索引器错误的原因是什么：列表索引超出范围？

Question

所以，我的教授让我创建一个程序，使用龙格库塔方法（二阶）来解决这个问题。但是我得到了一个索引器：列表索引超出范围，这可以追溯到我的第一个函数。我不完全明白这里的错误是什么，如果有人向我解释，我会很高兴的

import numpy as np

def f(a, rec, inf, j, b, q):
    f = b*(1-inf[j]-rec[j])*(a[j][0]*inf[0]+a[j][1]*inf[1]+a[j][2]*inf[2]+a[j][3]*inf[3]+a[j][4]*inf[4]+a[j][5]*inf[5]+a[j][6]*inf[6]+a[j][7]*inf[7]+a[j][8]*inf[8]+a[j][9]*inf[9]) - q*inf[j]
    return f

def rk2a(I, R, b, q):
    beta = b
    sigma = q
    I.astype(float)
    R.astype(float)
    n = len(I)
    S = 1 - I - R
    A = np.array([[0,1,1,1,0,0,0,0,0,0], 
    [1,0,1,1,1,0,0,0,0,0],
    [1,1,0,1,1,1,0,0,0,0],
    [1,1,1,0,1,1,1,0,0,0],
    [0,1,1,1,0,1,1,1,0,0],
    [0,0,1,1,1,0,1,1,1,0],
    [0,0,0,1,1,1,0,1,1,1],
    [0,0,0,0,1,1,1,0,1,0],
    [0,0,0,0,0,1,1,1,0,1],
    [0,0,0,0,0,0,1,0,1,0]])
    y_coef = []
    old_I = []
    for i in range(0,50,1):
        print(I.transpose())
        for j in range(0,n-1,1):
            k1 = f(A,R,I,j,beta,sigma)
            y_coef.append(I[j])
            old_I.append(I[j])
            y_coef[j] = I[j]+(k1/2)
            k2 = f(A, R, y_coef, j, beta, sigma)
            I[j] = old_I[j] + k2
        return None

infect = np.array([0, 1, 0, 0, 0, 0, 0, 1, 0, 0]) 
recov = np.array([0,0, 0, 0, 0, 0, 0, 0, 0, 0]) 
rk2a(infect, recov, 0.3, 0.1)

Answer 1

完全忽略索引。所有numpy数组算术运算都是逐单元进行的。线性代数矩阵向量积是数组的numpy.dot或.dot方法。所以你可以重写

def f(inf, rec, A,b,q)
    susc = 1-inf-rec           # computes the susceptibles in every node
    trans = b*susc*A.dot(inf)  # newly infected in time span
    recov = q*inf              # recovered in time span
    return trans-recov, recov

然后迭代

for t in time[:-1]:
    k1i, k1r = f(inf,rec,A,b,q)
    k2i, k2r = f(inf+k1i*dt/2,rec+k1r*dt/2,A,b,q)
    inf, rec = inf + k2i*dt, rec+k2r*dt
    I.append(inf); R.append(rec);

Answer 2

尝试使用以下选项运行：

def f(a, rec, inf, j, b, q):
    print('   DEBUG   -')
    print('a', a, 'rec', rec, 'inf', inf, 'j', j, 'b', b, 'q', q, sep="\n")
    print('       -')
    f = b*(1-inf[j])*(a[j][0]*inf[0]+a[j][1]*inf[1]+a[j][2]*inf[2]+a[j][3]*inf[3]+a[j][4]*inf[4]+a[j][5]*inf[5]+a[j][6]*inf[6]+a[j][7]*inf[7]+a[j][8]*inf[8]+a[j][9]*inf[9]) - q*inf[j]
    return f

您传递的inf数组错误