sum_of_squares = lambda x, y: x**2 + y**2

这段代码中应该是要计算 `x` 和 `y` 的平方和，但是代码中的语法错误导致了无法正确执行。应该将 `x**2` 和 `y**2` 使用加号连接起来，代码如下：

sum_of_squares = lambda x, y: x**2 + y**2

这个函数可以接收两个参数 `x` 和 `y`，并返回它们的平方和。例如，如果调用 `sum_of_squares(3, 4)`，函数会返回 `25`。

相关问题

编写一个匿名函数，计算两个数的平方和。使用匿名函数和map函数计算两个数字列表（列表值可以直接给定）对应位置数值的平方和并输出。

可以参考以下代码实现：

num_list_1 = [1, 2, 3, 4, 5]
num_list_2 = [2, 4, 6, 8, 10]

# 定义匿名函数，计算两个数的平方和
sum_of_squares = lambda x, y: x**2 + y**2

# 使用 map 函数，计算两个数字列表对应位置数值的平方和并输出
result = list(map(sum_of_squares, num_list_1, num_list_2))
print(result)

输出结果为：

[5, 20, 45, 80, 125]

其中，`sum_of_squares` 是一个匿名函数，可以接收两个参数 x 和 y，返回它们的平方和。使用 `map` 函数，将 `sum_of_squares` 函数应用到 `num_list_1` 和 `num_list_2` 中对应位置的数值上，得到一个新的列表 `result`，其中每个元素都是对应位置上两个数的平方和。最后将 `result` 输出即可。

优化这段代码function parafitzzsd1 t=[1964 1985 1992 1997 1999 2004 2007 2020]; y=[1.0625 1.1333 1.4167 1.4407 1.4783 1.4783 1.5455 1.5455]; z=[0.6563 0.7188 0.8438 0.9375 0.8906 0.8125 0.9375 0.9375]; y0=1.0625; % Nonlinear least square estimate using lsqnonlin() k0=[0,0,0]; lb=[0,0,0];ub=[inf,inf,inf]; [k,resnorm,residual,exitflag,output,lambda,jacobian] = lsqnonlin(@Func,k0,lb,ub,[],t,y0); ci = nlparci(k,residual,jacobian); k; % result fprintf('\n Estimated Parameters by Lsqnonlin():\n') fprintf('\t k1 = %.4f ± %.4f\n',k(1),ci(1,2)-k(1)) fprintf('\t k2 = %.4f ± %.4f\n',k(2),ci(2,2)-k(2)) fprintf('\t k3 = %.4f ± %.4f\n',k(3),ci(3,2)-k(3)) fprintf(' The sum of the residual squares is: %.1e\n\n',sum(residual.^2)) % plot of fit results tspan = [1964 2050]; [tt yc] = ode45(@ModelEqs,tspan,y0,[],k); tc=linspace(1964,2050,400); yca = spline(tt,yc,tc); plot(t,y,'ro',tc,yca,'r-'); hold on xlabel('Time'); ylabel('y'); hold off % ======================================= function f1 = Func(k,t,y,y0) % Define objective function tspan =t; [tt yy] = ode45(@ModelEqs,tspan,y0,[],k); yc= spline(tt,yy,t); f1=y-yc; % ================================== function dydt = ModelEqs(t,y,k) dydt =k(1)*y-k(2)*y.^2+k(3)*y*z;

There are a few ways to optimize this code:

1. Vectorization: Instead of using a for loop to solve the ODE, you can use vectorization to solve it for all time points at once. This can significantly speed up the code.

2. Preallocation: Preallocate any arrays that are used in loops to avoid reallocation at each iteration.

3. Use anonymous functions: Instead of defining separate functions for the objective function and the ODE, you can use anonymous functions to define them inline. This can make the code more concise and easier to read.

Here's an optimized version of the code:

% Define data
t = [1964 1985 1992 1997 1999 2004 2007 2020];
y = [1.0625 1.1333 1.4167 1.4407 1.4783 1.4783 1.5455 1.5455];
z = [0.6563 0.7188 0.8438 0.9375 0.8906 0.8125 0.9375 0.9375];
y0 = 1.0625;

% Define objective function and ODE inline
objfun = @(k) y - spline(ode45(@(t,y) k(1)*y - k(2)*y.^2 + k(3)*y.*z, t, y0), t);
odefun = @(t,y,k) k(1)*y - k(2)*y.^2 + k(3)*y.*z;

% Nonlinear least square estimate using lsqnonlin()
k0 = [0, 0, 0];
lb = [0, 0, 0];
ub = [inf, inf, inf];
options = optimoptions('lsqnonlin', 'Display', 'off');
k = lsqnonlin(objfun, k0, lb, ub, options);

% Calculate confidence intervals
J = jacobianest(objfun, k);
ci = nlparci(k, objfun(k), J);

% Display results
fprintf('\n Estimated Parameters by Lsqnonlin():\n')
fprintf('\t k1 = %.4f ± %.4f\n', k(1), ci(1,2)-k(1))
fprintf('\t k2 = %.4f ± %.4f\n', k(2), ci(2,2)-k(2))
fprintf('\t k3 = %.4f ± %.4f\n', k(3), ci(3,2)-k(3))
fprintf('  The sum of the residual squares is: %.1e\n\n', sum(objfun(k).^2))

% Plot of fit results
tspan = [1964  2050];
[tt, yc] = ode45(odefun, tspan, y0, [], k);
tc = linspace(1964, 2050, 400);
yca = spline(tt, yc, tc);
plot(t, y, 'ro', tc, yca, 'r-');
xlabel('Time');
ylabel('y');