-
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, thenp(x)
is returned for each element ofx
. Ifx
is another polynomial then the composite polynomialp(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, an 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? inp
and the simplified result is returned. In addition, the type ofx
- 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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numpy.polyval()
2017-01-10 18:17:53
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