bzm3r
bzm3r

Reputation: 4596

How to get patches to draw with equal aspect ratio even though associated axis does not have equal aspect ratio?

Consider the following toy code:

# -*- coding: utf-8 -*-
import numpy as np
import matplotlib.pyplot as plt
import matplotlib.patches as mpatches

def draw_circle_arrangement(ax, drawing_origin, drawing_space, scale, num_circles, box_height, box_width):
    bw = drawing_space*(box_width*scale)

    drawing_origin[0] = drawing_origin[0] - bw*0.5

    circle_diameter = drawing_space*scale
    circle_radius = 0.5*circle_diameter
    y_delta = np.array([0., circle_diameter])
    x_delta = np.array([circle_diameter, 0.])

    cell_origin = drawing_origin + np.array([circle_radius, circle_radius])
    y_delta_index = 0
    x_delta_index = 0

    for ci in range(num_circles):
        cell_patch = mpatches.Circle(cell_origin + y_delta_index*y_delta + x_delta_index*x_delta, radius=circle_radius, color='k', fill=False, ls='solid', clip_on=False)
        ax.add_artist(cell_patch)

        if y_delta_index == box_height - 1:
            y_delta_index = 0
            x_delta_index += 1
        else:
            y_delta_index += 1


fig, ax = plt.subplots()

# each tuple is: number of circles, height of box containing circles, width of box containing circle
circle_arrangements = [(10, 2, 5), (3, 1, 3), (1, 1, 1)]
data = np.random.rand(3)
ax.set_ylim([0, 1])
ax.plot(np.arange(3) + 1, data, marker='o')
ax.get_xaxis().set_ticklabels([])
scale = 1./10.

for i, ca in enumerate(circle_arrangements):
    do = np.array([1.0 + i, -0.2])
    nc, bh, bw = ca
    draw_circle_arrangement(ax, do, 0.8, scale, nc, bh, bw)

When run, it produces output like so: enter image description here

As you can see, the circles are not circles! They are squished. One way to fix this is by setting ax.set_aspect('equal'), but if I don't necessarily want the axis aspect ratio to be equal, how can I still have the patches produced that way?

Upvotes: 2

Views: 366

Answers (1)

ImportanceOfBeingErnest
ImportanceOfBeingErnest

Reputation: 339170

It is not possible to only set part of the axes aspect to equal. The solution would be either to draw the circles in a different coordinate system, or, by using a different box that uses a different coordinate system.

I find the latter solution preferable. That would involve using a DrawingArea which can be placed inside an AnnotationBbox.

import numpy as np
import matplotlib.pyplot as plt
import matplotlib.patches as mpatches
from matplotlib.offsetbox import  DrawingArea, AnnotationBbox

def draw_circle_arrangement(ax, drawing_origin, radius, 
                            num_circles, box_height, box_width):
    y_delta_index = 0
    x_delta_index = 0

    origin = np.array([radius,radius])
    y_delta = np.array([0., 2*radius])
    x_delta = np.array([2*radius, 0.])

    da = DrawingArea(box_width*2*radius, box_height*2*radius, 0, 0)

    for ci in range(num_circles):
        cell_patch = mpatches.Circle(origin+y_delta_index*y_delta + x_delta_index*x_delta, 
                                     radius=radius, color='k', fill=False, ls='solid',clip_on=False)
        da.add_artist(cell_patch)

        if y_delta_index == box_height - 1:
            y_delta_index = 0
            x_delta_index += 1
        else:
            y_delta_index += 1
    ab = AnnotationBbox(da, xy=(drawing_origin[0],0),
                        xybox=drawing_origin,
                        xycoords=("data", "axes fraction"),
                        boxcoords=("data", "axes fraction"),
                        box_alignment=(0.5,0.5), frameon=False)

    ax.add_artist(ab)


fig, ax = plt.subplots()
# each tuple is: number of circles, height of box containing circles, width of 
#box containing circle
circle_arrangements = [(10, 2, 5), (3, 1, 3), (1, 1, 1)]
data = np.random.rand(3)
ax.set_ylim([0, 1])
ax.plot(np.arange(3) + 1, data, marker='o')
ax.get_xaxis().set_ticklabels([])


for i, ca in enumerate(circle_arrangements):
    do = np.array([1.0 + i, -0.06])
    nc, bh, bw = ca
    draw_circle_arrangement(ax, do, 5, nc, bh, bw)

plt.show()

enter image description here

The units of the DrawingArea are points. E.g. in the first case we create a DrawingArea of 5*2*5 = 50 points width and place 5 circles with a radius of 5 points into it (the 5 circles fill the complete 50 points then). A point is ~1.4 pixels. The position of the AnnotationBox is given in data coordinates for the x position and in axes fraction for the y position. This can be changed by using the boxcoords argument. Same for the xy argument, where we make sure that the y coordinate is at 0, and thus inside the axes (such that it AnnotationsBox is shown).

Upvotes: 3

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