Hug*_*ugo 1 python arrays numpy scipy
我有 3 个 numpy 数组,其中包含长、纬度和雨量值:
[-7.4989786799999996, -7.5215012000000003, -7.4764561599999997, -7.4989786799999996, -7.5215012000000003, -7.4764561599999997]
[41.90415308, 41.90415308, 41.90415308, 41.881630559999998, 41.881630559999998, 41.881630559999998]
[0.020928397800000002, 0.0299166963, 0.0171956848, 0.0340920761, 0.0429551788, 0.0301877651]
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Scipy 插值需要 1 个具有 3 个不同 Lons 的数组,另一个具有 2 个不同 Lons 的数组和一个 3x2 降雨值数组。我怎样才能重塑它们?
听起来你想要一个二维插值。
例如,让我们使用一些与您类似的随机数据并将其绘制出来:
import numpy as np
import matplotlib.pyplot as plt
np.random.seed(1) # Make example exactly reproducable
num_points = 20
lon = 0.1 * np.random.random(num_points) - 7.5
lat = 0.1 * np.random.random(num_points) + 41.8
z = 0.05 * np.random.random(num_points)
fig, ax = plt.subplots()
artist = ax.scatter(lon, lat, c=z, s=200, cmap='gist_earth')
fig.colorbar(artist)
ax.axis('tight')
plt.show()
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对于这种情况,插值算法的一个很好的选择是某种径向基函数(例如样条曲线)。在 scipy 中,那就是scipy.interpolate.Rbf:
import scipy.interpolate
interp = scipy.interpolate.Rbf(lon, lat, z, function='linear')
# 20x20 grid of points to interpolate on
yy, xx = np.mgrid[lat.min():lat.max():20j, lon.min():lon.max():20j]
zi = interp(xx, yy)
# Plot the results
fig, ax = plt.subplots()
artist = ax.scatter(lon, lat, c=z, s=100, cmap='gist_earth',
vmin=zi.min(), vmax=zi.max())
ax.pcolormesh(xx, yy, zi, cmap='gist_earth')
ax.axis('tight')
plt.show()
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