scipy.interpolate.UnivariateSpline

class scipy.interpolate.UnivariateSpline(x, y, w=None, bbox=[None, None], k=3, s=None, ext=0)

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. s specifies 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.

See also

InterpolatedUnivariateSpline

Subclass with smoothing forced to 0

LSQUnivariateSpline

Subclass in which knots are user-selected instead of being set by smoothing condition

splrep

An older, non object-oriented wrapping of FITPACK

splev, sproot, splint, spalde

BivariateSpline

A similar class for two-dimensional spline interpolation

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()

(Source code)

Methods

__call__(x[, nu, ext])	Evaluate spline (or its nu-th derivative) at positions x.
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.