在 Python 中计算多项式？

Question

I really don't have code to post with this, since I'm pretty stuck on how to write it. I have to give a list of positive ints and an x value to replicate the following example:

>>> poly([1, 2, 1], 2)
9
>>> poly([1, 0, 1, 0, 1], 2)
21
>>> poly([1, 0, 1, 0, 1], 3)
91

The equation I have is p(x) = a0 + a1x + a2x**2 + a3x**3 + ... + anx**n , so an idea I had was checking the length of the list and making it so that it automatically determined how many calculations it had to do, then just replacing x with whatever value was outside the list. Unfortunately I don't know how to write that or where to start really.

Answer 1

Even better if you have numpy:

>>> from numpy import polyval
>>> polyval([1, 2, 1], 2)
9
>>> polyval([1, 0, 1, 0, 1], 2)
21

I think you have to reverse the a_list first though. (it happens to work since a_list is palindromic )

Answer 2

Here is how you can implement poly :

def poly(l, x):
    sum = 0
    xp =1 
    for a in l:
        sum += a *xp #add next term
        xp = x* xp #xp is x^p
    return sum

print poly([1,2,1],2)
print poly([1,0,1,0,1],3)

Answer 3

def poly(a_list, x):
    ans = 0
    for n,a in enumerate(a_list):
        ans += a*x**n
    return ans

The enumerate function returns a tuple containing the index and value of each element in the list. So you can iterate easily through a list using "for index,value in enumerate(list)".

Answer 4

You could use itertools starmap

>>> from itertools import starmap
>>> def poly(a_list, x):
...     val = lambda p, a: a*x**p
...     return sum(starmap(val, enumerate(a_list)))
...
>>> poly([1, 2, 1], 2)
9
>>> poly([1, 0, 1, 0, 1], 2)
21