loss = -np.mean(np.log(y_hat[range(num_samples), y]))没看懂
时间: 2024-04-28 07:21:46 浏览: 10
这是一个计算交叉熵损失的公式,其中y_hat是模型的预测值,y是实际标签。在这个公式中,首先使用range(num_samples)生成一个长度为num_samples的数组,然后通过y_hat[range(num_samples), y]得到预测值中对应y标签的概率值。接着,使用np.log函数取对数,再使用np.mean函数求平均值,最后加上负号得到交叉熵损失。这个公式的具体解释可以参考交叉熵损失函数的定义。
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优化 import numpy as np import open3d as o3d from sklearn.cluster import DBSCAN # 读取点云数据 pcd = o3d.io.read_point_cloud("laser.pcd") points = np.asarray(pcd.points) # DBSCAN聚类 dbscan = DBSCAN(eps=0.2, min_samples=10) dbscan.fit(points) labels = dbscan.labels_ # 获取可行驶区域点云数据 drivable_mask = labels != -1 drivable_points = points[drivable_mask] # 获取路沿点云数据 curb_mask = np.logical_and(labels != -1, points[:, 1] < 0) curb_points = points[curb_mask] # 获取车道线点云数据 line_mask = np.logical_and(labels != -1, points[:, 1] >= 0) line_points = points[line_mask] # 可视化结果 drivable_pcd = o3d.geometry.PointCloud() drivable_pcd.points = o3d.utility.Vector3dVector(drivable_points) o3d.visualization.draw_geometries([drivable_pcd]) curb_pcd = o3d.geometry.PointCloud() curb_pcd.points = o3d.utility.Vector3dVector(curb_points) o3d.visualization.draw_geometries([curb_pcd]) line_pcd = o3d.geometry.PointCloud() line_pcd.points = o3d.utility.Vector3dVector(line_points) o3d.visualization.draw_geometries([line_pcd]) 加上预处理
import numpy as np
import open3d as o3d
from sklearn.cluster import DBSCAN
# 读取点云数据
pcd = o3d.io.read_point_cloud("laser.pcd")
points = np.asarray(pcd.points)
# 预处理:去除离群点
mean = np.mean(points, axis=0)
std = np.std(points, axis=0)
inlier_mask = np.all(np.abs(points - mean) < 2 * std, axis=1)
points = points[inlier_mask]
# DBSCAN聚类
dbscan = DBSCAN(eps=0.2, min_samples=10)
dbscan.fit(points)
labels = dbscan.labels_
# 获取可行驶区域点云数据
drivable_mask = labels != -1
drivable_points = points[drivable_mask]
# 获取路沿点云数据
curb_mask = np.logical_and(labels != -1, points[:, 1] < 0)
curb_points = points[curb_mask]
# 获取车道线点云数据
line_mask = np.logical_and(labels != -1, points[:, 1] >= 0)
line_points = points[line_mask]
# 可视化结果
drivable_pcd = o3d.geometry.PointCloud()
drivable_pcd.points = o3d.utility.Vector3dVector(drivable_points)
o3d.visualization.draw_geometries([drivable_pcd])
curb_pcd = o3d.geometry.PointCloud()
curb_pcd.points = o3d.utility.Vector3dVector(curb_points)
o3d.visualization.draw_geometries([curb_pcd])
line_pcd = o3d.geometry.PointCloud()
line_pcd.points = o3d.utility.Vector3dVector(line_points)
o3d.visualization.draw_geometries([line_pcd])
#预测因子(海温) #nino3.4赤道东太平洋(190-220,-5-5) a22=sst_djf.sel(lon=slice(190,220),lat=slice(5,-5)).mean(axis=1).mean(axis=1) a2=(a22-a22.mean())/a22.std() #赤道印度洋(50-80,-5-5) a33=sst_djf.sel(lon=slice(50,100),lat=slice(5,-5)).mean(axis=1).mean(axis=1) a3=(a33-a33.mean())/a33.std() #预测因子(环流场) #南欧(30-40,35-45) b11=hgt_djf.sel(lon=slice(30,40),lat=slice(45,35)).mean(axis=1).mean(axis=1) b1=(b11-b11.mean())/b11.std() #太平洋副高(120-180,-10-10) b22=hgt_djf.sel(lon=slice(120,180),lat=slice(10,-10)).mean(axis=1).mean(axis=1) b2=(b22-b22.mean())/b22.std() #印度洋(60-80,-10-10) b33=hgt_djf.sel(lon=slice(60,80),lat=slice(10,-10)).mean(axis=1).mean(axis=1) b3=(b33-b33.mean())/b33.std() x=np.vstack([(a2,a3,b1,b2,b3)]).T x2=np.vstack([(a2,b1)]).T y=pre_standard #多元线性回归 res=np.linalg.lstsq(x,y,rcond=None) n=res[0] ##各项系数 y_fit=(n.T*x).sum(axis=1) #拟合数据 res2=np.linalg.lstsq(x2,y,rcond=None) n2=res2[0] ##各项系数 y_fit2=(n2.T*x2).sum(axis=1) #拟合数据 #可视化 time=np.arange(1961,2017,1) fig = plt.figure(figsize=[16, 5]) ax = fig.add_subplot() ax.plot(time, y,marker='o', color='gray', markersize=5) ax.plot(time, y_fit,marker='*', color='b', markersize=5) ax.plot(time, y_fit2,marker='^', color='r', markersize=5) ax.set_title('model',fontsize=20,fontweight='bold') ax.set_xlabel('Time') ax.set_ylabel('Pre') plt.legend(['Source data','Fitted1','Fitted2'],frameon=False,loc='best') plt.show()选做剔除一年的交叉检验,独立试报
好的,我理解你的意思是要对这段代码进行交叉验证和独立测试,以评估模型的性能。交叉验证可以用来评估模型的泛化能力,独立测试可以用来测试模型的预测能力。以下是一个简单的交叉验证和独立测试的示例代码,仅供参考:
交叉验证:
```python
from sklearn.model_selection import KFold
# 将数据集分成 5 份,进行 5 折交叉验证
kf = KFold(n_splits=5, shuffle=True, random_state=1)
# 存储每次交叉验证的评估结果
scores = []
for train_idx, test_idx in kf.split(x):
# 划分训练集和验证集
x_train, x_valid = x[train_idx], x[test_idx]
y_train, y_valid = y[train_idx], y[test_idx]
# 训练模型
res = np.linalg.lstsq(x_train, y_train, rcond=None)
n = res[0]
# 在验证集上进行预测,计算评估指标
y_pred = (n.T * x_valid).sum(axis=1)
score = np.sqrt(((y_pred - y_valid) ** 2).mean())
scores.append(score)
# 计算交叉验证的平均评估指标
mean_score = np.mean(scores)
print('交叉验证 RMSE:', mean_score)
```
独立测试:
```python
# 划分训练集和测试集
train_size = 0.8
n_samples = x.shape[0]
n_train = int(train_size * n_samples)
x_train, x_test = x[:n_train], x[n_train:]
y_train, y_test = y[:n_train], y[n_train:]
# 训练模型
res = np.linalg.lstsq(x_train, y_train, rcond=None)
n = res[0]
# 在测试集上进行预测,计算评估指标
y_pred = (n.T * x_test).sum(axis=1)
test_score = np.sqrt(((y_pred - y_test) ** 2).mean())
print('独立测试 RMSE:', test_score)
```
需要注意的是,这里使用的是简单的线性回归模型,可能存在欠拟合或过拟合的问题,需要根据实际情况进行调整。另外,数据预处理和特征工程也是影响模型性能的重要因素,需要进行合理的处理。