我正在尝试制作一个“有利于”较低值的颜色图,即从较深的颜色到浅色需要更长的时间。目前我正在使用它作为颜色图:
cmap = clr.LinearSegmentedColormap.from_list('custom blue', ['#ffff00','#002266'], N=256)
我正在围绕一个圆柱体绘制它以查看效果(参见帖子末尾的圆柱体代码),这是运行代码时发生的情况:
如您所见,这是非常“线性”的。颜色开始在圆柱体的一半左右发生变化。有没有办法增加颜色开始快速变化的阈值? IE。我只希望非常高的数字具有最亮的黄色。谢谢。
from matplotlib import cm
from matplotlib import pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
from scipy.linalg import norm
from mpl_toolkits.mplot3d.art3d import Poly3DCollection
import numpy as np
import math
import mpl_toolkits.mplot3d.art3d as art3d
import matplotlib.colors as clr
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
origin = [0,0,0]
#radius = R
p0 = np.array(origin)
p1 = np.array([8, 8, 8])
origin = np.array(origin)
R = 1
#vector in direction of axis
v = p1 - p0
#find magnitude of vector
mag = norm(v)
#unit vector in direction of axis
v = v / mag
#make some vector not in the same direction as v
not_v = np.array([1, 0, 0])
if (v == not_v).all():
not_v = np.array([0, 1, 0])
#make vector perpendicular to v
n1 = np.cross(v, not_v)
#normalize n1
n1 /= norm(n1)
#make unit vector perpendicular to v and n1
n2 = np.cross(v, n1)
#surface ranges over t from 0 to length of axis and 0 to 2*pi
t = np.linspace(0, mag, 600)
theta = np.linspace(0, 2 * np.pi, 100)
#use meshgrid to make 2d arrays
t, theta = np.meshgrid(t, theta)
#generate coordinates for surface
X, Y, Z = [p0[i] + v[i] * t + R * np.sin(theta) * n1[i] + R * np.cos(theta) * n2[i] for i in [0, 1, 2]]
#THIS IS WHERE THE COLOR MAP IS
cmap = clr.LinearSegmentedColormap.from_list('custom blue', ['#ffff00','#002266'], N=256)
col1 = cmap(np.linspace(0,1,600)) # linear gradient along the t-axis
col1 = np.repeat(col1[np.newaxis,:, :], 100, axis=0) # expand over the theta- axis
ax.plot_surface(X, Y,Z, facecolors = col1, shade = True,edgecolors = "None", alpha = 0.9, linewidth = 0)
ax.view_init(15,-40)
plt.show()
最佳答案
使用 LinearSegmentedColormap.from_list
制作颜色图时,您可以提供形式为 (value, color) 的元组列表(而不是简单的颜色列表),其中值对应于相对颜色的位置。值的范围必须从 0
到 1
,因此您必须提供中间颜色。在你的情况下,我可以试试这个,
cmap = clr.LinearSegmentedColormap.from_list('custom blue',
[(0, '#ffff00'),
(0.25, '#002266'),
(1, '#002266')], N=256)
并调整颜色/值直到满意为止。归功于https://stackoverflow.com/a/25000108/5285918
关于python - 如何将 'LinearSegmentedColormap' 更改为不同的颜色分布?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/38147997/