制作散射轮廓

Question

在python中,如果我有一组数据

x, y, z

我可以散布

import matplotlib.pyplot as plt
plt.scatter(x,y,c=z)

我如何得到plt.contourf(x,y,z)分散？

Answer 1

使用以下函数转换为contourf所需的格式:

import matplotlib.tri as tri
import matplotlib.pyplot as plt

plt.tricontour(x, y, z, 15, linewidths=0.5, colors='k')
plt.tricontourf(x, y, z, 15)

现在你可以这样做:

from numpy import linspace, meshgrid
from matplotlib.mlab import griddata

def grid(x, y, z, resX=100, resY=100):
    "Convert 3 column data to matplotlib grid"
    xi = linspace(min(x), max(x), resX)
    yi = linspace(min(y), max(y), resY)
    Z = griddata(x, y, z, xi, yi)
    X, Y = meshgrid(xi, yi)
    return X, Y, Z

Answer 2

解决方案将取决于数据的组织方式.

常规网格数据

如果x和y数据已经定义了网格,则可以轻松地将它们重新整形为四边形网格.例如

#x  y  z
 4  1  3
 6  1  8
 8  1 -9
 4  2 10
 6  2 -1
 8  2 -8
 4  3  8
 6  3 -9
 8  3  0
 4  4 -1
 6  4 -8
 8  4  8 

可以绘制为contour使用

import matplotlib.pyplot as plt
import numpy as np
x,y,z = np.loadtxt("data.txt", unpack=True)
plt.contour(x.reshape(4,3), y.reshape(4,3), z.reshape(4,3))

任意数据

一个.插值

如果数据不是生活在四边形网格上,则可以在网格上插入数据.matplotlib本身提供了一种方法,使用scipy.interpolate.griddata.不推荐使用此方法.替代方案:使用tricontour

import numpy as np
from scipy.interpolate import griddata

xi = np.linspace(4, 8, 10)
yi = np.linspace(1, 4, 10)
zi = griddata((x, y), z, (xi[None,:], yi[:,None]), method='linear')
plt.contour(xi, yi, zi)

湾 非网格轮廓

最后,可以在不使用四边形网格的情况下完全绘制轮廓.这可以使用x.

plt.tricontour(x,y,z)

在matplotlib页面上可以找到比较后两种方法的示例.

Answer 3

contour期望定期网格化数据.因此,您需要首先插入数据:

import numpy as np
from scipy.interpolate import griddata
import matplotlib.pyplot as plt
import numpy.ma as ma
from numpy.random import uniform, seed
# make up some randomly distributed data
seed(1234)
npts = 200
x = uniform(-2,2,npts)
y = uniform(-2,2,npts)
z = x*np.exp(-x**2-y**2)
# define grid.
xi = np.linspace(-2.1,2.1,100)
yi = np.linspace(-2.1,2.1,100)
# grid the data.
zi = griddata((x, y), z, (xi[None,:], yi[:,None]), method='cubic')
# contour the gridded data, plotting dots at the randomly spaced data points.
CS = plt.contour(xi,yi,zi,15,linewidths=0.5,colors='k')
CS = plt.contourf(xi,yi,zi,15,cmap=plt.cm.jet)
plt.colorbar() # draw colorbar
# plot data points.
plt.scatter(x,y,marker='o',c='b',s=5)
plt.xlim(-2,2)
plt.ylim(-2,2)
plt.title('griddata test (%d points)' % npts)
plt.show()

请注意,我无耻地从优秀的matplotlib食谱中偷走了这段代码