scipy.interpolate.UnivariateSpline — SciPy v0.15.1 Reference Guide (original) (raw)
class scipy.interpolate.UnivariateSpline(x, y, w=None, bbox=[None, None], k=3, s=None, ext=0)[source]¶
One-dimensional smoothing spline fit to a given set of data points.
Fits a spline y = spl(x) of degree k to the provided x, y data. sspecifies the number of knots by specifying a smoothing condition.
| Parameters: | x : (N,) array_like 1-D array of independent input data. Must be increasing. y : (N,) array_like 1-D array of dependent input data, of the same length as x. w : (N,) array_like, optional Weights for spline fitting. Must be positive. If None (default), weights are all equal. bbox : (2,) array_like, optional 2-sequence specifying the boundary of the approximation interval. If None (default), bbox=[x[0], x[-1]]. k : int, optional Degree of the smoothing spline. Must be <= 5. Default is k=3, a cubic spline. s : float or None, optional Positive smoothing factor used to choose the number of knots. Number of knots will be increased until the smoothing condition is satisfied: sum((w[i] * (y[i]-spl(x[i])))**2, axis=0) <= s If None (default), s = len(w) which should be a good value if 1/w[i] is an estimate of the standard deviation of y[i]. If 0, spline will interpolate through all data points. ext : int or str, optional Controls the extrapolation mode for elements not in the interval defined by the knot sequence. if ext=0 or ‘extrapolate’, return the extrapolated value. if ext=1 or ‘zeros’, return 0 if ext=2 or ‘raise’, raise a ValueError if ext=3 of ‘const’, return the boundary value. The default value is 0. |
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Notes
The number of data points must be larger than the spline degree k.
NaN handling: If the input arrays contain nan values, the result is not useful, since the underlying spline fitting routines cannot deal with nan . A workaround is to use zero weights for not-a-number data points:
w = np.isnan(y) y[w] = 0. spl = UnivariateSpline(x, y, w=~w)
Notice the need to replace a nan by a numerical value (precise value does not matter as long as the corresponding weight is zero.)
Examples
import matplotlib.pyplot as plt from scipy.interpolate import UnivariateSpline x = np.linspace(-3, 3, 50) y = np.exp(-x**2) + 0.1 * np.random.randn(50) plt.plot(x, y, 'ro', ms=5)
Use the default value for the smoothing parameter:
spl = UnivariateSpline(x, y) xs = np.linspace(-3, 3, 1000) plt.plot(xs, spl(xs), 'g', lw=3)
Manually change the amount of smoothing:
spl.set_smoothing_factor(0.5) plt.plot(xs, spl(xs), 'b', lw=3) plt.show()
Methods
| __call__(x[, nu, ext]) | Evaluate spline (or its nu-th derivative) at positions x. |
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| antiderivative([n]) | Construct a new spline representing the antiderivative of this spline. |
| derivative([n]) | Construct a new spline representing the derivative of this spline. |
| derivatives(x) | Return all derivatives of the spline at the point x. |
| get_coeffs() | Return spline coefficients. |
| get_knots() | Return positions of (boundary and interior) knots of the spline. |
| get_residual() | Return weighted sum of squared residuals of the spline approximation: sum((w[i] * (y[i]-spl(x[i])))**2, axis=0). |
| integral(a, b) | Return definite integral of the spline between two given points. |
| roots() | Return the zeros of the spline. |
| set_smoothing_factor(s) | Continue spline computation with the given smoothing factor s and with the knots found at the last call. |