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推测一下下述代码在高斯研团模型的应用：def gaussnmdl_ins(qt, t, tr, Umean, hs, sigmax, sigmay, sigmaz, x, y, z, inverse, hi=10e50): Gy = 1 / (np.sqrt(2 * np.pi) * sigmay) * np.exp(-0.5 * (y / sigmay) ** 2) if inverse == 0: Gz = 1 / (np.sqrt(2 * np.pi) * sigmaz) * ( np.exp(-0.5 * ((z - hs) / sigmaz) ** 2) + np.exp(-0.5 * ((z + hs) / sigmaz) ** 2)) else: GZ = 0 for ii in range(1, 6): GZ = GZ + np.exp(-0.5 * ((z - 2 * ii * hi - hs) / sigmaz) ** 2) + np.exp( -0.5 * ((z + 2 * ii * hi - hs) / sigmaz) ** 2) + np.exp( -0.5 * ((z - 2 * ii * hi + hs) / sigmaz) ** 2) + np.exp(-0.5 * ((z + 2 * ii * hi + hs) / sigmaz) ** 2) Gz = GZ + 1 / (np.sqrt(2 * np.pi) * sigmaz) * ( np.exp(-0.5 * ((z - hs) / sigmaz) ** 2) + np.exp(-0.5 * ((z + hs) / sigmaz) ** 2)) dist = qt / Umean * Gy * Gz if t <= tr: c = dist / 2 * (erf(x / (sigmax * np.sqrt(2))) - erf((x - Umean * t) / (sigmax * np.sqrt(2)))) else: c = dist / 2 * (erf((x - Umean * (t - tr)) / (sigmax * np.sqrt(2))) - erf( (x - Umean * t) / (sigmax * np.sqrt(2)))) return c 其中x, y = np.meshgrid(Xx, Yy)，Xx、Yy都为np.linspace(-1500, 1500 1500)

这段代码是一个高斯研团模型的实现，用于描述油藏中油的运移过程。...而x, y = np.meshgrid(Xx, Yy)，Xx、Yy都为np.linspace(-1500, 1500 1500)是为了生成一个二维的网格，方便对各个位置上的油层单元进行计算。

def gaussnmdl_ins(qt, t, tr, Umean, hs, sigmax, sigmay, sigmaz, x, y, z, inverse, hi=10e50): Gy = 1 / (np.sqrt(2 * np.pi) * sigmay) * np.exp(-0.5 * (y / sigmay) ** 2) if inverse == 0: Gz = 1 / (np.sqrt(2 * np.pi) * sigmaz) * ( np.exp(-0.5 * ((z - hs) / sigmaz) ** 2) + np.exp(-0.5 * ((z + hs) / sigmaz) ** 2)) else: GZ = 0 for ii in range(1, 6): GZ = GZ + np.exp(-0.5 * ((z - 2 * ii * hi - hs) / sigmaz) ** 2) + np.exp( -0.5 * ((z + 2 * ii * hi - hs) / sigmaz) ** 2) + np.exp( -0.5 * ((z - 2 * ii * hi + hs) / sigmaz) ** 2) + np.exp(-0.5 * ((z + 2 * ii * hi + hs) / sigmaz) ** 2) Gz = GZ + 1 / (np.sqrt(2 * np.pi) * sigmaz) * ( np.exp(-0.5 * ((z - hs) / sigmaz) ** 2) + np.exp(-0.5 * ((z + hs) / sigmaz) ** 2)) dist = qt / Umean * Gy * Gz if t <= tr: c = dist / 2 * (erf(x / (sigmax * np.sqrt(2))) - erf((x - Umean * t) / (sigmax * np.sqrt(2)))) else: c = dist / 2 * (erf((x - Umean * (t - tr)) / (sigmax * np.sqrt(2))) - erf( (x - Umean * t) / (sigmax * np.sqrt(2)))) return c这段代码是高斯烟团模型计算浓度扩散的实现，帮我分析一下执行慢的原因

这段代码执行慢的主要原因是循环和if语句的存在，每次计算都需要进行一次循环和if语句判断，导致程序的执行速度变慢。此外，代码中使用了大量的函数调用，也会增加程序的执行时间。另外，如果数据量较大，每次调用...

import numpy as np def sigmoid(z): return 1 / (1 + np.exp(-z)) def cost_function(theta, X, y): m = len(y) h = sigmoid(X @ theta) J = -(1/m) * (y.T @ np.log(h) + (1-y).T @ np.log(1-h)) grad = (1/m) * X.T @ (h - y) return J, grad def trust_region_newton_method(X, y, max_iter=100, eta=0.05, delta=0.1): n = X.shape[1] theta = np.zeros((n,1)) J, grad = cost_function(theta, X, y) H = np.eye(n) for i in range(max_iter): # solve trust region subproblem p = np.linalg.solve(H, -grad) if np.linalg.norm(p) <= delta: d = p else: d = delta * p / np.linalg.norm(p) # compute actual reduction and predicted reduction J_new, grad_new = cost_function(theta+d, X, y) actual_reduction = J - J_new predicted_reduction = -grad.T @ d - 0.5 * d.T @ H @ d # update trust region radius rho = actual_reduction / predicted_reduction if rho < 0.25: delta = 0.25 elif rho > 0.75 and np.abs(np.linalg.norm(d) - delta) < 1e-8: delta = min(2delta, eta*np.linalg.norm(theta)) # update parameters if rho > 0: theta += d J, grad = J_new, grad_new H += (grad_new - grad) @ (grad_new - grad).T / ((grad_new - grad).T @ d) # check convergence if np.linalg.norm(grad) < 1e-5: break return theta 修改此代码，让他运行出来

delta = min(2*delta, eta*np.linalg.norm(theta)) # update parameters if rho > 0: theta += d J, grad = J_new, grad_new H += (grad_new - grad) @ (grad_new - grad).T / ((grad_new - grad).T @ d) #...

import numpy as np def sigmoid(z): return 1 / (1 + np.exp(-z)) def cost_function(theta, X, y): m = len(y) h = sigmoid(X @ theta) J = -(1/m) * (y.T @ np.log(h) + (1-y).T @ np.log(1-h)) grad = (1/m) * X.T @ (h - y) return J, grad def trust_region_newton_method(X, y, max_iter=100, eta=0.05, delta=0.1): n = X.shape[1] theta = np.zeros((n,1)) J, grad = cost_function(theta, X, y) H = np.eye(n) for i in range(max_iter): # solve trust region subproblem p = np.linalg.solve(H, -grad) if np.linalg.norm(p) <= delta: d = p else: d = delta * p / np.linalg.norm(p) # compute actual reduction and predicted reduction J_new, grad_new = cost_function(theta+d, X, y) actual_reduction = J - J_new predicted_reduction = -grad.T @ d - 0.5 * d.T @ H @ d # update trust region radius rho = actual_reduction / predicted_reduction if rho < 0.25: delta = 0.25 elif rho > 0.75 and np.abs(np.linalg.norm(d) - delta) < 1e-8: delta = min(2delta, eta*np.linalg.norm(theta)) # update parameters if rho > 0: theta += d J, grad = J_new, grad_new H += (grad_new - grad) @ (grad_new - grad).T / ((grad_new - grad).T @ d) # check convergence if np.linalg.norm(grad) < 1e-5: break return theta ——修正代码中的问题，实现信赖域算法求解逻辑回归问题

delta = min(2*delta, eta*np.linalg.norm(theta)) # update parameters if rho > 0: theta += d J, grad = J_new, grad_new H += (grad_new - grad) @ (grad_new - grad).T / ((grad_new - grad).T @ d) ...

详细解释一下import numpy as np import matplotlib.pyplot as plt def ricker(f, length=0.128, dt=0.001): t = np.linspace(-length/2, (length-dt)/2, int(length/dt)) y = (1-2*(np.pi**2)*(f**2)*(t**2))*np.exp(-(np.pi**2)*(f**2)*(t**2)) return t, y f = 20 # 设置Ricker子波频率为20Hz t, wavelet = ricker(f) # 生成20Hz Ricker子波 plt.figure() plt.plot(t, wavelet) plt.xlabel("Time (s)") plt.ylabel("Amplitude") plt.title("Ricker Wavelet (f = %d Hz)" % f) plt.show()

在函数内部，使用np.linspace生成一个时间序列t，然后计算Ricker子波的振幅序列y，并返回时间序列t和振幅序列y。接着，设置Ricker子波的频率为20Hz，调用ricker函数生成20Hz Ricker子波的时间序列和振幅...

# 定义昂贵的函数 def expensive_func(t): return np.sum(t**2 - 10np.cos(2np.pit) + 10) # 定义高斯核函数 def gaussian_kernel(x, y, theta): return np.exp(-theta cdist(x, y)2) # 定义对数似然函数 def log_likelihood(params, x, y): theta, sigma = params k = gaussian_kernel(x, x, theta) + sigma2 * np.eye(len(x)) try: L = np.linalg.cholesky(k) except np.linalg.LinAlgError: return -np.inf alpha = np.linalg.solve(L.T, np.linalg.solve(L, y)) return -0.5y.T.dot(alpha) - np.sum(np.log(np.diag(L))) - 0.5len(x)np.log(2np.pi) # 定义预测函数 def predict(x, y, x0, theta, sigma): k = gaussian_kernel(x, x, theta) + sigma**2 * np.eye(len(x)) k0 = gaussian_kernel(x, x0.reshape(1, -1), theta) k00 = gaussian_kernel(x0.reshape(1, -1), x0.reshape(1, -1), theta) try: L = np.linalg.cholesky(k) except np.linalg.LinAlgError: return np.nan, np.nan alpha = np.linalg.solve(L.T, np.linalg.solve(L, y)) mu = k0.T.dot(alpha) v = k00 - k0.T.dot(np.linalg.solve(L.T, np.linalg.solve(L, k0))) return mu, v # 生成随机数据 np.random.seed(666) X = np.random.uniform(-20, 20, size=(200, 10)) y = np.array([expensive_func(x) for x in X]) # 优化超参数 initial_params = [1, 1] bounds = [(1e-5, None), (1e-5, None)] res = minimize(lambda params: -log_likelihood(params, X, y), initial_params, bounds=bounds) theta, sigma = res.x # 在随机点上进行预测 x0 = np.random.uniform(-20, 20, size=(1, 10)) mu, v = predict(X, y, x0, theta, sigma) # 计算误差 exact_val = expensive_func(x0) error = (exact_val - mu)**2 print("预测误差:", error) print("预测方差:", v)注释一下

这段代码主要实现了使用高斯过程进行回归分析。其中定义了一个昂贵的函数 expensive_func，该函数实现了在给定输入的情况下进行昂贵计算的功能。然后定义了一个高斯核函数 gaussian_kernel，用于计算输入数据的...

from scipy.optimize import root import numpy as np import matplotlib.pyplot as plt # 方程1为2Vo - CEce'=0.1/Vm # 2: Vo + Oox = 2/Vm # 3: CEce' +CEcex = 0.9/Vm # 4: CEce'方·Vo·0.97的二分之一次方 = exp（H/RT）exp(S/R) # Vo:带正电的氧空位，设为var[0], Oox：不带电的氧缺陷，设为var[1] # CEce‘：带一个负电荷的铈缺陷，设为var[2],CEcex：不带电的铈缺陷，设为var[3] # 定义方程组 Vm = 24.3*10**(-6) PO2 = np.array([]) T = np.array([700.0, 800.0, 900.0]) for temp in T: def fun(var): x, y, z, w = var[0], var[1], var[2], var[3] eq1 = 2 * x - y - 0.1 / Vm eq2 = z + w - 0.9 / Vm eq3 = x + y - 2 / Vm eq4 = (z ** 2) * x * (PO2 ** 0.5) - np.exp(436.8 / (8.314 * temp)) * np.exp(0.08 / 8.314) return [eq1, eq2, eq3, eq4] print(temp) # 求解方程组 guess = [0, 0, 0, 0] x0 = np.array(guess) sol = root(fun, x0=x0, method='lm') print('T', temp) print('x', sol.x[0]) print('z', sol.x[2]) 但是现在需要将PO2的值改成10的负24次方到10的负6次方我该如何修改代码

eq4 = (z ** 2) * x * (PO2 ** 0.5) - np.exp(436.8 / (8.314 * temp)) * np.exp(0.08 / 8.314) return [eq1, eq2, eq3, eq4] # 求解方程组 guess = [0, 0, 0, 0] x0 = np.array(guess) sol = root(fun, x0=...

import numpy as np import matplotlib.pyplot as plt def ricker(f, length, dt): t = np.arange(-length/2,(length-dt)/2, dt) y = (1.0 - 2.0*(np.pi**2)*(f**2)*(t**2)) * np.exp(-(np.pi**2)*(f**2)*(t**2)) return t, y def plot_data(ax, data, time, title, ylabel=''): ax.plot(data, time) ax.xaxis.set_ticks_position('top') ax.invert_yaxis() ax.set_title(title, fontsize=12) ax.set_ylabel(ylabel, fontsize=12) # Parameters Frequency = 20 length = 0.128 dt = 0.001 t0, w0 = ricker(Frequency, length, dt) rho = np.array([1.6, 1.8, 2.3, 2.4, 2.8, 2.6, 3.0, 3.4]) v = np.array([1200, 1600, 1300, 2000, 4400, 4000, 5500, 8000]) depth = np.array([100, 200, 500, 1000, 1500, 2000, 3000, 5000]) L = np.zeros(8); Z = rho * v L = np.diff(Z) / (Z[:-1] + Z[1:]) t1 = np.arange(0, 2.75, dt) L1 = np.zeros(np.size(t1)) t = np.cumsum(np.concatenate(([depth[0]2/v[0]], np.diff(depth)2/v[1:]))) L1[np.round(t/dt).astype(int)] = L syn = np.convolve(L1, w0, 'same') fig, axes = plt.subplots(1, 3, figsize=(20, 15), dpi=300) plot_data(axes[0], w0, t0, "Amplitude", "Time(s)") plot_data(axes[1], L1, t1, "Reflection coefficient", "Two-way travel time(s)") plot_data(axes[2], syn, t1, "Amplitude", "Two-way travel time(s)") fig.suptitle('Eight-layer synthetic seismogram', fontsize=18) plt.tight_layout() plt.show()中出现了 L1[np.round(t/dt).astype(int)] = L ValueError: shape mismatch: value array of shape (7,) could not be broadcast to indexing result of shape (8,)帮我改正

y = (1.0 - 2.0*(np.pi**2)*(f**2)*(t**2)) * np.exp(-(np.pi**2)*(f**2)*(t**2)) return t, y def plot_data(ax, data, time, title, ylabel=''): ax.plot(data, time) ax.xaxis.set_ticks_position('top'...

得到函数y_rec的函数表达式：import numpy as np import matplotlib.pyplot as plt def gen_data(x1, x2): y_sample = np.sin(np.pi * x1 / 2) + np.cos(np.pi * x1 / 3) y_all = np.sin(np.pi * x2 / 2) + np.cos(np.pi * x2 / 3) return y_sample, y_all def kernel_interpolation(y_sample, x1, sig): gaussian_kernel = lambda x, c, h: np.exp(-(x - x[c]) ** 2 / (2 * (h ** 2))) num = len(y_sample) w = np.zeros(num) int_matrix = np.asmatrix(np.zeros((num, num))) for i in range(num): int_matrix[i, :] = gaussian_kernel(x1, i, sig) w = int_matrix.I * np.asmatrix(y_sample).T return w def kernel_interpolation_rec(w, x1, x2, sig): gkernel = lambda x, xc, h: np.exp(-(x - xc) ** 2 / (2 * (h ** 2))) num = len(x2) y_rec = np.zeros(num) for i in range(num): for k in range(len(w)): y_rec[i] = y_rec[i] + w[k] * gkernel(x2[i], x1[k], sig) return y_rec

函数y_rec的函数表达式为： y_rec = np.zeros(num) for i in range(num): for k in range(len(w)): y_rec[i] = y_rec[i] + w[k] * np.exp(-(x2[i] - x1[k]) ** 2 / (2 * (sig ** 2)))

我需要你使用手动实现的BO算法实现y=x**2+108在10-15上的最优值寻找

sqdist = np.sum(X1**2, 1).reshape(-1, 1) + np.sum(X2**2, 1) - 2 * np.dot(X1, X2.T) return sigma_f**2 * np.exp(-0.5 / l**2 * sqdist) # Define the acquisition function def acquisition(X, X_sample, Y_...

import numpy as np rows = int(input("输入行数：")) cols = int(input("输入列数：")) # 输入数据元素 data = [] for i in range(rows): row = [] for j in range(cols): value = float(input(f"输入元素[{i}, {j}]: ")) row.append(value) data.append(row) X0 = data m, n = X0.shape X1 = np.cumsum(X0) X1 = np.cumsum(X0) X2 = np.zeros((n-2, n)) for i in range(1, n-1): X2[i-1, i:] = X1[i] + X1[i+1] B = -0.5 * X2 t = np.ones((n-1, 1)) B = np.concatenate((B, t), axis=1) YN = X0[1:] P_t = YN / X1[:n-1] A = np.dot(np.dot(np.linalg.inv(np.dot(B.T, B)), B.T), YN.T) a = A[0] u = A[1] c = u / a b = X0[0] - c X = str(b) + 'exp(' + str(-a) + 'k)' + str(c) equation = 'X(k+1)=' + X print(equation) k = np.arange(len(X0)) Y_k_1 = b * np.exp(-a * k) + c Y = Y_k_1[1:] - Y_k_1[:-1] XY = np.concatenate(([Y_k_1[0]], Y)) print(XY) CA = np.abs(XY - X0) Theta = CA XD_Theta = CA / X0 AV = np.mean(CA) R_k = (np.min(Theta) + 0.5 * np.max(Theta)) / (Theta + 0.5 * np.max(Theta)) P = 0.5 R = np.sum(R_k) / len(R_k) print(R) Temp0 = (CA - AV) 2 Temp1 = np.sum(Temp0) / len(CA) S2 = np.sqrt(Temp1) AV_0 = np.mean(X0) Temp_0 = (X0 - AV_0) 2 Temp_1 = np.sum(Temp_0) / len(CA) S1 = np.sqrt(Temp_1) TempC = S2 / S1 * 100 C = str(TempC) + '%' SS = 0.675 * S1 Delta = np.abs(CA - AV) TempN = np.where(Delta <= SS)[0] N1 = len(TempN) N2 = len(CA) TempP = N1 / N2 * 100 P = str(TempP) + '%'

这段代码是用来计算指数平滑模型的预测值和评估模型拟合优度的指标。具体来说，根据输入的行数和列数创建一个矩阵，并根据用户输入的元素值进行填充。然后，根据指数平滑模型的公式计算预测值，并将其打印出来。...

import numpy as np from scipy.optimize import minimize from scipy.stats import norm # 定义测试函数 def test_func(t): return np.sum(t**2 - 10 * np.cos(2 * np.pi * t) + 10) # 生成200个随机数据点 np.random.seed(42) X = np.random.uniform(low=-20, high=20, size=(200, 10)) y = np.apply_along_axis(test_func, 1, X) # 定义高斯模型 class GaussianProcess: def init(self, kernel, noise=1e-10): self.kernel = kernel self.noise = noise def fit(self, X, y): self.X = X self.y = y self.K = self.kernel(X, X) + self.noise * np.eye(len(X)) self.K_inv = np.linalg.inv(self.K) def predict(self, X_star): k_star = self.kernel(self.X, X_star) y_mean = k_star.T @ self.K_inv @ self.y y_var = self.kernel(X_star, X_star) - k_star.T @ self.K_inv @ k_star return y_mean, y_var # 定义高斯核函数 def rbf_kernel(X1, X2, l=1.0, sigma_f=1.0): dist = np.sum(X12, 1).reshape(-1, 1) + np.sum(X22, 1) - 2 * np.dot(X1, X2.T) return sigma_f**2 * np.exp(-0.5 / l**2 * dist) # 训练高斯模型 gp = GaussianProcess(kernel=rbf_kernel) gp.fit(X, y) # 预测新数据点 X_star = np.random.uniform(low=-20, high=20, size=(1, 10)) y_mean, y_var = gp.predict(X_star) # 计算精确值 y_true = test_func(X_star) # 输出结果 print("预测均值：", y_mean) print("预测方差：", y_var) print("精确值：", y_true) print("预测误差：", (y_true - y_mean)**2) print("预测方差是否一致：", np.isclose(y_var, gp.kernel(X_star, X_star)))不能实现每次运行都是不同的结果

如果你希望每次运行都得到不同的结果，可以在生成随机数据点时设置不同的随机数...y = np.apply_along_axis(test_func, 1, X) 这样每次运行程序时，种子都会不同，生成的随机数据点也会不同，从而得到不同的结果。

解释一下这个代码 #%%In[1] import numpy as np import matplotlib.pyplot as plt #%%In[2] def ricker(f, length, dt): t = np.arange(-length/2,(length-dt)/2, dt) #t = np.arange(0,(length-dt)/2, dt) y = (1.0 - 2.0*(np.pi**2)*(f**2)*(t**2)) * np.exp(-(np.pi**2)*(f**2)*(t**2)) return t,y i = 0 ; Frequency = 20;length = 0.128;dt = 0.001 t0, w0 = ricker(Frequency, length, dt) #%% rho = np.array([1.6, 2.2]) v = np.array([2000, 2500]) depth = 50 Z = rhov L = (Z[1]-Z[0])/(Z[1]+Z[0]) t1 = np.arange(0, 0.2, dt) L1 = np.zeros(np.size(t1)) t = depth2/v[0] L1[int(np.round(t/dt))] = L #%% syn = np.convolve(L1, w0, 'same') #%% fig = plt.figure(num=1, figsize=(20,15),dpi=300) ax1 = fig.add_subplot(1, 3 , 1) ax1.plot(w0, t0) ax1.xaxis.set_ticks_position('top') ax1.invert_yaxis() ax1.set_title("Amplitude", fontsize = 12) ax1.set_ylabel("Time(s)",fontsize = 12) ax2 = fig.add_subplot(1, 3, 2) ax2.plot(L1, t1) ax2.xaxis.set_ticks_position('top') ax2.invert_yaxis() ax2.set_title("Reflection coefficient", fontsize = 12) ax2.set_ylabel("Two-way travel time(s)",fontsize = 12) ax3 = fig.add_subplot(1, 3, 3) ax3.plot(syn, t1) ax3.xaxis.set_ticks_position('top') ax3.invert_yaxis() ax3.set_title("Amplitude", fontsize = 12) ax3.set_ylabel("Two-way travel time(s)",fontsize = 12) fig.suptitle('Two-layer synthetic seismogram', fontsize = 18) plt.tight_layout()

函数返回子波的时间序列t和振幅序列y。 3. i = 0 ; Frequency = 20;length = 0.128;dt = 0.001定义了一些参数，包括采样间隔dt、子波长度length和主频Frequency。 4. t0, w0 = ricker(Frequency, ...

python多项式拟合之np.polyfit 和 np.polyld详解

主要介绍了python多项式拟合之np.polyfit 和 np.polyld的实例代码，python数据拟合主要可采用numpy库，库的安装可直接用pip install numpy等,需要的朋友跟随小编一起学习吧

y = t**2*np.exp(-t**2)

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