y = np.array([[1, 2, 1, 2], ... [2, 2, 0, 0], ... [5, 3, 4, 4]]) ... >>> argrelextrema(y, np.less, axis=1)

s1x = np.array([sector_x 0, sector_x 1]) s1y = np.array([sector_u 0, sector_y 1]) s1z = np.array([sector_z 0, sector_z 1]) s2x = np.array([sector_x 1, sector_x 2]) s2y = np.array([sector_x 1, sector_y 2]) s2z = np.array([sector_x 1, sector_z 2]) s3x = np.array([sector_x 2, sector_x 3]) s3y = np.array([sector_x 2, sector_y 3]) s3z = np.array([sector_x 2, sector_z 3])

s1y = np.array([0, sector_y]) s1z = np.array([0, sector_z]) s2x, s2y, s2z = s1x + sector_x, s1y + sector_y, s1z + sector_z s3x, s3y, s3z = s2x + sector_x, s2y + sector_y, s2z + sector_z 这样可以...

import sympy from sympy import diff from sympy import hessian import numpy as np import pandas as pd def f(x1,x2): return 2x12+x22 def f1(x1,x2): x=diff(f(x1,x2),x1) return x def f2(x1,x2): y=diff(f(x1,x2),x2) return y def niudun(x1,x2,e): x3=np.array([x1,x2]) x1,x2=sympy.symbols('x1 x2') k=1 grad=np.array([f1(x1,x2),f2(x1,x2)]) heisai=hessian(f(x1,x2),(x1,x2)) heisai=np.array(heisai,dtype='float') niheisai=np.linalg.inv(heisai) d=-1np.dot(niheisai,grad) fan=np.linalg.norm(np.array(d.astype(float)),ord=2) while abs(fan)>e: x4=np.array(x3)+np.array(d) k+=1 x3=x4 x1=x3[0] x2=x3[1] grad=np.array([f1(x1,x2),f2(x1,x2)]) heisai=hessian(f(x1,x2),(x1,x2)) heisai=np.array(heisai,dtype='float') niheisai=np.linalg.inv(heisai) d=-1*np.dot(np.array(niheisai),grad) d=pd.DataFrame(d,dtype=np.float16) fan=np.linalg.norm(np.array(d.astype(float)),ord=2) return ("运行次数为："+k+','+"极值点为："+x4) x1,x2 = map(float,input("请输入初点：").split(' ')) e= eval(input("请输入精度：")) print(niudun(x1,x2,e))

d = -1 * np.dot(np.array(niheisai), grad) d = pd.DataFrame(d, dtype=np.float16) fan = np.linalg.norm(np.array(d.astype(float)), ord=2) return "运行次数为：" + str(k) + ',' + "极值点为：" + str(x4)...

如果要等价实现下列代码，应该如何更改def edge_detection(image, type): if type == 'roberts': roberts_x = np.array([[-1, 0], [0, 1]]) roberts_y = np.array([[0, -1], [1, 0]]) # roberts 算子计算x和y方向的梯度 gradient_x_roberts = cv2.filter2D(image, -1, roberts_x) gradient_y_roberts = cv2.filter2D(image, -1, roberts_y) edges_roberts = cv2.add(np.abs(gradient_x_roberts), np.abs(gradient_y_roberts)) edges_roberts = np.uint8(edges_roberts) return edges_roberts elif type == 'prewitt': prewitt_x = np.array([[-1, 0, 1], [-1, 0, 1], [-1, 0, 1]]) prewitt_y = np.array([[-1, -1, -1], [0, 0, 0], [1, 1, 1]]) # prewitt 算子计算x和y方向的梯度 gradient_x_prewitt = cv2.filter2D(image, -1, prewitt_x) gradient_y_prewitt = cv2.filter2D(image, -1, prewitt_y) edges_prewitt = cv2.add(np.abs(gradient_x_prewitt), np.abs(gradient_y_prewitt)) edges_prewitt = np.uint8(edges_prewitt) return edges_prewitt elif type == 'sobel': # Sobel算子 sobel_x = np.array([[-1, 0, 1], [-2, 0, 2], [-1, 0, 1]]) sobel_y = np.array([[-1, -2, -1], [0, 0, 0], [1, 2, 1]]) # Sobel 算子计算x和y方向的梯度 gradient_x_sobel = cv2.filter2D(image, -1, sobel_x) gradient_y_sobel = cv2.filter2D(image, -1, sobel_y) edges_sobel = cv2.add(np.abs(gradient_x_sobel), np.abs(gradient_y_sobel)) edges_sobel = np.uint8(edges_sobel) return edges_sobel elif type == 'canny': edges = cv2.Canny(image, threshold1=50, threshold2=100) return edges else: raise NotImplementedError

prewitt_y = np.array([[-1, -1, -1], [0, 0, 0], [1, 1, 1]]) gradient_x_prewitt = conv2d(image, prewitt_x) gradient_y_prewitt = conv2d(image, prewitt_y) edges_prewitt = np.abs(gradient_x_prewitt) ...

datawin1, datawin2, datawin3, Currentout, Timeout = [], [], [], [], [] for i in range(look_back, len(dataset_feature)-look_back, 1): window = dataset_feature[i-look_back:i, 0:-1] window = np.reshape(window, (1, 6look_back)) window = np.array(window) datawin1.append(window) datawin2.append(dataY[i-1]) Timeout.append(Time[i-1]) Currentout.append(Current[i-1]) datawin1 = np.array(datawin1) datawin2 = np.array(datawin2) datawin2 = np.reshape(datawin2, (datawin2.shape[0], datawin2.shape[1])) datawin1 = np.reshape(datawin1, (datawin1.shape[0], 6look_back)) Timeout = np.array(Timeout) Currentout = np.array(Currentout) Timeout = np.reshape(Timeout, (Timeout.shape[0], 1)) Currentout = np.reshape(Currentout, (Currentout.shape[0], 1)) dataset_featureout = np.concatenate((datawin1, datawin2), axis=1)

这段代码看起来是在进行数据窗口化的操作，将原始数据集的某个时间段内的数据作为一个窗口，然后将这些窗口分别存储在 datawin1 和 datawin2 中。其中，datawin1 存储的是特征数据，datawin2 存储的是对应的...

逐行解释np.random.seed(0) X = np.random.randint(0, 10, size=(6, 2)) y = np.array([0, 0, 0, 1, 1, 1]) data = pd.DataFrame(np.concatenate([X, y.reshape(-1, 1)], axis=1), columns=["x1", "x2", "y"]) print(data)

- y = np.array([0, 0, 0, 1, 1, 1])：生成一个包含6个元素的NumPy数组y，其中前三个元素为0，后三个元素为1。 - data = pd.DataFrame(np.concatenate([X, y.reshape(-1, 1)], axis=1), columns=["x1", "x2", "y...

theta_t = np.array([-2,-1,1,2]).reshape(4,1) X_t =np.array([np.linspace(0.1,1.5,15)]).reshape(3,5).T X_t = np.hstack((np.ones((5,1)), X_t)) y_t = np.array([1,0,1,0,1]).reshape(5,1)在python中的意思

这是一个线性回归的例子，其中：...- y_t 是一个 5 行 1 列的数组，表示输出数据。这个例子中的数据可以用来训练一个线性回归模型，使用最小二乘法来求解参数 theta_t，使得模型的预测值与实际值的平方差最小化。

import numpy as np import matplotlib.pyplot as plt from sklearn import linear_model import pandas as pd from sklearn import preprocessing def data_input(): data=pd.read_excel('22AI1.xlsx') data=data.dropna(axis=0) data = data.reset_index(drop=True) X=data.身高 Y=data.体重 X=np.array(X).reshape(-1,1) Y=np.array(Y).reshape(-1,1) return X,Y X,Y=data_input()[0],data_input()[1] clf=linear_model.LinearRegression() clf.fit(X,Y) X2=np.array([[160],[165],[170]]) Y2=clf.predict(X2) res=clf.predict(np.array(190).reshape(-1,1))[0] print('预测身高190的学生体重为:{:.2f}'.format(res[0])) # 画图部分 plt.plot(X,Y,'ks') plt.plot(X2,Y2,'g-') plt.show()我的代码如上,请再次给出解决办法

Y=np.array(Y).reshape(-1,1) return X, Y # 标准化身高数据 X = preprocessing.scale(data_input()[0]) Y=data_input()[1] clf=linear_model.LinearRegression() clf.fit(X,Y) # 标准化待预测的身高数据 X2 = ...

输入并运行下列程序，查看运行结果并提交，自行查阅文档并写出各部分代码的功能： import matplotlib.pyplot as plt import numpy as np #plot 1: xpoints = np.array([0, 6]) ypoints = np.array([0, 100]) plt.subplot(1, 2, 1) plt.plot(xpoints,ypoints) plt.title("plot 1") #plot 2: x = np.array([1, 2, 3, 4]) y = np.array([1, 4, 9, 16]) plt.subplot(1, 2, 2) plt.plot(x,y) plt.title("plot 2") plt.suptitle("绘制子图") plt.show()

y = np.array([1, 4, 9, 16]) plt.subplot(1, 2, 2) # 设置子图的位置 plt.plot(x, y) # 绘制线条 plt.title("plot 2") # 设置子图标题 plt.suptitle("绘制子图") # 设置整个图的标题 plt.show() # 展示图像 ...

x = np.array([0, 1, 2, 3, 4, 5]) y = np.array([1, 2, 3, 2, 5, 6])，计算线性插值时如何对已有的点也计算，例如x=2时插值结果为y=2

y = np.array([1, 2, 3, 2, 5, 6]) # 将已有点也包含在输入数据中 x_all = np.array([0, 1, 2, 3, 4, 5]) y_all = np.array([1, 2, 3, 2, 5, 6]) # 创建一个线性插值函数并对已有点也进行插值计算 f = interp1d(x_...

程序执行报错ValueError: too many values to unpack (expected 2)，im1 = Image.open('skeleton_median.bmp') im2 = Image.open('binary_high.bmp') arr1 = np.array(im1) arr2 = np.array(im2) # 获取矩阵形状 h, w = arr1.shape # 创建坐标矩阵 x, y = np.meshgrid(np.arange(w), np.arange(h)) coords = np.hstack((x.reshape(-1, 1), y.reshape(-1, 1))) # 将两个矩阵合并为一个矩阵 mat = np.hstack((coords, arr1.reshape(-1, 1), arr2.reshape(-1, 1))) # 保存矩阵为txt文件 np.savetxt('mat.txt', mat, fmt='%.2f')

coords = np.hstack((x.reshape(-1, 1), y.reshape(-1, 1))) mat = np.hstack((coords, arr1.reshape(-1, 1), arr2.reshape(-1, 1))) np.savetxt('mat.txt', mat, fmt='%.2f') 如果arr1.shape的值不是一个长度...

import numpy as np import matplotlib.pyplot as plt import sympy from scipy.interpolate import interp1d gamma = 1.2 R = 8.314 T0 = 500 Q = 50 * R * T0 a0 = np.sqrt(gamma * R * T0) M0 = 6.216 P_P0 = sympy.symbols('P_P0') num = 81 x0 = np.linspace(0,1,num) t_t0 = np.linspace(0,15,num) x = x0[1:] T_T0 = t_t0[1:] h0 = [] h1 = []#创建拉姆达为1的空数组 r = [] t = [] c = [] s = [] i = 0 for V_V0 in x: n1 = sympy.solve(1 / (gamma-1) * (P_P0 * V_V0 - 1) - 0.5 * (P_P0 + 1) * (1 - V_V0)- gamma * 0 * Q / a0 ** 2,P_P0)#lamuda=0的Hugoniot曲线方程 n2 = sympy.solve(1 / (gamma-1) * (P_P0 * V_V0 - 1) - 0.5 * (P_P0 + 1) * (1 - V_V0)- gamma * 1 * Q / a0 ** 2,P_P0)#lamuda=1的Hugoniot曲线方程 n3 = sympy.solve(-1 * P_P0 + 1 - gamma * M0 ** 2 * (V_V0 - 1),P_P0)#Reyleigh曲线方程 n4 = 12.014556 / V_V0#等温线 n5 = sympy.solve((P_P0 - 1 / (gamma+1) )* (V_V0-gamma / (gamma + 1)) - gamma / ((gamma + 1) ** 2),P_P0)#声速线 n6 = 10.6677 / np.power(V_V0,1.2)#等熵线 h0.append(n1) h1.append(n2) r.append(n3) t.append(n4) c.append(n5) s.append(n6) i = i+1 h0 = np.array(h0) h1 = np.array(h1) r = np.array(r) t = np.array(t) c = np.array(c) s = np.array(s) plt.plot(x,r,label='Rayleigh') plt.plot(x,t,color='purple',label='isothermal') plt.plot(x,s,color='skyblue',label='isentropic') a = np.where(h0 < 0) b = np.where(c < 0) h0 = np.delete(h0,np.where(h0 < 0)[0],axis = 0)#去除解小于0的值 h1 = np.delete(h1,np.where(h1 < 0)[0],axis = 0)#去除解小于0的值 c = np.delete(c,np.where(c < 0)[0],axis = 0)#去除解小于0的值 x0 = np.delete(x,a,axis = 0)#对应去除x轴上错误值的坐标 x1 = np.delete(x,b,axis = 0) plt.plot(x0,h0,label='Hugoniot(lambda=0)') plt.plot(x0,h1,label='Hugoniot(lambda=1)') plt.plot(x1,c,color='yellow',label='soniclocus') plt.ylim((0,50)) plt.legend() # 显示图例 plt.xlabel('V/V0') plt.ylabel('P/P0') f1 = interp1d(x1, c.T, kind='cubic') f2 = interp1d(x,r.T,kind='cubic') f3 = interp1d(x, t.T, kind='cubic') epsilon = 0.0001 x0 = 0.56 y0 = f1(x0) - f2(x0) while abs(y0) > epsilon: df = (f1(x0 + epsilon) - f2(x0 + epsilon) - y0) / epsilon x0 -= y0 / df y0 = f1(x0) - f2(x0) plt.scatter(x0, y0, 50, color ='red') plt.show()

这是一个 Python 代码，我可以解读出来。这段代码主要是用来绘制气体动力学中的一些...这段代码还包括了一些数据处理的操作，比如删除数组中小于0的值、插值等等。最后，这段代码还用牛顿迭代法求解了两条曲线的交点。

请给下面代码添加注释 import numpy as npdef least_squares(X, y): X = np.array(X) y = np.array(y) w = np.linalg.inv(X.T @ X) @ X.T @ y return wX = np.array([[1, 2], [1, 3], [1, 4]])y = np.array([2, 3, 4])w = least_squares(X, y)print(w)

好的，我来为代码添加注释。...X = np.array([[1, 2], [1, 3], [1, 4]]) y = np.array([2, 3, 4]) # 调用最小二乘法函数并获取系数 w = least_squares(X, y) # 打印系数 print(w) 希望这些注释能帮助理解代码。

请为为这个函数作图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.pif*t)**2) * np.exp(-(np.pift)**2) return t,y i = 0 Frequency = 20 length = 0.128 dt = 0.001# 八层介质 rho = np.array([1.5, 1.8, 2.2, 2.4, 2.6, 2.8, 3.0, 3.2]) v = np.array([1500, 1700, 2000, 2200, 2400, 2600, 2800, 3000]) depth = np.array([0, 50, 100, 150, 200, 250, 300, 350]) Z = rho * v L = (Z[1:] - Z[:-1]) / (Z[1:] + Z[:-1]) t1 = np.arange(0, depth[-1]/v[0]*2, dt) L1 = np.zeros(np.size(t1)) for i in range(1, np.size(depth)): t = depth[i]/v[i-1] + depth[i]/v[i] L1[int(np.round(t/dt))] = L[i-1] t0, w0 = ricker(Frequency, length, dt) syn = np.convolve(L1, w0, 'same')

首先，让我们来理解一下这个函数的作用。该函数定义了一个 Ricker ...这将绘制一个时间范围为 0 到深度最大值的地震记录，其中深度由介质速度和每个介质层的深度计算得出。您可以根据需要调整图形的大小和样式。

import numpy as np import matplotlib.pyplot as plt #1、序列的相加和相乘： n1=np.linspace(0,3,4,dtype=int) x1=np.array([2,0.5,0.9,1]).reshape(1,4) n2=np.linspace(0,7,8,dtype=int) x2=np.linspace(0,0.7,8,dtype=float) n=np.linspace(0,7,8,dtype=int) x1=np.append(x1,np.z

x1 = np.array([2, 0.5, 0.9, 1]).reshape(1, 4) n2 = np.linspace(0, 7, 8, dtype=int) x2 = np.linspace(0, 0.7, 8, dtype=float) n = np.linspace(0, 7, 8, dtype=int) x1 = np.append(x1, np.zeros((1, 4)), ...

翻译代码import numpy as np from cvxopt import matrix, solvers solvers.options['show_progress'] = False # 市场出清，考虑网络阻塞 def market_clearing(alpha): # 供给曲线的截距和斜率 a_real = np.array([15.0, 18.0]) b_real = np.array([0.01, 0.008]) # 需求曲线的截距和斜率 c_real = np.array([40.0, 40.0]) * -1 d_real = np.array([0.08, 0.06]) # 机组功率上下限 p_min = np.array([0.0, 0.0]) p_max = np.array([500.0, 500.0]) # 负荷需求上下限 q_min = np.zeros(2) q_max = np.array([500.0, 666.666666666667]) J_g = ([[-0.333333333333333, -0.333333333333333, -0.666666666666667], [0.333333333333334, -0.666666666666667, -0.333333333333333], [0, 0, 0]]) J = np.array([[-0.333333333333333, 0.0, 0.333333333333333, -0.333333333333334], [-0.333333333333333, 0.0, 0.333333333333333, 0.666666666666667], [-0.666666666666667, 0.0, 0.666666666666667, 0.333333333333333]]) J_max = np.array([25.0, 1000.0, 1000.0, 25.0, 1000.0, 1000.0]) P = matrix(np.diag(np.append(b_real, d_real))) q = matrix(np.append(alpha, c_real)) G = matrix(np.vstack((J, -J, np.diag(-np.ones(4)), np.diag(np.ones(4))))) h = matrix(np.hstack((J_max, -p_min, -q_min, p_max, q_max))) A = matrix(np.hstack((-np.ones(2), np.ones(2)))).T b = matrix(0.0) sv = solvers.qp(P, q, G, h, A, b) miu1 = sv['z'][0:3] miu2 = sv['z'][3:6] nodal_price = (np.ones((3, 1)) * sv['y'][0] - np.dot(J_g, miu1 - miu2)).squeeze() nodal_price_g = np.array([nodal_price[0], nodal_price[2]]) mc_amount = np.array(sv['x'][:2]).squeeze() cost_real = 0.5 * b_real * mc_amount ** 2 + a_real * mc_amount cost_declare = mc_amount * np.transpose(nodal_price_g) profit = cost_declare - cost_real return nodal_price_g, profit if name == 'main': alpha = np.array([20.29, 22.98]) print(market_clearing(alpha))

- a_real和b_real：表示供给曲线的截距和斜率，是一个长度为2的数组。 - c_real和d_real：表示需求曲线的截距和斜率，同样是一个长度为2的数组。 - p_min和p_max：表示机组功率的上下限，也是一个长度为2的数组。 - ...

y = np.array([[1, 2, 1, 2], ... [2, 2, 0, 0], ... [5, 3, 4, 4]]) ... >>> argrelextrema(y, np.less, axis=1)

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逐行解释np.random.seed(0) X = np.random.randint(0, 10, size=(6, 2)) y = np.array([0, 0, 0, 1, 1, 1]) data = pd.DataFrame(np.concatenate([X, y.reshape(-1, 1)], axis=1), columns=["x1", "x2", "y"]) print(data)

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x = np.array([0, 1, 2, 3, 4, 5]) y = np.array([1, 2, 3, 2, 5, 6])，计算线性插值时如何对已有的点也计算，例如x=2时插值结果为y=2

请给下面代码添加注释 import numpy as npdef least_squares(X, y): X = np.array(X) y = np.array(y) w = np.linalg.inv(X.T @ X) @ X.T @ y return wX = np.array([[1, 2], [1, 3], [1, 4]])y = np.array([2, 3, 4])w = least_squares(X, y)print(w)

import numpy as np import matplotlib.pyplot as plt #1、序列的相加和相乘： n1=np.linspace(0,3,4,dtype=int) x1=np.array([2,0.5,0.9,1]).reshape(1,4) n2=np.linspace(0,7,8,dtype=int) x2=np.linspace(0,0.7,8,dtype=float) n=np.linspace(0,7,8,dtype=int) x1=np.append(x1,np.z

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