我正在尝试编写代码以使用数组。我有用于变量的数组，并且试图从这些变量中调用某些数字以定义函数。我附上了我要完成的任务的简化片段。当我只将变量作为一个数字时，代码运行良好，但是当我合并数组时，它就不起作用。

import numpy as np

noe = 3

# E of members ksi

E = 29000*np.ones((1, noe))

# Area of members in^2

A = 1000*np.array([20, 30, 20])

# length of members in

L = np.array([144, 240, 144])

# moment of inertia of members in^4

I = np.array([600, 1200, 600])

def k_frame_local_6x6(E, I, A, L):

return (E/L)*np.array([[A, 0, 0, -A, 0, 0], [0, (12*I)/(L**2), (6*I)/L, 0, (-12*I)/(L**2), (6*I)/L], [0, (6*I)/L, 4*I, 0, (-6*I)/L, 2*I], [-A, 0, 0, A, 0, 0], [0, (-12*I)/(L**2), (-6*I)/L, 0, (12*I)/(L**2), (-6*I)/L], [0, (6*I)/L, 2*I, 0, (-6*I)/L, 4*I]])

m=k_frame_local_6x6(E[0, 1], I[0, 1], A[0, 1], L[0, 1])

print(m)

我收到的错误是“ IndexError：数组的索引过多”

当我手动输入值时，我试图让函数读取它的功能，如下所示：

def k_frame_local_6x6(E, I, A, L):

return (E/L)*np.array([[A, 0, 0, -A, 0, 0], [0, (12*I)/(L**2), (6*I)/L, 0, (-12*I)/(L**2), (6*I)/L], [0, (6*I)/L, 4*I, 0, (-6*I)/L, 2*I], [-A, 0, 0, A, 0, 0], [0, (-12*I)/(L**2), (-6*I)/L, 0, (12*I)/(L**2), (-6*I)/L], [0, (6*I)/L, 2*I, 0, (-6*I)/L, 4*I]])

m=k_frame_local_6x6(29000, 1200, 30000, 240)

print(m)

我得到的结果是：

[[ 3.62500000e+06 0.00000000e+00 0.00000000e+00 -3.62500000e+06

0.00000000e+00 0.00000000e+00]

[ 0.00000000e+00 3.02083333e+01 3.62500000e+03 0.00000000e+00

-3.02083333e+01 3.62500000e+03]

[ 0.00000000e+00 3.62500000e+03 5.80000000e+05 0.00000000e+00

-3.62500000e+03 2.90000000e+05]

[ -3.62500000e+06 0.00000000e+00 0.00000000e+00 3.62500000e+06

0.00000000e+00 0.00000000e+00]

[ 0.00000000e+00 -3.02083333e+01 -3.62500000e+03 0.00000000e+00

3.02083333e+01 -3.62500000e+03]

[ 0.00000000e+00 3.62500000e+03 2.90000000e+05 0.00000000e+00

-3.62500000e+03 5.80000000e+05]]

我刚刚注意到的一件事是，我的一个数组有两组括号：

E = 29000*np.ones((1, noe))

结果是：

array([[ 29000., 29000., 29000.]])

但是，对于其余的数组，我只能得到一个括号集：

A = 1000*np.array([20, 30, 20])

给我：

array([20000, 30000, 20000])

让我知道是否需要其他说明。谢谢你们！！！

解决方案

您的k_frame_local_6x6-很难看，对普通人来说几乎是不可读的，当给定4个数字时，它会生成6x6数组。

但据我所知，这些术语都不适合用于数组：

def k_frame_local_6x6(E, I, A, L):

temp = np.array([[A, 0, 0, -A, 0, 0],

[0, (12*I)/(L**2), (6*I)/L, 0, (-12*I)/(L**2), (6*I)/L],

[0, (6*I)/L, 4*I, 0, (-6*I)/L, 2*I],

[-A, 0, 0, A, 0, 0],

[0, (-12*I)/(L**2), (-6*I)/L, 0, (12*I)/(L**2), (-6*I)/L],

[0, (6*I)/L, 2*I, 0, (-6*I)/L, 4*I]

])

return (E/L)*temp

尝试，例如最简单的条款

np.array([-A，0,0，A，0,0])

如果A是数字，那很好。但是，如果A是数组，则不起作用。很好，它确实有效，但不会返回有意义的信息

In [105]: A = 1000*np.array([20, 30, 20])

In [106]: np.array([A, 0, 0, -A, 0, 0])

Out[106]:

array([array([20000, 30000, 20000]), 0, 0, array([-20000, -30000, -20000]),

0, 0], dtype=object)

你试图通过A[0, 1]。你试过那种表情了吗？

In [107]: A[0,1]

---------------------------------------------------------------------------

IndexError Traceback (most recent call last)

in ()

----> 1 A[0,1]

IndexError: too many indices

In [108]: A.shape

Out[108]: (3,)

It does not work because A is defined as 1d 3 element array. You can't index it with 2 values.

E is defined as 2d array, but all the rest are 1d. Is that intentional, or a mistake?

It looks like you are jumping into defining a complex array without understanding the basics of numpy array creation and indexing. Or are you coming from a MATLAB world where everything is 2d (or more)? In numpy arrays can be 1d.