for i = 1 : 499 v_yijpadding = padarray(v_yij, [1,1], 'replicate') %填充,按照边界填充一圈 dv_xij = (conv2(v_yijpadding, T, 'valid') - v_xij) * dt %计算变化值 v_xij = v_xij + dv_xij %更新 v_yij = 0.5 * (abs(v_xij + 1) - abs(v_xij - 1)) %计算输出 vx22 = [vx22 v_xij(2,2)] %扩充输入 vy22 = [vy22 v_yij(2,2)] %扩充输出 end 分析这段代码
时间: 2024-04-17 10:27:18 浏览: 13
这段代码是一个循环,循环变量i从1到499。主要目的是对变量v_xij和v_yij进行迭代更新,并将每次迭代的结果存储在vx22和vy22数组中。
首先,在每次循环中,使用padarray函数将v_yij矩阵进行边界填充,使其在边界上扩充一圈,并将结果存储在v_yijpadding中。
接下来,使用conv2函数对v_yijpadding和T进行卷积运算,得到的结果再减去v_xij,并乘以dt,得到变化值dv_xij。然后,将变化值dv_xij加到v_xij上,更新v_xij的值。
然后,根据公式0.5 * (abs(v_xij + 1) - abs(v_xij - 1))计算出v_yij的值,并将结果存储在v_yij中。
最后,将v_xij(2,2)和v_yij(2,2)的值分别添加到vx22和vy22数组中。
整个过程会重复执行499次,直到循环结束。最终,vx22和vy22数组将包含499个迭代后的v_xij(2,2)和v_yij(2,2)的值。
相关问题
把下列这段Matlab代码转换成python代码:runn = 10; for i = 1:runn disp(['Run: ', num2str(i)]); [M, per] = M.Regression(X, T); %% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Error = per.Error; %%% O1 = M.GetOutput(X); %%% O2 = M.GetOutput(X2); %%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% if i == 1 Train_Error(i,:) = Error; elseif i>1 && i<=runn if size(Error,2) > size(Train_Error,2) Train_Error = padarray(Train_Error, [0, (size(Error,2)-size(Train_Error,2))], 'post'); else Error = padarray(Error, [0, (size(Train_Error,2)-size(Error,2))], 'post'); end Train_Error(i,:) = Error; end O1_(i,:) = O1'; O2_(i,:) = O2'; end
以下是Python代码的转换:
```python
runn = 10
Train_Error = np.zeros((runn, 1))
O1_ = np.zeros((runn, len(X)))
O2_ = np.zeros((runn, len(X2)))
for i in range(runn):
print('Run: ' + str(i+1))
M, per = M.Regression(X, T)
Error = per.Error
O1 = M.GetOutput(X)
O2 = M.GetOutput(X2)
if i == 0:
Train_Error[i,:] = Error
elif i>0 and i<runn:
if Error.shape[1] > Train_Error.shape[1]:
Train_Error = np.pad(Train_Error, [(0,0), (0, Error.shape[1]-Train_Error.shape[1])], 'constant')
else:
Error = np.pad(Error, [(0,0), (0, Train_Error.shape[1]-Error.shape[1])], 'constant')
Train_Error[i,:] = Error
O1_[i,:] = O1.T
O2_[i,:] = O2.T
```
请注意,在Python中,你需要导入numpy库使用np.zeros和np.pad函数。此外,Python从0开始索引列表,因此在迭代器中需要添加1。
把下列这段Matlab代码转换成python代码:runn = 10; for i = 1:runn disp(['Run: ', num2str(i)]); [M, per] = M.Regression(X, T); Error = per.Error; O1 = M.GetOutput(X); O2 = M.GetOutput(X2); if i == 1 Train_Error(i,:) = Error; elseif i>1 && i<=runn if size(Error,2) > size(Train_Error,2) Train_Error = padarray(Train_Error, [0, (size(Error,2)-size(Train_Error,2))], 'post'); else Error = padarray(Error, [0, (size(Train_Error,2)-size(Error,2))], 'post'); end Train_Error(i,:) = Error; end O1_(i,:) = O1'; O2_(i,:) = O2'; end
以下是Python代码的转换:
```python
runn = 10
Train_Error = np.zeros((runn, 1))
O1_ = np.zeros((runn, len(X)))
O2_ = np.zeros((runn, len(X2)))
for i in range(runn):
print('Run: ' + str(i+1))
M, per = M.Regression(X, T)
Error = per.Error
O1 = M.GetOutput(X)
O2 = M.GetOutput(X2)
if i == 0:
Train_Error[i,:] = Error
elif i>0 and i<runn:
if Error.shape[1] > Train_Error.shape[1]:
Train_Error = np.pad(Train_Error, [(0,0), (0, Error.shape[1]-Train_Error.shape[1])], 'constant')
else:
Error = np.pad(Error, [(0,0), (0, Train_Error.shape[1]-Error.shape[1])], 'constant')
Train_Error[i,:] = Error
O1_[i,:] = O1.T
O2_[i,:] = O2.T
```
请注意,在Python中,你需要导入numpy库使用np.zeros和np.pad函数。此外,Python从0开始索引列表,因此在迭代器中需要添加1。