SciPy

numpy.polyval

numpy.polyval(p, x)[source]

Evaluate a polynomial at specific values.

If p is of length N, this function returns the value:

p[0]*x**(N-1) + p[1]*x**(N-2) + ... + p[N-2]*x + p[N-1]

If x is a sequence, then p(x) is returned for each element of x. If x is another polynomial then the composite polynomial p(x(t)) is returned.

Parameters :

p : array_like or poly1d object

1D array of polynomial coefficients (including coefficients equal to zero) from highest degree to the constant term, or an instance of poly1d.

x : array_like or poly1d object

A number, a 1D array of numbers, or an instance of poly1d, “at” which to evaluate p.

Returns :

values : ndarray or poly1d

If x is a poly1d instance, the result is the composition of the two polynomials, i.e., x is “substituted” in p and the simplified result is returned. In addition, the type of x - array_like or poly1d - governs the type of the output: x array_like => values array_like, x a poly1d object => values is also.

See also

poly1d
A polynomial class.

Notes

Horner’s scheme [R65] is used to evaluate the polynomial. Even so, for polynomials of high degree the values may be inaccurate due to rounding errors. Use carefully.

References

[R65](1, 2) I. N. Bronshtein, K. A. Semendyayev, and K. A. Hirsch (Eng. trans. Ed.), Handbook of Mathematics, New York, Van Nostrand Reinhold Co., 1985, pg. 720.

Examples

>>> np.polyval([3,0,1], 5)  # 3 * 5**2 + 0 * 5**1 + 1
76
>>> np.polyval([3,0,1], np.poly1d(5))
poly1d([ 76.])
>>> np.polyval(np.poly1d([3,0,1]), 5)
76
>>> np.polyval(np.poly1d([3,0,1]), np.poly1d(5))
poly1d([ 76.])

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