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Learning

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Analysis

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Numerical

Analysis

TENTH

EDITION

Richard

Burden

Youngstown

University

Douglas

Faires

Youngstown

University

Annette

Burden

Youngstown

University

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优化下列MATLAB代码：% 定义计算区间和网格数 xStart = -0.5; xEnd = 0.5; tStart = 0; tEnd = 10; Nx = 100; Nt = 10000; % 定义CFL数值和速度a CFL = 0.5; a = 1; % 定义网格步长 dx = (xEnd - xStart) / Nx; dt = CFL / a / Nx; % 定义初始条件 u0 = zeros(1, Nx); u0(xStart <= -0.5 & -0.5 < xStart + (0:Nx-1)dx) = 0; u0(-0.25 <= xStart + (0:Nx-1)dx & xStart + (0:Nx-1)dx <= 0.25) = 1; u0(0.25 < xStart + (0:Nx-1)dx & xStart + (0:Nx-1)dx <= 0.5) = 0; % 定义周期边界条件 u = zeros(Nt, Nx); u(1,:) = u0; u(:,1) = u(:,Nx); u(:,Nx+1) = u(:,2); % 进行数值计算 for n = 1:Nt-1 % 一阶迎风格式 u(n+1,2:end-1) = u(n,2:end-1) - CFL (u(n,2:end-1) - u(n,1:end-2)); u(n+1,1) = u(n,1) - CFL * (u(n,1) - u(n,Nx)); u(n+1,Nx) = u(n,Nx) - CFL * (u(n,Nx) - u(n,Nx-1)); % 更新周期边界条件 u(n+1,1) = u(n+1,Nx+1); u(n+1,Nx+2) = u(n+1,2); end % 绘制数值解图像 x = xStart + (0:Nx-1)dx; t = tStart + (0:Nt-1)dt; figure plot(x, u(1,:), 'b', x, u(round(Nt/10),:), 'r', x, u(end,:), 'g') legend('t=0', 't=1', 't=10') xlabel('x') ylabel('u') title('Numerical Solution of Linear Convection Equation') % 绘制数值解在间断附近的行为图像 figure plot(x, u(round(Nt/10),:), 'r-', x, u(end,:), 'g-') legend('t=1', 't=10') xlabel('x') ylabel('u') title('Numerical Solution Near Discontinuity')

legend('t=0', 't=1', 't=10') xlabel('x') ylabel('u') title('Numerical Solution of Linear Convection Equation') % 绘制数值解在间断附近的行为图像 figure plot(x, u(round(Nt/10),:), 'r-', x, u(end,:), 'g...

kmeans++ matlab

K-means is a clustering algorithm used in data analysis and machine learning. It is used to group similar data points together in a given dataset. The algorithm works by dividing the dataset into k ...

根据论文“Analysis of the first variation and a numerical gradient flow for integral Menger curvature”中的计算 integral Menger curvature的公式编写计算空间曲线integral Menger curvature的matlab代码

根据论文中的公式，计算空间曲线integral Menger ... Analysis of the first variation and a numerical gradient flow for integral Menger curvature. Computer Aided Geometric Design, 29(7):488-501, 2012.

numerical recipes c语言实现

Numerical Recipes是一套著名的数值计算库，它提供了多种数值计算方法的实现，例如：线性求解、非线性求解、求解ODE、FFT、数据拟合等等。该库最初是基于Fortran编写，在1992年推出了C语言版本。 Numerical Recipes...

numerical indexing

Numerical indexing refers to accessing elements or values in an array, list, or matrix using numerical indices or positions. In other words, each element in the data structure is assigned a unique ...

numerical recipes 3rd source code

Numerical Recipes是一个领先的数值计算，在科学和工程领域广泛使用的计算机程序文库。这个库包含了各种常用数值计算算法的源代码和相关的有用文档。Numerical Recipes 3rd Source Code 是 Numerical Recipes 的第三...

解决方法：ValueError Traceback (most recent call last) <ipython-input-4-1b29fd882b35> in <module> 23 y = [1, 1, 0, 0, 0, 1, 0, 0, 1, 1]# 生成决策树 24 clf = DecisionTreeClassifier(random_state=0) ---> 25 clf.fit(X_numerical, y)# 画出决策树 26 plot_tree(clf, feature_names=['日志密度', '好友密度', '是否使用真实头像', '账号是否真实'], class_names=['不被封号', '被封号'], filled=True) D:\Anaconda\lib\site-packages\sklearn\tree\_classes.py in fit(self, X, y, sample_weight, check_input, X_idx_sorted) 896 """ 897 --> 898 super().fit( 899 X, y, 900 sample_weight=sample_weight, D:\Anaconda\lib\site-packages\sklearn\tree\_classes.py in fit(self, X, y, sample_weight, check_input, X_idx_sorted) 279 280 if len(y) != n_samples: --> 281 raise ValueError("Number of labels=%d does not match " 282 "number of samples=%d" % (len(y), n_samples)) 283 if not 0 <= self.min_weight_fraction_leaf <= 0.5: ValueError: Number of labels=10 does not match number of samples=1

请确保y的长度与X_numerical的长度相同，即10。你可以修改代码如下： ```python # 将数据数值化 X = [['s', 's', 'no', 'no'], ['s', 'l', 'yes', 'yes'], ['l', 'm', 'yes', 'yes'], ['m', 'm', 'yes', 'yes'...

numerical recipe in c++

Numerical Recipes in C是一本介绍数值计算方法和算法的经典书籍，它以C语言为载体，详细介绍了数值计算中的常见问题和解决方法。该书的作者是William H. Press等人，由于其深入浅出的讲解和实用的代码示例，成为了...

优化这段代码：import numpy as np import matplotlib.pyplot as plt from ipywidgets import interact, interactive, fixed # 定义模型参数 c = 1.0 # 平流速度 L = 130.0 # 空间范围 T = 40.0 # 时间范围 dx = 1 # 空间步长 dt = 0.05 # 时间步长 # 定义空间和时间的网格 xgrid = np.arange(-30.0, 100.0+dx, dx) tgrid = np.arange(0.0, T+dt, dt) # 定义初始条件 u0 = np.zeros_like(xgrid) u0[(xgrid >= -10.0) & (xgrid <= 10.0)] = 20.0 # 定义边界条件 def bc(u, t): u[0] = u[-1] return u # 定义迎风格式 def upwind(u, dx, dt, c): unew = u.copy() for i in range(1, len(u)): if c > 0: unew[i] = u[i] - cdt/dx(u[i] - u[i-1]) else: unew[i] = u[i] - cdt/dx(u[i+1] - u[i]) return unew # 计算数值解 u = u0.copy() for n in range(len(tgrid)-1): u = bc(u, tgrid[n]) u = upwind(u, dx, dt, c) # 绘制数值解 def plot_numerical_solution(t): n = int(t/dt) plt.plot(xgrid, u0, 'b--', label='Initial Condition') plt.plot(xgrid, u, 'r-', label='Numerical Solution') plt.title('Numerical Solution at t = {:.2f}'.format(tgrid[n])) plt.xlabel('x') plt.ylabel('u') plt.legend() plt.show(),优化后能有下一个时刻的数值解曲线

unew[0] = u[0] unew[1:] = u[1:] - c * dt / dx * (u[1:] - u[:-1]) else: unew[-1] = u[-1] unew[:-1] = u[:-1] - c * dt / dx * (u[1:] - u[:-1]) return unew # 计算数值解 u = u0.copy() for n in range...

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