我希望找到以下Matlab语句的Python等价物:
vq interp1(x,y, xq,'nearest','extrap')
看起来interp(xq, x, y)
非常适合线性插值/外推。
我也看了
F = scipy.interpolate.interp1d(x, y, kind='nearest')
适用于最近的方法,但不会执行外推。
还有什么我忽略了吗?感谢。
答案 0 :(得分:6)
对于使用最近插值进行外推的线性插值,请使用numpy.interp
。它默认情况下会这样做。
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
与上面最近的插值示例之间的差异:
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()
答案 1 :(得分:0)
在更高版本的SciPy(至少v0.19.1 +)中,scipy.interpolate.interp1d
具有选项fill_value = “extrapolate”
。
例如:
import pandas as pd
>>> s = pd.Series([1, 2, 3])
Out[1]:
0 1
1 2
2 3
dtype: int64
>>> t = pd.concat([s, pd.Series(index=s.index + 0.1)]).sort_index()
Out[2]:
0.0 1.0
0.1 NaN
1.0 2.0
1.1 NaN
2.0 3.0
2.1 NaN
dtype: float64
>>> t.interpolate(method='nearest')
Out[3]:
0.0 1.0
0.1 1.0
1.0 2.0
1.1 2.0
2.0 3.0
2.1 NaN
dtype: float64
>>> t.interpolate(method='nearest', fill_value='extrapolate')
Out[4]:
0.0 1.0
0.1 1.0
1.0 2.0
1.1 2.0
2.0 3.0
2.1 3.0
dtype: float64