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用python绘制4d图

plot matplotlib python

Beliaev Maksim · 技术社区 · 8 年前

我知道我在打一匹死马,但我找不到合适的答案。

我想用大数据绘制一个图:

X-Coord   Y-Coord   Z-Coord   Value
20'000 rows

当我调用

X, Y = np.meshgrid(X, Y)

我得到一个错误:

  Traceback (most recent call last):
  File "new 1.py", line 17, in <module>
    X, Y = np.meshgrid(X, Y)  # <-- returns a 2D grid from initial 1D arrays
  File "C:\Program Files (x86)\lib\site-packages\numpy\lib\function_base.py", line 4698, in meshgrid
    output = [x.copy() for x in output]
  File "C:\Program Files (x86)\lib\site-packages\numpy\lib\function_base.py", line 4698, in <listcomp>
    output = [x.copy() for x in output]
MemoryError

实际上,我只想从x y z坐标得到一个三维曲面,然后从这个坐标中的列值绘制值。请给我一个建议

UPD: 数据示例

          X                       Y                       Z                     Value
-3.6296815834229800E+13 9.0179395964544800E+13  4.3243022996875400E+13  2.3293827867020395e-03
-3.6546185417114900E+13 8.9339697982272400E+13  4.3845054348437700E+13  2.2153085734286245e-03
-3.6546185417114900E+13 9.2024057357272400E+13  4.3845054348437700E+13  2.6335681277863542e-03
-3.6795554999999900E+13 8.8500000000000000E+13  4.4447085699999900E+13  2.2448110225069475e-03
-3.6795554999999900E+13 9.1184359374999900E+13  4.4447085699999900E+13  2.3661800082893664e-03
-3.6795554999999900E+13 9.3868718749999900E+13  4.4447085699999900E+13  3.1766708204588683e-03
-3.6296815834229800E+13 9.0179395964544800E+13  4.3243022996875400E+13  2.3293827867020395e-03
-3.6546185417114900E+13 9.2024057357272400E+13  4.3845054348437700E+13  2.6335681277863542e-03
-3.6337452147547400E+13 9.0857342861310100E+13  4.3341127722985300E+13  2.4227047423936087e-03
-3.6795554999999900E+13 9.3868718749999900E+13  4.4447085699999900E+13  3.1766708204588683e-03
-3.6586821730432500E+13 9.2702004254037600E+13  4.3943159074547600E+13  2.8272105071883709e-03
-3.6378088460865000E+13 9.1535289758075300E+13  4.3439232449095200E+13  2.5588155147357474e-03
-3.6378088460865000E+13 9.1535289758075300E+13  4.3439232449095200E+13  2.6341575316456271e-03
-3.6586821730432500E+13 9.2702004254037600E+13  4.3943159074547600E+13  2.8628186103490019e-03
-3.6491956423152800E+13 9.3434980585604900E+13  4.3714133992590100E+13  3.2129310022084552e-03
-3.6795554999999900E+13 9.3868718749999900E+13  4.4447085699999900E+13  3.1710978217960057e-03
-3.6700689692720300E+13 9.4601695081567300E+13  4.4218060618042400E+13  3.5478526100425675e-03
-3.6605824385440600E+13 9.5334671413134600E+13  4.3989035536084900E+13  3.9315260648101454e-03
-3.6196152400000000E+13 8.8500000000000000E+13  4.3000000000000000E+13  5.1978166985800038e-03
-3.6495853699999900E+13 8.8500000000000000E+13  4.3723542849999900E+13  4.8689291409766524e-03

UPD2: 感谢@importanceofbeingernest提供的良好解决方案

最后,为了在分散循环期间加快数据处理速度,我减少了阵列并使用了另一个后端。还为结果添加了颜色条。

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

import matplotlib
matplotlib.use('svg')

x,y,z,v = np.loadtxt("datafile.txt", skiprows=2, unpack=True)
if len(x) > 400000:
  x = x[::50] 
  y = y[::50] 
  z = z[::50]
  v = v[::50] 
elif len(x) > 200000:
  x = x[::20] 
  y = y[::20] 
  z = z[::20]
  v = v[::20] 
elif len(x) > 100000:
  x = x[::10] 
  y = y[::10] 
  z = z[::10]
  v = v[::10]   

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


ax.scatter(x,y,z,c=v, s=10, cmap=cm.rainbow)

m = cm.ScalarMappable(cmap=cm.rainbow)

m.set_array(v)
cbar = plt.colorbar(m)

plt.show()

1 回复 | 直到 8 年前

ImportanceOfBeingErnest 8 年前

您可以使用散点图根据表的最后一列绘制带颜色的点;然后 plot_trisurf 可以用来得到曲面交叉的概念。

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

x,y,z,v = np.loadtxt("datafile.txt", skiprows=1, unpack=True)

fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
ax.plot_trisurf(x,y,z, edgecolor="gray", color="None")
ax.scatter(x,y,z,c=v, s=100)