Question

我在嵌套for循环中遇到了索引问题。 Python正在吐出一个索引错误，告诉我我的索引超出范围。

以下是我的代码和后续错误：

from math import *
import numpy as np
from scipy import integrate
import matplotlib.pyplot as plt
import os

croot = 1
ctip = 1
span = 1
thetaroot = 0
thetatip = 0
a0root = 0.11
a0tip = 0.11
alpha = 0
alpha0root = -2.5
alpha0tip = -2.5
thetaroot = thetaroot * arctan(1.) / 45.
thetatip = thetatip * arctan(1.) / 45.
alpha = alpha * arctan(1.) / 45.
alpha0root = alpha0root * arctan(1.) / 45.
alpha0tip = alpha0tip * arctan(1.) / 45.
n = 10
theta = np.empty(n, dtype = object)
y = np.empty(n, dtype = object)
c = np.empty(n, dtype = object)
cl = np.empty(n, dtype = object)
alp = np.empty(n, dtype = object)
a = np.empty(n, dtype = object)
rhs = np.empty(n, dtype = object)
b = np.empty(n, dtype = object)
a = np.empty(n, dtype = object)
rhs = rhs[:,None]
b = b[:,None]
a = a[:,None]
#
# Define properties at n span stations
#
pi = 4. * arctan(1.)
for i in range(0,n):
    theta[i] = i * pi / (2. * n)
    y[i] = span * 0.5 * cos(theta[i])
    c[i] = croot + (ctip - croot) * y[i] * 2. / span
    alp[i] = alpha + thetaroot - (alpha0root + (alpha0tip - alpha0root + thetaroot - thetatip) * y[i] * 2. / span)
    a[i] = a0root + (a0tip - a0root) * y[i] * 2. / span

pi = 4. * arctan(1.)
# Set up 2n x 2n system of equations for A1, A3 , ... A2n-1
for j in range(0,n):
    mu = c[j] * a[j] / (4. * span); print('mu=',mu)
    rhs[j,0] = alp[j] * sin(theta[j]) * c[j] * a[j] / (4 * span)
    for i in range(0,n):
        l = 2 * i - 1
        b[j,i] = sin(l * theta[j]) * (mu * l + sin(theta[j]))

然后我收到错误：

---------------------------------------------------------------------------
IndexError                                Traceback (most recent call last)
<ipython-input-70-b5bd95e69bb5> in <module>()
     45     for i in range(0,n):
     46         l = 2 * i - 1
---> 47         b[j,i] = sin(l * theta[j]) * (mu * l + sin(theta[j]))
     48 
     49 

IndexError: index 1 is out of bounds for axis 1 with size 1

如何有效地调出两个指数？在MATLAB中，b（j，i）是正常的语法。

感谢任何帮助，谢谢！

Answer 1

使用b = np.empty(n, dtype = object)的{{1}}调用会生成一维数组，但您正在将其编入索引（n=10），就像它是一个二维数组一样。

要初始化10 x 10阵列，请拨打b[j,i]。

编辑：我没有注意到这项任务：b = np.empty([n, n], dtype = object) 这创造了这个：

b = b[:,None]

这是一个二维数组，但是尝试索引内部数组的第一个元素（仅包含None）会导致错误。

嵌套的FOR循环索引问题 - Python

1 个答案: