在Python中使用'nearest'方法进行外部开发

Question

我想找到以下Matlab语句的Python等价物:

vq interp1(x,y, xq,'nearest','extrap')

它看起来好像interp(xq, x, y)适用于线性插值/外推.

我也看了看

F = scipy.interpolate.interp1d(x, y, kind='nearest')

它适用于最近的方法,但不会执行外推.

还有什么我忽略的吗？谢谢.

Answer 1

对于使用最近插值进行外推的线性插值,请使用numpy.interp.它默认情况下这样做.

例如:

yi = np.interp(xi, x, y)

否则,如果您只想在最近处进行最近插值,如您所述,您可以用简短但效率低下的方式进行:(如果需要,可以将其设为单行插值)

def nearest_interp(xi, x, y):
    idx = np.abs(x - xi[:,None])
    return y[idx.argmin(axis=1)]

或者以更有效的方式使用searchsorted:

def fast_nearest_interp(xi, x, y):
    """Assumes that x is monotonically increasing!!."""
    # Shift x points to centers
    spacing = np.diff(x) / 2
    x = x + np.hstack([spacing, spacing[-1]])
    # Append the last point in y twice for ease of use
    y = np.hstack([y, y[-1]])
    return y[np.searchsorted(x, xi)]

为了说明numpy.interp上面和最近的插值示例之间的区别:

import numpy as np
import matplotlib.pyplot as plt

def main():
    x = np.array([0.1, 0.3, 1.9])
    y = np.array([4, -9, 1])
    xi = np.linspace(-1, 3, 200)

    fig, axes = plt.subplots(nrows=2, sharex=True, sharey=True)
    for ax in axes:
        ax.margins(0.05)
        ax.plot(x, y, 'ro')

    axes[0].plot(xi, np.interp(xi, x, y), color='blue')
    axes[1].plot(xi, nearest_interp(xi, x, y), color='green')

    kwargs = dict(x=0.95, y=0.9, ha='right', va='top')
    axes[0].set_title("Numpy's $interp$ function", **kwargs)
    axes[1].set_title('Nearest Interpolation', **kwargs)

    plt.show()

def nearest_interp(xi, x, y):
    idx = np.abs(x - xi[:,None])
    return y[idx.argmin(axis=1)]

main()