=0.1
));

fig.update_layout(width =600, height=400,
           xaxis_range=[-5.2, 5.2], 
           yaxis_range=[-3.2, 3.2],       
    shapes=[
           dict(type="path",
           path= ellipse_arc(a=5, b=3, start_angle=-np.pi/6, end_angle=3*np.pi/2, N=60),
           line_color="RoyalBlue"),
           dict(type="path",
                path = ellipse_arc(x_center=2, a= 0.5, b= 1.5, closed=True),
                fillcolor="LightPink",
                line_color="Crimson")
        ]
        );
fig.show()

If a=b you’ll get an arc of circle (if layout.width=layout.height), otherwise an arc of ellipse.

[mutia]

hello If I want to change your script to giving the parameter x, y, width, height with(width and height as diameter) like

matplotlib ellipse function

https://matplotlib.org/api/_as_gen/matplotlib.patches.Ellipse.html

what is the modification that I need to do?

I tried to replicate this matplotlib function to plotly but I don’t know how, I’ve tried to see the source code but there is no formula to draw the ellipse in there

[empet]

Hi @mutia

width and height in the matplotlib ellipse definition are 2*a, 2*b while xy =[x_center, y_center].