31

这个问题与这个有关。

我想知道的是如何将建议的解决方案应用于一堆数据（4列），例如：

0.1 0 0.1 2.0
0.1 0 1.1 -0.498121712998
0.1 0 2.1 -0.49973005075
0.1 0 3.1 -0.499916082038
0.1 0 4.1 -0.499963726586
0.1 1 0.1 -0.0181405895692
0.1 1 1.1 -0.490774988618
0.1 1 2.1 -0.498653742846
0.1 1 3.1 -0.499580747953
0.1 1 4.1 -0.499818696063
0.1 2 0.1 -0.0107079119572
0.1 2 1.1 -0.483641823093
0.1 2 2.1 -0.497582061233
0.1 2 3.1 -0.499245863438
0.1 2 4.1 -0.499673749657
0.1 3 0.1 -0.0075248589089
0.1 3 1.1 -0.476713038166
0.1 3 2.1 -0.49651497615
0.1 3 3.1 -0.498911427589
0.1 3 4.1 -0.499528887295
0.1 4 0.1 -0.00579180003048
0.1 4 1.1 -0.469979974092
0.1 4 2.1 -0.495452458086
0.1 4 3.1 -0.498577439505
0.1 4 4.1 -0.499384108904
1.1 0 0.1 302.0
1.1 0 1.1 -0.272727272727
1.1 0 2.1 -0.467336140806
1.1 0 3.1 -0.489845926622
1.1 0 4.1 -0.495610916847
1.1 1 0.1 -0.000154915998165
1.1 1 1.1 -0.148803329865
1.1 1 2.1 -0.375881358454
1.1 1 3.1 -0.453749548548
1.1 1 4.1 -0.478942841849
1.1 2 0.1 -9.03765566114e-05
1.1 2 1.1 -0.0972702806613
1.1 2 2.1 -0.314291859842
1.1 2 3.1 -0.422606253083
1.1 2 4.1 -0.463359353084
1.1 3 0.1 -6.31234088628e-05
1.1 3 1.1 -0.0720095219203
1.1 3 2.1 -0.270015786897
1.1 3 3.1 -0.395462300716
1.1 3 4.1 -0.44875793248
1.1 4 0.1 -4.84199181874e-05
1.1 4 1.1 -0.0571187054704
1.1 4 2.1 -0.236660992042
1.1 4 3.1 -0.371593983211
1.1 4 4.1 -0.4350485869
2.1 0 0.1 1102.0
2.1 0 1.1 0.328324567994
2.1 0 2.1 -0.380952380952
2.1 0 3.1 -0.462992178846
2.1 0 4.1 -0.48400342421
2.1 1 0.1 -4.25137933034e-05
2.1 1 1.1 -0.0513190921508
2.1 1 2.1 -0.224866151101
2.1 1 3.1 -0.363752470126
2.1 1 4.1 -0.430700436658
2.1 2 0.1 -2.48003822279e-05
2.1 2 1.1 -0.0310025255124
2.1 2 2.1 -0.158022037087
2.1 2 3.1 -0.29944612818
2.1 2 4.1 -0.387965424205
2.1 3 0.1 -1.73211484062e-05
2.1 3 1.1 -0.0220466245862
2.1 3 2.1 -0.12162780064
2.1 3 3.1 -0.254424041889
2.1 3 4.1 -0.35294082311
2.1 4 0.1 -1.32862131387e-05
2.1 4 1.1 -0.0170828002197
2.1 4 2.1 -0.0988138417802
2.1 4 3.1 -0.221154587294
2.1 4 4.1 -0.323713596671
3.1 0 0.1 2402.0
3.1 0 1.1 1.30503380917
3.1 0 2.1 -0.240578771191
3.1 0 3.1 -0.41935483871
3.1 0 4.1 -0.465141248676
3.1 1 0.1 -1.95102493785e-05
3.1 1 1.1 -0.0248114638773
3.1 1 2.1 -0.135153019304
3.1 1 3.1 -0.274125336409
3.1 1 4.1 -0.36965644171
3.1 2 0.1 -1.13811197906e-05
3.1 2 1.1 -0.0147116366819
3.1 2 2.1 -0.0872950700627
3.1 2 3.1 -0.202935925412
3.1 2 4.1 -0.306612285308
3.1 3 0.1 -7.94877050259e-06
3.1 3 1.1 -0.0103624783432
3.1 3 2.1 -0.0642253568271
3.1 3 3.1 -0.160970897235
3.1 3 4.1 -0.261906474418
3.1 4 0.1 -6.09709039262e-06
3.1 4 1.1 -0.00798626913355
3.1 4 2.1 -0.0507564081263
3.1 4 3.1 -0.133349565782
3.1 4 4.1 -0.228563754423
4.1 0 0.1 4202.0
4.1 0 1.1 2.65740045079
4.1 0 2.1 -0.0462153115214
4.1 0 3.1 -0.358933906213
4.1 0 4.1 -0.439024390244
4.1 1 0.1 -1.11538537794e-05
4.1 1 1.1 -0.0144619860317
4.1 1 2.1 -0.0868190343718
4.1 1 3.1 -0.203767982755
4.1 1 4.1 -0.308519215265
4.1 2 0.1 -6.50646078271e-06
4.1 2 1.1 -0.0085156584289
4.1 2 2.1 -0.0538784714494
4.1 2 3.1 -0.140215240068
4.1 2 4.1 -0.23746380125
4.1 3 0.1 -4.54421180079e-06
4.1 3 1.1 -0.00597669061814
4.1 3 2.1 -0.038839789599
4.1 3 3.1 -0.106675396816
4.1 3 4.1 -0.192922262523
4.1 4 0.1 -3.48562423225e-06
4.1 4 1.1 -0.00459693165308
4.1 4 2.1 -0.0303305231375
4.1 4 3.1 -0.0860368842133
4.1 4 4.1 -0.162420599686

初始问题的解决方案是：

# Python-matplotlib Commands
from mpl_toolkits.mplot3d import Axes3D
from matplotlib import cm
import matplotlib.pyplot as plt
import numpy as np

fig = plt.figure()
ax = fig.gca(projection='3d')
X = np.arange(-5, 5, .25)
Y = np.arange(-5, 5, .25)
X, Y = np.meshgrid(X, Y)
R = np.sqrt(X**2 + Y**2)
Z = np.sin(R)
Gx, Gy = np.gradient(Z) # gradients with respect to x and y
G = (Gx**2+Gy**2)**.5  # gradient magnitude
N = G/G.max()  # normalize 0..1
surf = ax.plot_surface(
    X, Y, Z, rstride=1, cstride=1,
    facecolors=cm.jet(N),
    linewidth=0, antialiased=False, shade=False)
plt.show()

据我所知，这适用于所有 matplotlib-demos，变量 X、Y 和 Z 都准备好了。在实际情况下，情况并非总是如此。

想法如何用任意数据重用给定的解决方案？

4

5 回答 5

57

好问题 Tengis，所有数学家都喜欢炫耀带有给定函数的浮华表面图，而忽略处理现实世界的数据。您提供的示例代码使用梯度，因为变量的关系是使用函数建模的。对于这个例子，我将使用标准正态分布生成随机数据。

无论如何，这里是您如何快速绘制 4D 随机（任意）数据，前三个变量在轴上，第四个是颜色：

from mpl_toolkits.mplot3d import Axes3D
import matplotlib.pyplot as plt
import numpy as np

fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')

x = np.random.standard_normal(100)
y = np.random.standard_normal(100)
z = np.random.standard_normal(100)
c = np.random.standard_normal(100)

img = ax.scatter(x, y, z, c=c, cmap=plt.hot())
fig.colorbar(img)
plt.show()

注意：第 4 维使用具有热配色方案（黄色到红色）的热图

结果：

] 1

于 2015-07-22T23:21:46.013 回答
15

我知道这个问题很老了，但我想提出这个替代方案，而不是使用“散点图”，我们有一个 3D 表面图，其中颜色基于第四维。就我个人而言，在“散点图”的情况下，我并没有真正看到空间关系，因此使用 3D 表面可以帮助我更容易地理解图形。

主要思想与公认的答案相同，但我们有一个表面的 3D 图，可以更好地直观地看到点之间的距离。这里的下面代码主要是根据这个问题给出的答案。

import numpy as np
from mpl_toolkits.mplot3d import Axes3D
import matplotlib.pyplot as plt
import matplotlib.tri as mtri

# The values ​​related to each point. This can be a "Dataframe pandas" 
# for example where each column is linked to a variable <-> 1 dimension. 
# The idea is that each line = 1 pt in 4D.
do_random_pt_example = True;

index_x = 0; index_y = 1; index_z = 2; index_c = 3;
list_name_variables = ['x', 'y', 'z', 'c'];
name_color_map = 'seismic';

if do_random_pt_example:
    number_of_points = 200;
    x = np.random.rand(number_of_points);
    y = np.random.rand(number_of_points);
    z = np.random.rand(number_of_points);
    c = np.random.rand(number_of_points);
else:
    # Example where we have a "Pandas Dataframe" where each line = 1 pt in 4D.
    # We assume here that the "data frame" "df" has already been loaded before.
    x = df[list_name_variables[index_x]]; 
    y = df[list_name_variables[index_y]]; 
    z = df[list_name_variables[index_z]]; 
    c = df[list_name_variables[index_c]];
#end
#-----

# We create triangles that join 3 pt at a time and where their colors will be
# determined by the values ​​of their 4th dimension. Each triangle contains 3
# indexes corresponding to the line number of the points to be grouped. 
# Therefore, different methods can be used to define the value that 
# will represent the 3 grouped points and I put some examples.
triangles = mtri.Triangulation(x, y).triangles;

choice_calcuation_colors = 1;
if choice_calcuation_colors == 1: # Mean of the "c" values of the 3 pt of the triangle
    colors = np.mean( [c[triangles[:,0]], c[triangles[:,1]], c[triangles[:,2]]], axis = 0);
elif choice_calcuation_colors == 2: # Mediane of the "c" values of the 3 pt of the triangle
    colors = np.median( [c[triangles[:,0]], c[triangles[:,1]], c[triangles[:,2]]], axis = 0);
elif choice_calcuation_colors == 3: # Max of the "c" values of the 3 pt of the triangle
    colors = np.max( [c[triangles[:,0]], c[triangles[:,1]], c[triangles[:,2]]], axis = 0);
#end
#----------
# Displays the 4D graphic.
fig = plt.figure();
ax = fig.gca(projection='3d');
triang = mtri.Triangulation(x, y, triangles);
surf = ax.plot_trisurf(triang, z, cmap = name_color_map, shade=False, linewidth=0.2);
surf.set_array(colors); surf.autoscale();

#Add a color bar with a title to explain which variable is represented by the color.
cbar = fig.colorbar(surf, shrink=0.5, aspect=5);
cbar.ax.get_yaxis().labelpad = 15; cbar.ax.set_ylabel(list_name_variables[index_c], rotation = 270);

# Add titles to the axes and a title in the figure.
ax.set_xlabel(list_name_variables[index_x]); ax.set_ylabel(list_name_variables[index_y]);
ax.set_zlabel(list_name_variables[index_z]);
plt.title('%s in function of %s, %s and %s' % (list_name_variables[index_c], list_name_variables[index_x], list_name_variables[index_y], list_name_variables[index_z]) );

plt.show();

具有随机值的结果示例

对于我们绝对希望每个点具有第 4 维的原始值的情况，另一种解决方案是简单地使用“散点图”结合 3D 表面图，该图将简单地链接它们以帮助您查看之间的距离他们。

name_color_map_surface = 'Greens';  # Colormap for the 3D surface only.

fig = plt.figure(); 
ax = fig.add_subplot(111, projection='3d');
ax.set_xlabel(list_name_variables[index_x]); ax.set_ylabel(list_name_variables[index_y]);
ax.set_zlabel(list_name_variables[index_z]);
plt.title('%s in fcn of %s, %s and %s' % (list_name_variables[index_c], list_name_variables[index_x], list_name_variables[index_y], list_name_variables[index_z]) );

# In this case, we will have 2 color bars: one for the surface and another for 
# the "scatter plot".
# For example, we can place the second color bar under or to the left of the figure.
choice_pos_colorbar = 2;

#The scatter plot.
img = ax.scatter(x, y, z, c = c, cmap = name_color_map);
cbar = fig.colorbar(img, shrink=0.5, aspect=5); # Default location is at the 'right' of the figure.
cbar.ax.get_yaxis().labelpad = 15; cbar.ax.set_ylabel(list_name_variables[index_c], rotation = 270);

# The 3D surface that serves only to connect the points to help visualize 
# the distances that separates them.
# The "alpha" is used to have some transparency in the surface.
surf = ax.plot_trisurf(x, y, z, cmap = name_color_map_surface, linewidth = 0.2, alpha = 0.25);

# The second color bar will be placed at the left of the figure.
if choice_pos_colorbar == 1: 
    #I am trying here to have the two color bars with the same size even if it 
    #is currently set manually.
    cbaxes = fig.add_axes([1-0.78375-0.1, 0.3025, 0.0393823, 0.385]);  # Case without tigh layout.
    #cbaxes = fig.add_axes([1-0.844805-0.1, 0.25942, 0.0492187, 0.481161]); # Case with tigh layout.

    cbar = plt.colorbar(surf, cax = cbaxes, shrink=0.5, aspect=5);
    cbar.ax.get_yaxis().labelpad = 15; cbar.ax.set_ylabel(list_name_variables[index_z], rotation = 90);

# The second color bar will be placed under the figure.
elif choice_pos_colorbar == 2: 
    cbar = fig.colorbar(surf, shrink=0.75, aspect=20,pad = 0.05, orientation = 'horizontal');
    cbar.ax.get_yaxis().labelpad = 15; cbar.ax.set_xlabel(list_name_variables[index_z], rotation = 0);
#end
plt.show();

第二种方法的随机值样本结果

最后，还可以使用“plot_surface”，我们定义将用于每个面的颜色。在这种情况下，我们每个维度有 1 个值向量，问题是我们必须对这些值进行插值以获得 2D 网格。在第 4 维插值的情况下，只根据 XY 定义，不考虑 Z。结果，颜色代表 C (x, y) 而不是 C (x, y, z)。以下代码主要基于以下响应：plot_surface 每个维度都有一个 1D 向量；plot_surface 为每个表面选择颜色。请注意，与以前的解决方案相比，计算量很大，显示可能需要一点时间。

import matplotlib
from scipy.interpolate import griddata

# X-Y are transformed into 2D grids. It's like a form of interpolation
x1 = np.linspace(x.min(), x.max(), len(np.unique(x))); 
y1 = np.linspace(y.min(), y.max(), len(np.unique(y)));
x2, y2 = np.meshgrid(x1, y1);

# Interpolation of Z: old X-Y to the new X-Y grid.
# Note: Sometimes values ​​can be < z.min and so it may be better to set 
# the values too low to the true minimum value.
z2 = griddata( (x, y), z, (x2, y2), method='cubic', fill_value = 0);
z2[z2 < z.min()] = z.min();

# Interpolation of C: old X-Y on the new X-Y grid (as we did for Z)
# The only problem is the fact that the interpolation of C does not take
# into account Z and that, consequently, the representation is less 
# valid compared to the previous solutions.
c2 = griddata( (x, y), c, (x2, y2), method='cubic', fill_value = 0);
c2[c2 < c.min()] = c.min(); 

#--------
color_dimension = c2; # It must be in 2D - as for "X, Y, Z".
minn, maxx = color_dimension.min(), color_dimension.max();
norm = matplotlib.colors.Normalize(minn, maxx);
m = plt.cm.ScalarMappable(norm=norm, cmap = name_color_map);
m.set_array([]);
fcolors = m.to_rgba(color_dimension);

# At this time, X-Y-Z-C are all 2D and we can use "plot_surface".
fig = plt.figure(); ax = fig.gca(projection='3d');
surf = ax.plot_surface(x2, y2, z2, facecolors = fcolors, linewidth=0, rstride=1, cstride=1,
                       antialiased=False);
cbar = fig.colorbar(m, shrink=0.5, aspect=5);
cbar.ax.get_yaxis().labelpad = 15; cbar.ax.set_ylabel(list_name_variables[index_c], rotation = 270);
ax.set_xlabel(list_name_variables[index_x]); ax.set_ylabel(list_name_variables[index_y]);
ax.set_zlabel(list_name_variables[index_z]);
plt.title('%s in fcn of %s, %s and %s' % (list_name_variables[index_c], list_name_variables[index_x], list_name_variables[index_y], list_name_variables[index_z]) );
plt.show();

第三个解决方案的随机值结果示例

于 2019-08-24T14:30:36.477 回答
9

我想加我的两分钱。给定一个每个条目代表一个特定数量的三维矩阵，我们可以使用 Numpy 的unravel_index()函数结合 Matplotlib 的scatter()方法创建一个伪四维图。

import numpy as np
import matplotlib.pyplot as plt


def plot4d(data):
    fig = plt.figure(figsize=(5, 5))
    ax = fig.add_subplot(projection="3d")
    ax.xaxis.pane.fill = False
    ax.yaxis.pane.fill = False
    ax.zaxis.pane.fill = False
    mask = data > 0.01
    idx = np.arange(int(np.prod(data.shape)))
    x, y, z = np.unravel_index(idx, data.shape)
    ax.scatter(x, y, z, c=data.flatten(), s=10.0 * mask, edgecolor="face", alpha=0.2, marker="o", cmap="magma", linewidth=0)
    plt.tight_layout()
    plt.savefig("test_scatter_4d.png", dpi=250)
    plt.close(fig)


if __name__ == "__main__":
    X = np.arange(-10, 10, 0.5)
    Y = np.arange(-10, 10, 0.5)
    Z = np.arange(-10, 10, 0.5)
    X, Y, Z = np.meshgrid(X, Y, Z, indexing="ij")
    density_matrix = np.sin(np.sqrt(X**2 + Y**2 + Z**2))
    plot4d(density_matrix)

在此处输入图像描述

于 2021-04-04T09:45:04.927 回答
0

一种可能性是使用颜色空间，例如 RGBA 或 HSVA，它们是 4 维的，但很好地显示 alpha（透明度）可能是个问题。

其他可能性是带有滑块的动态图。其中一个维度将由滑块表示。

不过，我不确定这是否是您要问的。

于 2013-02-23T06:37:14.210 回答
0

我还想加我的两分钱！

不久前，我为我正在教授的一门多变量微积分课程创建了一个程序，并认为我会将它包括在这里。

我的方法与已发布的其他方法相似（使用散点图），但是，我进行了一些更改以真正使图形流行。

具体来说，我包含了一个函数来删除 Alpha 通道范围的一部分，以使部分范围透明。这由下面代码中的函数 f_AlphaControl 控制。

我在演示中使用的函数是函数 f(x,y,z)=x y z*exp(-x^2-y^2-z^2)。它有 4 个局部最大值和 4 个局部最小值，所有这些都显示在下图中。我认为结果不言自明，所以请看看它们，让我知道你的想法。

2700 点： epsilon=2 , epsilon=1 , epsilon=.5 , epsilon=.1 , epsilon=5Andx0=-1

1000000 点：epsilon=5，epsilon=1

此代码允许创建与 Mayavi-VTK 和/或 OpenGL 相媲美的等体积图，但无需付出所有努力。我在 Mayavi 中编写了类似的例程，但是，由于 OP 刚刚询问了 matplotlib，我想展示它的强大功能。


import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
from matplotlib import cm

from matplotlib.colors import ListedColormap

## The following code creates the figure plotting function
def MakePlot(xx,yy,zz,ww,cmapO=cm.jet): 
    ##Create Custom Colormap with Alpha varying depending on the functions behavior. 
    ## This produces very nice isovolume plots similar to Mayavi and OpenCV
    
    ##Preallocate new colormap
    my_cmapN=cmapO(np.arange(cmapO.N))
    
    #set Alpha of new colormap to be small in the middle of the colormap range using a bump function
    # this can be changed to emphasize different areas of the range that are of interest.   
    nA=cmapO.N    
    xA=np.linspace(-1,1,nA)
    epsilon=1.5 #Width of range to exclude from alpha channel
    x_0=0#Center of range to exclude from alpha channel
    
    def f_AlphaControl(x):
        u=(x-x_0)/epsilon
        return 1-np.exp(-u**2/(1-u**2))*(np.abs(u)<1.)

    yA=f_AlphaControl(xA) 
    plt.plot(xA,f_AlphaControl(xA))
    plt.xlim([-1,1])
    plt.ylim([0,1])
   
    my_cmapN[:,-1]=yA
    
    fig = plt.figure(dpi=200) 
    
    # Create new colormap
    my_cmap = ListedColormap(my_cmapN) 
    
    
    plt.style.use('dark_background')
    fig = plt.figure(dpi=200) 
    ax = fig.add_subplot(projection='3d')
    points=ax.scatter(xx,yy,zz,c=ww,cmap=my_cmap)   
    cbar=fig.colorbar(points)
    # cbar.solids.set_rasterized(True)
    
    cbar.set_alpha(1)
    cbar.draw_all()
    
    ## Make Title for plot
    ax.set_title(r'Plot of $f:\mathbb{R}^3\rightarrow \mathbb{R}$'+'\n'+r'$w=f(x,y,z)$')
      
    ##Plot x, y, and z axis useful for visual referencing when viewing the plot
    eps=.3
    tt=np.linspace(-(1+eps)*L,(1+eps)*L,2)
    ax.plot(tt,0*tt,0*tt,c='magenta',linewidth=2);
    ax.plot(0*tt,tt,0*tt,c='magenta',linewidth=2);
    ax.plot(0*tt,0*tt,tt,c='magenta',linewidth=2)
      
    ## Set viewing angle
    xang=-76;pang=12;
    ax.view_init(pang, xang)
    
    # Set axis limits
    ax.set_xlim([-(1+eps)*L,(1+eps)*L]);ax.set_ylim([-(1+eps)*L,(1+eps)*L]);ax.set_zlim([-(1+eps)*L,(1+eps)*L]);
       
    #Set axis labels
    ax.set_xlabel('$x$');ax.set_ylabel('$y$');ax.set_zlabel('$z$')
    plt.savefig("ScatterPlotVaryingAlpha.png",dpi=200)

if __name__ == "__main__":

    #Set Plot Grid 
    
    L=1.5
    x_C=0.0;y_C=0.0;z_C=0.0;
    # Set XYZ Plotting Grid
    a1=x_C-L;b1=x_C+L;
    a2=y_C-L;b2=y_C+L;
    a3=z_C-L;b3=z_C+L;
    
    n=30;
    NT=n**3
    
    
    
    ##The following if statement determines whether you want to use a random grid R=1 
    ## or a uniform grid R=0
    
    grid_flag=1
    
    if grid_flag==1:
       ## Random Grid
       xx=np.random.uniform(a1,b1,NT);
       yy=np.random.uniform(a2,b2,NT);
       zz=np.random.uniform(a3,b3,NT)
    else:
        ## Even Grid
        x = np.linspace(a1,b1,n);
        y = np.linspace(a2,b2,n);
        z = np.linspace(a3,b3,n);
    
        X, Y, Z = np.meshgrid(x, y, z, indexing='ij', sparse=False)
        
        xx=X.reshape(X.size);yy=Y.reshape(Y.size);zz=Z.reshape(Z.size);
    
    
    
    ## The following code defines the function of 3 variables that we wish to visualize
    ## This can be replaced with the flattened data array that you wish to plot
    
    def f(x,y,z):
        return x*y*z*np.exp(-(x**2+y**2+z**2))
    
    ww=f(xx,yy,zz)
    ww[np.isinf(ww)]=np.nan
    
  
    MakePlot(xx,yy,zz,ww)
    plt.show()
于 2021-11-13T23:45:03.880 回答