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        "value": "Suppose we have global data on a coarse grid\n\n"
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        "value": "import numpy as np\nlats = np.linspace(10, 170, 9) * np.pi / 180.\nlons = np.linspace(0, 350, 18) * np.pi / 180.\ndata = np.dot(np.atleast_2d(90. - np.linspace(-80., 80., 18)).T,\n              np.atleast_2d(180. - np.abs(np.linspace(0., 350., 9)))).T\n",
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        "value": "\nWe want to interpolate it to a global one-degree grid\n\n"
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        "value": "new_lats = np.linspace(1, 180, 180) * np.pi / 180\nnew_lons = np.linspace(1, 360, 360) * np.pi / 180\nnew_lats, new_lons = np.meshgrid(new_lats, new_lons)\n",
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        "value": "\nWe need to set up the interpolator object\n\n"
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        "value": "from scipy.interpolate import RectSphereBivariateSpline\nlut = RectSphereBivariateSpline(lats, lons, data)\n",
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        "value": "\nFinally we interpolate the data.  The `RectSphereBivariateSpline` object\nonly takes 1-D arrays as input, therefore we need to do some reshaping.\n\n"
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        "value": "data_interp = lut.ev(new_lats.ravel(),\n                     new_lons.ravel()).reshape((360, 180)).T\n",
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        "value": "\nLooking at the original and the interpolated data, one can see that the\ninterpolant reproduces the original data very well:\n\n"
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        "value": "import matplotlib.pyplot as plt\nfig = plt.figure()\nax1 = fig.add_subplot(211)\n",
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        "value": "\nChoosing the optimal value of ``s`` can be a delicate task. Recommended\nvalues for ``s`` depend on the accuracy of the data values.  If the user\nhas an idea of the statistical errors on the data, she can also find a\nproper estimate for ``s``. By assuming that, if she specifies the\nright ``s``, the interpolator will use a spline ``f(u,v)`` which exactly\nreproduces the function underlying the data, she can evaluate\n``sum((r(i,j)-s(u(i),v(j)))**2)`` to find a good estimate for this ``s``.\nFor example, if she knows that the statistical errors on her\n``r(i,j)``-values are not greater than 0.1, she may expect that a good\n``s`` should have a value not larger than ``u.size * v.size * (0.1)**2``.\n\nIf nothing is known about the statistical error in ``r(i,j)``, ``s`` must\nbe determined by trial and error.  The best is then to start with a very\nlarge value of ``s`` (to determine the least-squares polynomial and the\ncorresponding upper bound ``fp0`` for ``s``) and then to progressively\ndecrease the value of ``s`` (say by a factor 10 in the beginning, i.e.\n``s = fp0 / 10, fp0 / 100, ...``  and more carefully as the approximation\nshows more detail) to obtain closer fits.\n\nThe interpolation results for different values of ``s`` give some insight\ninto this process:\n\n"
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        "value": "fig2 = plt.figure()\ns = [3e9, 2e9, 1e9, 1e8]\n",
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        "value": "for idx, sval in enumerate(s, 1):\n    lut = RectSphereBivariateSpline(lats, lons, data, s=sval)\n    data_interp = lut.ev(new_lats.ravel(),\n                         new_lons.ravel()).reshape((360, 180)).T\n    ax = fig2.add_subplot(2, 2, idx)\n    ax.imshow(data_interp, interpolation='nearest')\n    ax.set_title(f\"s = {sval:g}\")\n",
        "execution_status": "failure"
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