Question

考虑以下玩具代码：

# -*- 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)

运行时，会产生如下输出：

如您所见，圆圈不是圆圈！他们被压扁了。解决这个问题的一种方法是设置ax.set_aspect('equal')，但如果我不一定希望轴纵横比相等，我怎么还能以这种方式生成补丁呢？

Answer 1

不能仅将轴方面的一部分设置为相等。解决方案是在不同的坐标系中绘制圆，或者使用使用不同坐标系的不同框。

我发现后一种解决方案更可取。这将涉及使用可以放在DrawingArea内的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()

DrawingArea的单位是积分。例如。在第一种情况下，我们创建一个5 * 2 * 5 = 50点宽度的DrawingArea，并在其中放置5个半径为5个点的圆圈（5个圆圈填满整个50个点）。一个点是~1.4像素。 AnnotationBox的位置在x位置的数据坐标中给出，在y位置的轴分数中给出。可以使用boxcoords参数更改此设置。对于xy参数也是如此，我们确保y坐标为0，因此在轴内（这样就显示了AnnotationsBox）。

即使关联的轴没有相等的宽高比，如何以相同的宽高比绘制补丁？

1 个答案: