在Matplotlib中调整坐标轴夹角至60度的实现问题

你之前的代码仅调整了坐标轴的显示变换，但散点坐标未应用同样的规则，导致点位置不变。要实现原点、(1,0)、(0,1)构成等边三角形，需要让散点坐标和坐标轴按相同逻辑转换。

方法一：手动转换坐标（直观易控） #

直接将原始(x,y)坐标映射为符合60度夹角的笛卡尔坐标，再绘制自定义斜坐标轴：

import matplotlib.pyplot as plt
import numpy as np

# 原始数据
x = [0.41729137, 0.31693662, 0.25544209, 0.13843966, 0.31218805,0.01035702, 0.4847026 , 0.11104309, 0.25144685, 0.18332651]
y = [0.20372773, 0.5395082 , 0.6020569 , 0.19787512, 0.57743819,0.93317626, 0.43035489, 0.8235019 , 0.36000727, 0.24365175]

# 坐标转换规则：让(0,1)映射到(0.5, √3/2)，保证与(1,0)、原点构成等边三角形
cos60 = 0.5
sin60 = np.sqrt(3)/2
new_x = [xi + yi * cos60 for xi, yi in zip(x, y)]
new_y = [yi * sin60 for yi in y]

fig, ax = plt.subplots()
ax.scatter(new_x, new_y, color='#1f77b4')

# 绘制自定义斜坐标轴
ax.plot([0, 1], [0, 0], 'k-', linewidth=1.5)  # x轴
ax.plot([0, cos60], [0, sin60], 'k-', linewidth=1.5)  # y轴

# 标注坐标轴
ax.text(1.02, 0, 'x', ha='left', va='center')
ax.text(cos60, sin60+0.02, 'y', ha='center', va='bottom')

# 设置显示范围
ax.set_xlim(-0.15, 1.15)
ax.set_ylim(-0.15, 1.0)

# 隐藏默认笛卡尔坐标轴
ax.spines[['top', 'right', 'left']].set_visible(False)
ax.yaxis.set_visible(False)

plt.show()

方法二：利用FloatingAxes的变换（保留网格刻度） #

修正你原有的代码，给散点坐标应用定义好的变换矩阵：

import matplotlib.pyplot as plt
from matplotlib.transforms import Affine2D
import mpl_toolkits.axisartist.floating_axes as floating_axes
import numpy as np

fig = plt.figure()
plot_extents = 0, 1, 0, 1

# 定义正确的变换矩阵：等价于x'=x+0.5y，y'=y*(√3/2)
cos60 = 0.5
sin60 = np.sqrt(3)/2
transform = Affine2D.from_values(1, cos60, 0, sin60, 0, 0)

helper = floating_axes.GridHelperCurveLinear(transform, plot_extents)
ax = floating_axes.FloatingSubplot(fig, 111, grid_helper=helper)
fig.add_axes(ax)

# 原始数据
x = [0.41729137, 0.31693662, 0.25544209, 0.13843966, 0.31218805,0.01035702, 0.4847026 , 0.11104309, 0.25144685, 0.18332651]
y = [0.20372773, 0.5395082 , 0.6020569 , 0.19787512, 0.57743819,0.93317626, 0.43035489, 0.8235019 , 0.36000727, 0.24365175]

# 将原始坐标应用变换后绘制
x_transformed, y_transformed = transform.transform(np.array([x, y]).T).T
ax.scatter(x_transformed, y_transformed, color='#1f77b4')

# 设置显示范围
ax.set_xlim(-0.15, 1.15)
ax.set_ylim(-0.15, 1.15)

plt.show()

关键说明 #

你原代码的核心问题是：floating_axes只给坐标轴和网格应用了变换，但ax.scatter默认使用原始笛卡尔坐标系绘制点。必须将变换矩阵同时应用到散点坐标上，才能让点和坐标轴同步变换，形成目标等边三角形区域。

内容的提问来源于stack exchange，提问作者Matthew Leingang