在每次迭代中使用列表中的两个连续值

Question

我在下面贴了一部分我的代码。Newton() 函数会调用 Bezier() 函数。Bezier() 函数有一个列表，从这个列表里我获取 p0 和 p3。我的目标是，在第一次循环中，程序应该把列表里的第一个和第二个项目作为 p0 和 p3。然后在第二次循环中，p0 和 p3 就变成第二个和第三个项目，依此类推。每次循环，p0 和 p3 的值都应该改变。就像新的 p0 是旧的 p3。我在代码里没能正确实现这一点。谢谢。

import math

w = 1.0

def Newton(poly):
    """ Returns a root of the polynomial"""
    x = 0.5  # initial guess value
    counter = 0
    epsilon = 0.000000000001
    poly_diff = poly_differentiate(poly)

    while True:
        x_n = x - (float(poly_substitute(poly, x)) / float(poly_substitute(poly_diff, x)))
        counter += 1
        if abs(x_n - x) < epsilon :
            break
        x = x_n
        print "\tIteration " , counter , " : ", x_n

    print "u: ", x_n
    Bezier(x_n)

def Bezier(x_n) :
    """ Calculating sampling points using rational bezier curve equation"""
    u = x_n
    plist = [0.5, 0.1, 0.4, 0.35, 0.8, 0.6, 1.0, 0.2, 0.7, 0.9] # list of parameter values of the phonemes

    for i in range(len(plist) - 1) :
        p0 = plist[i]
        p3 = plist[i + 1] 
        p1 = p0
        p2 = p3
        print p0, p3   
        p_u = math.pow(1 - u, 3) * p0 + 3 * u * math.pow(1 - u, 2) * p1 \
            + 3 * (1 - u) * math.pow(u, 2) * p2 + math.pow(u, 3) * p3
        p_u = p_u * w
        d = math.pow(1 - u, 3) * w + 3 * u * w * math.pow(1 - u, 2) + 3 * (1 - u) * w * math.pow(u, 2) + math.pow(u, 3) * w
        p_u = p_u / d

    print "p(u): ", p_u
    return plist

if __name__ == "__main__" :

Answer 1

也许可以试试来自itertools库的pairwise迭代器，它可能会对你有帮助？

from itertools import izip, tee
def pairwise(iterable):
    "s -> (s0,s1), (s1,s2), (s2, s3), ..."
    a, b = tee(iterable)
    next(b, None)
    return izip(a, b)

Answer 2

>>> p = [1, 2, 3, 4, 5]
>>> for p1, p2 in zip(p, p[1:]):
...     print p1, p2
... 
1 2
2 3
3 4
4 5

这样做有帮助吗？

Answer 3

把你的 Bezier 函数的开头改成下面这个样子：

def Bezier(x_n, p0, p3) :
    """ Calculating sampling points using rational bezier curve equation"""
    u = x_n
    p1 = p0
    p2 = p3

这样就可以去掉那个循环和 plist 了。

接着，在你的 time 函数里（这是我之前回答你问题时提到的 如何从另一个列表获取项目的边界），把它改成下面这样：

def time() :
    tlist = [0.0, 0.12, 0.16, 0.2, 0.31, 0.34, 0.38, 0.46, 0.51] # list of start time for the phonemes
    plist = [0.5, 0.1, 0.4, 0.35, 0.8, 0.6, 1.0, 0.2, 0.7, 0.9] # list of parameter values of the phonemes

    total_frames = math.floor(tlist[-1] / 0.04)
    t_u = 0.0
    i = 0
    while i < len(tlist) - 1:
        # if the number is in the range
        # do the calculations and move to the next number
        if t_u > tlist[i] and t_u < tlist[i + 1] :
            print "\n The t_u value:", t_u, 'is between',
            print "start:", tlist[i], " and end: ", tlist[i+1]
            poly = poly_coeff(tlist[i], tlist[i + 1], t_u)
            Newton(poly, plist[i], plist[i+1])
            t_u = t_u + 0.04 # regular time interval

        # if the number is at the lower boundary of the range no need of calculation as u = 0
        elif t_u == tlist[i] :
            print "\n The t_u value:", t_u, 'is on the boundary of',
            print "start:", tlist[i], " and end: ", tlist[i+1]
            print "u : 0"
            Bezier(0, plist[i], plist[i+1])
            t_u = t_u + 0.04 # regular time interval

        # if the number is at the upper boundary of the range no need of calculation as u = 1
        elif t_u == tlist[i + 1] :
            print "\n The t_u value:", t_u, 'is on the boundary of',
            print "start:", tlist[i], " and end: ", tlist[i+1]
            print " u : 1"
            Bezier(1, plist[i], plist[i+1])
            t_u = t_u + 0.04 # regular time interval

        # if the number isn't in the range, move to the next range
        else :
            i += 1

唯一的变化就是把 plist 放到这里，并把 plist 的值传递给 Newton 和 Bezier。

你对 Newton 函数的唯一修改是把第一行改成：

def Newton(poly, p0, p3):

最后一行改成：

    Bezier(x_n, p0, p3)