Convert positions to integer indices然后Mask out points that are outside the considered spatial extent.然后Sort tensors so that those within the same voxel are consecutives.然后Project to bird's-eye view by summing voxels.然后Put channel in second position and remove z dimension

时间: 2024-04-04 08:35:15 浏览: 173
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MMATrans.rar_Grid map_The First_z map_z-map

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这句话看起来像是一系列的数据处理步骤，需要进一步的上下文才能确定它指的是什么。不过从字面意思来看，这句话的意思可能是： 1. 将位置转换为整数索引； 2. 掩盖掉在考虑的空间范围之外的点； 3. 对张量进行排序，使得在同一个体素内的点是连续的； 4. 通过对体素求和来将点投影到鸟瞰图上； 5. 将通道放在第二个位置，并去掉 z 维度。
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Convert positions to integer indices是什么意思？

"Convert positions to integer indices" 意为将位置转换为整数索引。在计算机科学和图形学中，通常需要将连续的位置信息转换为离散的整数索引，以便于在计算机中进行处理和存储。这种转换通常涉及到以下步骤： 1. ...

# Convert positions to integer indices geometry_b = ((geometry[b] - (self.bev_start_position - self.bev_resolution / 2.0)) / self.bev_resolution) geometry_b = geometry_b.view(N, 3).long()

这段代码的作用是将一个三维空间中的坐标位置转化为整数索引，其中geometry...然后，将geometry_b张量的数据类型转换为整型，得到的结果是一个新的形状为(N,3)的整型张量，表示N个三维位置向量在离散空间中的整数索引。

Convert positions to integer indices geometry_b = ((geometry[b] - (self.bev_start_position - self.bev_resolution / 2.0)) / self.bev_resolution)这段代码中geometry是三维点坐标，b是指batch

具体来说，它首先将每个点的坐标减去一个起始位置（bev_start_position）并除以一个分辨率（bev_resolution），然后将结果取整（转换为整数类型），得到在栅格中的相对位置。这样做的目的是将点的坐标转换为栅格中的...

# Convert positions to integer indices geometry_b = ((geometry[b] - (self.bev_start_position - self.bev_resolution / 2.0)) / self.bev_resolution) geometry_b = geometry_b.view(N, 3).long()这段代码什么意思？

具体地说，它将给定的三维坐标减去一个偏移量，然后除以一个分辨率值，以得到一个新的三维坐标。这个新坐标的每个维度都被转换为整数类型，并被视为一个三元组的索引。最后，这些索引被视为一个形状为(N, 3)的张量...

class VoxelsSumming(torch.autograd.Function): """Adapted from https://github.com/nv-tlabs/lift-splat-shoot/blob/master/src/tools.py#L193""" @staticmethod def forward(ctx, x, geometry, ranks): """The features x and geometry are ranked by voxel positions.""" # Cumulative sum of all features. x = x.cumsum(0) # Indicates the change of voxel. mask = torch.ones(x.shape[0], device=x.device, dtype=torch.bool) mask[:-1] = ranks[1:] != ranks[:-1] x, geometry = x[mask], geometry[mask] # Calculate sum of features within a voxel. x = torch.cat((x[:1], x[1:] - x[:-1])) ctx.save_for_backward(mask) ctx.mark_non_differentiable(geometry) return x, geometry @staticmethod def backward(ctx, grad_x, grad_geometry): (mask,) = ctx.saved_tensors # Since the operation is summing, we simply need to send gradient # to all elements that were part of the summation process. indices = torch.cumsum(mask, 0) indices[mask] -= 1 output_grad = grad_x[indices] return output_grad, None, None这个方法是如何计算BEV的？

在前向传递中，x进行了累加和操作，然后通过比较相邻的ranks元素创建了一个布尔掩码mask。使用掩码可以将x和geometry中相邻的体素合并，然后计算每个体素中所有特征值的总和。在后向传递过程中，由于操作...

class VoxelsSumming(torch.autograd.Function): """Adapted from https://github.com/nv-tlabs/lift-splat-shoot/blob/master/src/tools.py#L193""" @staticmethod def forward(ctx, x, geometry, ranks): """The features x and geometry are ranked by voxel positions.""" # Cumulative sum of all features. x = x.cumsum(0) # Indicates the change of voxel. mask = torch.ones(x.shape[0], device=x.device, dtype=torch.bool) mask[:-1] = ranks[1:] != ranks[:-1] x, geometry = x[mask], geometry[mask] # Calculate sum of features within a voxel. x = torch.cat((x[:1], x[1:] - x[:-1])) ctx.save_for_backward(mask) ctx.mark_non_differentiable(geometry) return x, geometry @staticmethod def backward(ctx, grad_x, grad_geometry): (mask,) = ctx.saved_tensors # Since the operation is summing, we simply need to send gradient # to all elements that were part of the summation process. indices = torch.cumsum(mask, 0) indices[mask] -= 1 output_grad = grad_x[indices] return output_grad, None, None这段代码中有判断是否是同一体素的语句吗？

具体来说，在前向传递中，第12行的语句mask[:-1] = ranks[1:] != ranks[:-1]用于创建一个掩码mask，以指示哪些张量属于同一体素。这个掩码的长度为张量列表中所有张量的数量。mask中的每个元素对应于张量列表...

There are several other sorting algorithms besides selection sort that makesense at your level of programming knowledge. Unfortunately, those algorithms donot offer any advantages over selection sort in terms of their algorithmperformance. Even so, coding these algorithms gives you more practice usingarrays.For example, you can sort an integer array from lowest to highest by applyingthe following procedure. Start by going through the array, looking at adjacent pairsof values. If the values forming the pair are correctly ordered, do nothing; if thevalues are out of order, swap them. In either case, move on the next pair ofvalues. The pairs overlap as you move through the list so that the second elementin one pair becomes the first element of the next pair. Repeat this operation untilyou make a complete pass in which you do not need to exchange any integers.This algorithm is called bubble sort, because the values seem to “bubble up” totheir eventual positions.Reimplement the SortIntegerArray function using the bubble sort algorithm.

Sure, I can help you with that. Here's the implementation of the SortIntegerArray function using the bubble sort algorithm: void SortIntegerArray(int arr[], int size) { for (int i = 0; i ; i++) ...

前向传播不是把排序好的x传入进去了吗？ def forward(ctx, x, geometry, ranks): """The features x and geometry are ranked by voxel positions.""" # Cumulative sum of all features. x = x.cumsum(0) # Indicates the change of voxel. mask = torch.ones(x.shape[0], device=x.device, dtype=torch.bool) mask[:-1] = ranks[1:] != ranks[:-1] x, geometry = x[mask], geometry[mask] # Calculate sum of features within a voxel. x = torch.cat((x[:1], x[1:] - x[:-1])) ctx.save_for_backward(mask) ctx.mark_non_differentiable(geometry) return x, geometry

具体来说，可以通过比较相邻的体素的位置信息，将相邻的体素分为同一组，然后将同一组体素内的数据进行求和，得到每个体素的特征。这个过程中，mask 变量被用来标记哪些数据对应于同一体素，从而方便后续的计算。

代码We now want to always redraw all the points that have ever been drawn in the panel, not just the last point. To do this, we must save the coordinates of all these points so that we can redraw them all one by one in the paintComponent method every time this method is called. To save the coordinates of the various mouse positions we click, replace the x and y instance variables of the MyPanel class with a single private instance variable called points of type ArrayList. The Point class is provided to you by Swing. In the constructor of MyPanel, initialize the points instance variable with a new arraylist object of the same type. In the mouseClicked method of the mouse listener, use the getPoint method of the mouse event object to get a Point object representing the position of the mouse click (that Point object internally stores both the x and y coordinates of the mouse click event). Then add this Point object to the arraylist using the arraylist’s add method. Then, in the paintComponent method, add a loop to draw in the panel all the points of the arraylist. You can get the number of elements in the arraylist by using the size method of the arraylist; you can access a specific element of the arraylist at index i by using the get(i) method of the arraylist (element indexes start at zero in an arraylist). The Point class has getX and getY methods to get the coordinates of the point (these two methods return values of type double so you need to cast the returned values into the int type before you can use them to draw a point).

private ArrayList<Point> points; public MyPanel() { points = new ArrayList(); addMouseListener(this); } public void paintComponent(Graphics g) { super.paintComponent(g); for (int i = 0; i ...

def forward(ctx, x, geometry, ranks): """The features x and geometry are ranked by voxel positions.""" # Cumulative sum of all features. x = x.cumsum(0) # Indicates the change of voxel. mask = torch.ones(x.shape[0], device=x.device, dtype=torch.bool) mask[:-1] = ranks[1:] != ranks[:-1] x, geometry = x[mask], geometry[mask] # Calculate sum of features within a voxel. x = torch.cat((x[:1], x[1:] - x[:-1])) ctx.save_for_backward(mask) ctx.mark_non_differentiable(geometry) return x, geometry这个函数的操作过程我还是不明白。ranks是什么？ranks是排好顺序的点云序列吗？x为什么要求和， x = torch.cat((x[:1], x[1:] - x[:-1]))这样计算能得到什么？

然后，根据ranks张量中相邻元素的不同，创建了一个掩码张量mask，用于指示哪些元素需要被保留。 接下来，根据掩码张量mask筛选出了需要保留的特征向量x'和它们对应的几何信息geometry。然后，通过计算相邻特征向量...

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在animate方法内部，我们首先将初始状态转换为列表，然后记录每个时间步的状态，并将其存储在result列表中。接下来，我们使用两个for循环分别处理向右移动和向左移动的粒子。对于向右移动的粒子，我们计算它们下一...

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然后，在下一行中给出N个不同的正整数。一行中的所有数字由空格分隔。 输出格式：对于每个测试用例，在一行中打印输入数字的相应位置（索引从0开始）。一行中的所有数字由空格分隔，行末不能有多余的空格。如果无法...

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You have a lock with an integer array a of length n written on it. The numbers in the array can be repeated. The lock opens only when the array is sorted in non-decreasing order. To change the state of the array there is only operation available which uses a permutation p of numbers from 1 to n. When this operation is applied the elements of the array change positions according to this permutation, meaning the number ai moves from position i to position Pi for each i.In the beginning the array on the lock was sorted in non-decreasing order but then this operation was applied to it once. Your goal is to unlock the lock again but there's a trick: you can independently restore any position of the array to the value that was on this position before. In general, you can do the following: 1. First, you record the current state of the array. 2. Then you apply the operation to the array moving its elements according to the permutation p. 3. If for every i there exists a previously recorded state in which a equals to the initial value of ai, you stop, otherwise you repeat the first two steps. How many iterations of this algorithm will be performed until you'll be able to restore the initial values? Notice that you strictly follow the algorithm, for example if the array is already sorted in non-decreasing order you stil perform one iteration before you stop.给出中文解释和c++代码

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然后，我们用一个数组cnt记录每个余数出现的次数。由于我们只需要计算满足条件的子序列数目，因此可以用cnt[i]中的余数为i的数目计算出余数为i的子序列的数量。 最后，我们用一个变量ans累加所有余数为i的子序列的...

We have HW cards arranged in a matrix with H rows and W columns. For each i=1,…,N, the card at the A i ​ -th row from the top and B i ​ -th column from the left has a number i written on it. The other HW−N cards have nothing written on them. We will repeat the following two kinds of operations on these cards as long as we can. If there is a row without a numbered card, remove all the cards in that row and fill the gap by shifting the remaining cards upward. If there is a column without a numbered card, remove all the cards in that column and fill the gap by shifting the remaining cards to the left. Find the position of each numbered card after the end of the process above. It can be proved that these positions are uniquely determined without depending on the order in which the operations are done.c++代码

接下来，对于没有数字的列，我们需要将该列删除，即将该列左边的所有列右移一列，然后将最左边一列清空。同样地，这一步操作也可以通过双重循环实现，注意到该操作可能会使得某些数字所在的行数发生改变，因此我们...
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