Question

此问题与此one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()

据我所见，这适用于所有matplotlib-demos，变量X，Y和Z已经很好地准备好了。在实际情况中，情况并非总是如此。

如何使用任意数据重用给定的解决方案？

Answer 1

很棒的问题Tengis，所有的数学家都喜欢炫耀带有功能的华丽表面图，同时不遗余力地处理现实世界的数据。您提供的示例代码使用渐变，因为变量的关系是使用函数建模的。对于此示例，我将使用标准正态分布生成随机数据。

无论如何，这里是如何快速绘制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：使用热色方案（黄色到红色）的热图用于第4维

结果：

] 1

Answer 2

我想加上我的两分钱。给定一个三维矩阵，其中每个条目代表一个特定的数量，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)

Answer 3

我知道这个问题很老了，但我想提出一个替代方案，在该示例中，我们使用3D表面图代替颜色散点图，其中颜色基于第4维。就“散点图”而言，我个人并没有真正看到空间关系，因此使用3D表面有助于我更轻松地理解图形。

主要思想与接受的答案相同，但是我们有一个3D图形的表面，可以从视觉上更好地看到两点之间的距离。以下代码主要基于对this question的回答。

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();

对于我们绝对希望每个点都具有第4维原始值的情况，另一种解决方案是简单地将“散点图”与3D表面图结合使用，只需将它们链接起来即可帮助您查看它们之间的距离。

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();

最后，也可以使用“ plot_surface”来定义将用于每个面的颜色。在这样的情况下，每个维有1个值向量，问题是我们必须对值进行插值才能获得2D网格。在第4维插补的情况下，仅根据X-Y定义，并且不考虑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 with a 1D vector for each dimension; plot_surface with a selected color for each surface。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();

Answer 4

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

其他可能性是带滑块的动态图。其中一个尺寸将由滑块表示。

但是，我不确定这是不是你问的问题。

如何使用matplotlib使用任意数据制作4d图

4 个答案: