Gre*_*reg 22 python matplotlib
到目前为止我有这个:
x,y,z = data.nonzero()
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
ax.scatter(x, y, z, zdir='z', c= 'red')
plt.savefig("plot.png")
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这创造了:

我想要做的就是伸展它以使Z轴高9倍并保持X和Y相同.我想保持相同的坐标.
到目前为止我试过这个家伙:
fig = plt.figure(figsize=(4.,35.))
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但这只是延伸了plot.png图像.
Chr*_*een 22
下面的代码示例提供了一种相对于其他轴缩放每个轴的方法.但是,要这样做,您需要修改Axes3D.get_proj函数.以下是基于matplot lib提供的示例的示例:http://matplotlib.org/1.4.0/mpl_toolkits/mplot3d/tutorial.html#line-plots
(这个答案的末尾有一个较短的版本)
from mpl_toolkits.mplot3d.axes3d import Axes3D
from mpl_toolkits.mplot3d import proj3d
import matplotlib as mpl
import numpy as np
import matplotlib.pyplot as plt
#Make sure these are floating point values:
scale_x = 1.0
scale_y = 2.0
scale_z = 3.0
#Axes are scaled down to fit in scene
max_scale=max(scale_x, scale_y, scale_z)
scale_x=scale_x/max_scale
scale_y=scale_y/max_scale
scale_z=scale_z/max_scale
#Create scaling matrix
scale = np.array([[scale_x,0,0,0],
[0,scale_y,0,0],
[0,0,scale_z,0],
[0,0,0,1]])
print scale
def get_proj_scale(self):
"""
Create the projection matrix from the current viewing position.
elev stores the elevation angle in the z plane
azim stores the azimuth angle in the x,y plane
dist is the distance of the eye viewing point from the object
point.
"""
relev, razim = np.pi * self.elev/180, np.pi * self.azim/180
xmin, xmax = self.get_xlim3d()
ymin, ymax = self.get_ylim3d()
zmin, zmax = self.get_zlim3d()
# transform to uniform world coordinates 0-1.0,0-1.0,0-1.0
worldM = proj3d.world_transformation(
xmin, xmax,
ymin, ymax,
zmin, zmax)
# look into the middle of the new coordinates
R = np.array([0.5, 0.5, 0.5])
xp = R[0] + np.cos(razim) * np.cos(relev) * self.dist
yp = R[1] + np.sin(razim) * np.cos(relev) * self.dist
zp = R[2] + np.sin(relev) * self.dist
E = np.array((xp, yp, zp))
self.eye = E
self.vvec = R - E
self.vvec = self.vvec / proj3d.mod(self.vvec)
if abs(relev) > np.pi/2:
# upside down
V = np.array((0, 0, -1))
else:
V = np.array((0, 0, 1))
zfront, zback = -self.dist, self.dist
viewM = proj3d.view_transformation(E, R, V)
perspM = proj3d.persp_transformation(zfront, zback)
M0 = np.dot(viewM, worldM)
M = np.dot(perspM, M0)
return np.dot(M, scale);
Axes3D.get_proj=get_proj_scale
"""
You need to include all the code above.
From here on you should be able to plot as usual.
"""
mpl.rcParams['legend.fontsize'] = 10
fig = plt.figure(figsize=(5,5))
ax = fig.gca(projection='3d')
theta = np.linspace(-4 * np.pi, 4 * np.pi, 100)
z = np.linspace(-2, 2, 100)
r = z**2 + 1
x = r * np.sin(theta)
y = r * np.cos(theta)
ax.plot(x, y, z, label='parametric curve')
ax.legend()
plt.show()
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标准输出:

按(1,2,3)缩放:

按(1,1,3)缩放:

我特别喜欢这种方法的原因,Swap z和x,缩放(3,1,1):

下面是代码的较短版本.
from mpl_toolkits.mplot3d.axes3d import Axes3D
from mpl_toolkits.mplot3d import proj3d
import matplotlib as mpl
import numpy as np
import matplotlib.pyplot as plt
mpl.rcParams['legend.fontsize'] = 10
fig = plt.figure(figsize=(5,5))
ax = fig.gca(projection='3d')
theta = np.linspace(-4 * np.pi, 4 * np.pi, 100)
z = np.linspace(-2, 2, 100)
r = z**2 + 1
x = r * np.sin(theta)
y = r * np.cos(theta)
"""
Scaling is done from here...
"""
x_scale=1
y_scale=1
z_scale=2
scale=np.diag([x_scale, y_scale, z_scale, 1.0])
scale=scale*(1.0/scale.max())
scale[3,3]=1.0
def short_proj():
return np.dot(Axes3D.get_proj(ax), scale)
ax.get_proj=short_proj
"""
to here
"""
ax.plot(z, y, x, label='parametric curve')
ax.legend()
plt.show()
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And*_*bis 16
请注意,下面的答案简化了补丁,但使用与@ChristianSarofeen的答案相同的基本原则.
正如其他答案中已经指出的那样,它不是当前在matplotlib中实现的功能.但是,由于您要求的只是一个3D变换,可以应用于matplotlib使用的现有投影矩阵,并且由于Python的精彩功能,这个问题可以通过一个简单的oneliner来解决:
ax.get_proj = lambda: np.dot(Axes3D.get_proj(ax), np.diag([scale_x, scale_y, scale_z, 1]))
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其中scale_x,scale_y和scale_z是的值从0到1,将相应地重新缩放沿每个轴的曲线图.ax只是可以获得的3D轴ax = fig.gca(projection='3d')
为了解释,该函数get_proj的Axes3D生成从当前视听位置的投影矩阵.将其乘以缩放矩阵:
scale_x, 0, 0
0, scale_y, 0
0, 0, scale_z
0, 0, 1
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包括缩放到渲染器使用的投影中.所以,我们在这里做的是get_proj用一个表达式替换原始函数,该表达式取原始结果get_proj并将其乘以缩放矩阵.
使用标准参数函数示例说明结果:
from mpl_toolkits.mplot3d import Axes3D
import numpy as np
import matplotlib.pyplot as plt
fig = plt.figure()
ax = fig.gca(projection='3d')
theta = np.linspace(-4 * np.pi, 4 * np.pi, 100)
z = np.linspace(-2, 2, 100)
r = z ** 2 + 1
x = r * np.sin(theta)
y = r * np.cos(theta)
# OUR ONE LINER ADDED HERE:
ax.get_proj = lambda: np.dot(Axes3D.get_proj(ax), np.diag([0.5, 0.5, 1, 1]))
ax.plot(x, y, z)
plt.show()
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对于价值观0.5, 0.5, 1,我们得到:

而对于价值观0.2, 1.0, 0.2,我们得到:

ero*_*3pi 16
就我而言,我想将 z 轴拉伸 2 倍以获得更好的点可见性
from mpl_toolkits import mplot3d
from mpl_toolkits.mplot3d import Axes3D
import matplotlib.pyplot as plt
# plt.rcParams["figure.figsize"] = (10,200)
# plt.rcParams["figure.autolayout"] = True
ax = plt.axes(projection='3d')
ax.set_box_aspect(aspect = (1,1,2))
ax.plot(dataX,dataY,dataZ)
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我看起来默认情况下,mplot3d会在很高的情节的顶部和底部留下相当多的空间.但是,您可以使用它来欺骗它fig.subplots_adjust,并将顶部和底部扩展到正常的绘图区域(即top > 1和bottom < 0).您的特定情节可能需要一些试验和错误.
我为x,y和z创建了一些随机数组,其限制类似于你的情节,并且发现下面的参数(bottom=-0.15,top = 1.2)似乎工作正常.
您可能还想更改ax.view_init以设置一个很好的视角.
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import axes3d
from numpy import random
# Make some random data with similar limits to the OP's example
x,y,z=random.rand(3,100)
z*=250
y*=800
y+=900
x*=350
x+=1200
fig=plt.figure(figsize=(4,35))
# Set the bottom and top outside the actual figure limits,
# to stretch the 3D axis
fig.subplots_adjust(bottom=-0.15,top=1.2)
ax = fig.add_subplot(111, projection='3d')
# Change the viewing angle to an agreeable one
ax.view_init(2,None)
ax.scatter(x, y, z, zdir='z', c= 'red')
plt.savefig("plot.png")
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