我是Python的新手,并感谢任何建设性的反馈。 我的任务是使用ODEINT解决PDE,我的空间网格定义如下:
L = [8.0, 4.0, 8.0] # Lenght of spatial zones
dN = 1.0e2 # Number of grid points
N = [int(l*dN) for l in L] # Gives number of grid points over each spatial zone
通过使用ODEINT函数,我必须为我的解决方案定义一个零数组,以及一组时间点。我正在努力找出如何为数组写出正确的大小。我试过了:
dydt = np.zeros(shape=N) # For solution
t = np.linspace(0, 2, 2000) # for time
IndexError Traceback(最近一次调用最后一次) in() 70 t = np.linspace(0,2,1N) 71 ---> 72 U = odeint(数字,y0,t,args =(h,D1,D2,N)) 73 74 Fexit = U / Np
in numeric(y,t,h,D1,D2,N) 39 [N [0] + 1]中的i为40:#800 =特殊2,必须从N的开始[1] ---> 41 dydt [i] = D2 *( - 3 * y [i] + 4 * y [i + 1] - y [i + 2] / h ** 2)#pluss 42 43为i在范围内(N [0] + 2,N [0] + N [1] - 2):#从801到1197,或2到397
IndexError:索引801超出了轴0,大小为800
Which seems to come from my loops:
# ODE setup
def numeric(y, t, h, D1, D2, N):
dydt = np.zeros((N))
# End grid points
dydt[0] = 0
dydt[-1] = 0
# Inner grid points
for i in range (1, N[0] - 2): # 1 to 797
dydt[i] = D1 * (y[i - 1] - 2*y[i]+ y[i + 1]) / h**2 # range i = j - 1
for i in [N[0] - 2]: # 798 = special 1
dydt[i] = D1 * (-3*y[i] - 4*y[i - 1] + y[i - 2]) / h**2 # minus
for i in [N[0] - 1]: # 799 = unknown, and last index in N[0]
dydt[i] = ((D1*(-3*y[i] + 4*y[i + 1] - y[i + 2])) - (D2*(-3*y[i] - 4*y[i - 1] + y[i - 2])))
for i in [N[0] + 1]: # 800 = special 2, must be from begining of N[1]
dydt[i] = D2 * (-3*y[i] + 4*y[i + 1] - y[i + 2] / h**2) # pluss
for i in range (N[0] + 2, N[0] + N[1] - 2): # from 801 to 1197, or 2 to 397
dydt[i] = D2 * (y[i - 1] - 2*y[i] + y[i + 1]) / h**2 # range
for i in [N[0] + N[1] - 1]: # at 1198, or 398 = special 3
dydt[i] = D2 * (-3*y[i] - 4*y[i - 1] + y[i - 2]) / h**2 # range
for i in [N[0] + N[1]]: # 1199, or 399 = unknown
dydt[i] = ((D1*(-3*y[i] + 4*y[i + 1] - y[i + 2])) - (D2*(-3*y[i] - 4*y[i - 1] + y[i - 2])))
for i in [N[0] + N[1] + 1]: # at 1200, or 1 = special 4
dydt[i] = D1 * (-3*y[i + 1] + 4*y[i + 2] - y[i + 3] / h**2) # pluss
for i in range (N[0] + N[1] + 2, N[0] + N[1] + N[2] - 1): # 1202 to 1999
dydt[i] = D1 * (y[i - 1] - 2*y[i]+ y[i + 1]) / h**2 # range
return dydt
答案 0 :(得分:0)
从快速看,似乎问题是for循环,而不是使用for i in [N[0]]
,你确定它不是for i in range(N[0])
吗?
for i in [N[0]]
相当多余,因为列表只包含一个元素,即。 N[0]
,因此当您尝试寻找dydt[800]
< - 时,大小 800 的列表将超出索引,因为最大索引为799
。
像这样:
In [9]: N
Out[9]: [800, 400, 800]
In [10]: for i in [N[0]]: # ie. i == N[0] == 800
....: dydt[i] = 0
....:
---------------------------------------------------------------------------
IndexError Traceback (most recent call last)
<ipython-input-10-04daa45ec403> in <module>()
1 for i in [N[0]]: # ie. i == N[0] == 800
----> 2 dydt[i] = 0
3
IndexError: index 800 is out of bounds for axis 0 with size 800
In [11]: for i in range(N[0]):
....: dydt[i] = 0
....:
答案 1 :(得分:0)
L = [8.0, 4.0, 8.0]
dN = 1.0e2
N = [int(l*dN) for l in L] # create array [800, 400, 800]
dydt = np.zeros(shape=N) # create 3 dimensional matrix
# 1d has size 800 (from 0 to 799 including)
# 2d has size 400 (from 0 to 399 including)
# 3d same as 1st
在此代码中:
for i in[N[0]]:
dydt[i] = 0 # Unknown point
我总是800,所以它实际上与dydt[ 800 ] = 0
相同。这比799还大。
它看起来像是以错误的方式使用numpy数组。
在低级别是的,它们表示为连续的整数或浮点数。但是在python级别上,np.array表示为数组数组。 (即python中第一个元素为3d矩阵mtx = np.zeros(shape=[800,400,800]
将为mtx[0][0][0]
))。
因此for i in[N[0]]:
之后的所有代码都无效。它试图访问超出矩阵范围的子数组。
更新:
例如你有3d矩阵(3x3x3)
mtx=np.zero(shape=[3,3,3])
print(mtx)
#[ [ [ 0. 0. 0.]
# [ 0. 0. 0.]
# [ 0. 0. 0.] ]
# [ [ 0. 0. 0.]
# [ 0. 0. 0.]
# [ 0. 0. 0.] ]
# [ [ 0. 0. 0.]
# [ 0. 0. 0.]
# [ 0. 0. 0.] ] ]
mtx[0] = 1
print( mtx )
#[ [ [ 1. 1. 1.]
# [ 1. 1. 1.]
# [ 1. 1. 1.] ]
# [ [ 0. 0. 0.]
# [ 0. 0. 0.]
# [ 0. 0. 0.] ]
# [ [ 0. 0. 0.]
# [ 0. 0. 0.]
# [ 0. 0. 0.] ] ]
mtx[0][1] = 2
#[ [ [ 1. 1. 1.]
# [ 2. 2. 2.]
# [ 1. 1. 1.] ]
# others are zeros
mtx[0][1][2] = 3
#[ [ [ 1. 1. 1.]
# [ 2. 2. 3.]
# [ 1. 1. 1.] ]
# others are zeros