Skip to main content
 首页 » 编程设计

python之如何在 matplotlib 中获得平滑的填充轮廓

2025年05月04日8jyk

您好,我正在尝试绘制不均匀数据的填充轮廓。在三个列表中。我的问题是我无法获得平滑的填充轮廓。我所做的是首先使用 griddata 将数据从不规则点更改为网格。

import numpy as np 
import matplotlib.pyplot as plt 
import time as time 
from scipy.interpolate import griddata 
x = [39, 39, 603, 603, 540.8578720591851, 586.349172503832, 373.99215228030187, 436.4554443169055, 125.7177128362948, 56.44720056160912, 453.35159098310174, 384.081128192362, 51.846094630755104, 121.11660875746472, 278.0734642496455, 211.33415130113278, 508.642428513517, 453.0506702655636, 455.66065332357397, 381.7443137710119, 211.08060937414135, 271.19278437560484, 301.7212739516758, 337.50499942076925, 237.27644459337762, 277.8143694411149, 89.76821876085899, 145.66110067318877, 151.97990283138796, 197.59696541916784, 398.0895764975718, 453.7365065456195] 
y = [-29, 394, -29, 394, 96.31199431392861, 96.31199431392861, 65.63484056949213, 65.63484056949213, 353.9802948050525, 353.99631296027843, 354.83809861715105, 354.75376513965614, 170.85745938538898, 170.85745938538898, 156.95287962269862, 156.95287962269862, 161.4804871844196, 160.98633822221555, 242.17985596076556, 241.74154302501933, 214.02665403095247, 214.02665403095247, 65.63484056949213, 65.63484056949213, 63.49457402918261, 63.49457402918261, 54.22008568784131, 54.22008568784131, 7.134221801031751, 7.134221801031751, 3.5671109005158756, 3.5671109005158756] 
z = [0, 0, 0, 0, 1, 1, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 1, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0] 
extent = (min(x), max(x), min(y), max(y)) 
xs,ys = np.mgrid[extent[0]:extent[1], extent[2]:extent[3]] 
resampled = griddata((x, y), z, (xs, ys)) 
plt.figure() 
plt.imshow(resampled.T, extent=(min(x), max(x), max(y), min(y))) 
plt.hold(True) 
plt.scatter(x,y,c=z) 
plt.show() 

给出的情节如下。 这看起来很奇怪。我怎样才能将颜色变化平滑到下一点。 提前致谢

请您参考如下方法:

您的采样网格太密集,以至于 imshow 根本不需要插值。尺寸大约为 30x30 点,它看起来好多了。此外,将 imshow 的插值方法设置为 'bicubic'。

编辑:我忽略了griddata的插值方法可以设置为'cubic'。这会产生更好的结果,您可以像以前一样保留采样大小。

Edit2:看起来最好的结果是用我原来的方法实现的。让 griddata 通过线性插值在网格上生成样本,并根据该数据使用 imshow 或 contour 进行三次插值,即:

import numpy as np 
import matplotlib.pyplot as plt 
import time as time 
from scipy.interpolate import griddata 
x = [39, 39, 603, 603, 540.8578720591851, 586.349172503832, 373.99215228030187,436.4554443169055, 125.7177128362948, 56.44720056160912, 453.35159098310174, 384.081128192362, 51.846094630755104, 121.11660875746472, 278.0734642496455, 211.33415130113278, 508.642428513517, 453.0506702655636, 455.66065332357397, 381.7443137710119, 211.08060937414135, 271.19278437560484, 301.7212739516758, 337.50499942076925, 237.27644459337762, 277.8143694411149, 89.76821876085899, 145.66110067318877, 151.97990283138796, 197.59696541916784, 398.0895764975718, 453.7365065456195] 
y = [-29, 394, -29, 394, 96.31199431392861, 96.31199431392861, 65.63484056949213, 65.63484056949213, 353.9802948050525, 353.99631296027843, 354.83809861715105, 354.75376513965614, 170.85745938538898, 170.85745938538898, 156.95287962269862, 156.95287962269862, 161.4804871844196, 160.98633822221555, 242.17985596076556, 241.74154302501933, 214.02665403095247, 214.02665403095247, 65.63484056949213, 65.63484056949213, 63.49457402918261, 63.49457402918261, 54.22008568784131, 54.22008568784131, 7.134221801031751, 7.134221801031751, 3.5671109005158756, 3.5671109005158756] 
z = [0, 0, 0, 0, 1, 1, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 1, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0] 
extent = (min(x), max(x), min(y), max(y)) 
xs,ys = np.mgrid[extent[0]:extent[1]:30j, extent[2]:extent[3]:30j] 
resampled = griddata((x, y), z, (xs, ys)) 
plt.figure(figsize=(8,8)) 
plt.imshow(resampled.T, extent=(min(x), max(x), max(y), min(y)),interpolation='bicubic') 
plt.contour(resampled.T, extent=(min(x), max(x), max(y), min(y)),interpolation='bicubic',origin='upper') 
plt.hold(True) 
plt.scatter(x,y,c=z) 
plt.show()