scipy.interpolate.Rbf和scipy.interpolate.griddata之间的妥协?

时间:2017-08-25 20:58:39

标签: python scipy interpolation

我需要在给定一些非结构化(x, y)坐标的情况下生成二维密度图,并使用一个z值作为分配给每个点的权重。

我可以使用scipy.interpolate.Rbfscipy.interpolate.griddata将值插入网格(请参阅下面的代码)。第一个非常准确,但非常慢,而第二个很多更快但明显不太准确。

请参阅下面的图片,其中Rfb位于左侧,griddata位于右侧:

enter image description here

有没有办法在这两个功能之间妥协?什么会产生一个与Rfb函数更相似但速度更接近griddata函数的结果?

# Data
N = 1000
x, y, z = np.random.uniform(0.0005, 0.03, N),\
    np.random.uniform(6., 10., N), np.random.uniform(1., 10., N)
xmin, xmax, ymin, ymax = min(x), max(x), min(y), max(y)

# Set up a regular grid of interpolation points
xi, yi = np.linspace(xmin, xmax, 200), np.linspace(ymin, ymax, 200)
xi, yi = np.meshgrid(xi, yi)
# Normalize data and grid.
x_new, xi_new = (x - xmin) / (xmax - xmin), (xi - xmin) / (xmax - xmin)
y_new, yi_new = (y - ymin) / (ymax - ymin), (yi - ymin) / (ymax - ymin)

# Interpolate new data with Rbf.    
s = time.clock()
rbf = scipy.interpolate.Rbf(x_new, y_new, z, function='linear')
zi = rbf(xi_new, yi_new)
print(time.clock() - s)
# Plot density map.
ax1 = plt.subplot(121)
plt.pcolormesh(xi, yi, zi, cmap=plt.get_cmap('GnBu_r'), zorder=2)
plt.contour(xi, yi, zi, 2, colors='#551a8b', linewidths=0.5, zorder=3)

# Interpolate new data with griddata.    
s = time.clock()
# Python 2.7
vals = zip(*[x_new, y_new])
# Python 3.6
# vals = np.array([x_new, y_new]).T
zi = scipy.interpolate.griddata(vals, z, (xi_new, yi_new), method='linear')
print(time.clock() - s)
# Plot density map.
ax2 = plt.subplot(122)
plt.imshow(
    zi, extent=(xmin, xmax, ymin, ymax), origin='lower',
    cmap=plt.get_cmap('GnBu_r'))
plt.contour(xi, yi, zi, 2, colors='#551a8b', linewidths=0.5, zorder=3)

plt.show()

0 个答案:

没有答案
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