Can't convert non-rectangular Python sequence to Tensor.

这个报错通常表示您尝试将一个非矩形的Python序列（如一个列表中含有不同长度的子列表）转换为Tensor，但是Tensor要求所有的子列表必须是相同长度的。

解决这个问题的方法是确保您的输入数据具有相同的形状。您可以使用`numpy.pad()`或其他方法来填充不同长度的子列表，以确保它们具有相同的形状。

例如，如果您有一个列表`my_list`，其中包含多个不同长度的子列表：

my_list = [[1, 2, 3], [4, 5], [6, 7, 8, 9]]

您可以使用`numpy.pad()`来将每个子列表填充到相同的长度：

import numpy as np

max_len = max(len(sub_list) for sub_list in my_list)
padded_list = [np.pad(sub_list, (0, max_len - len(sub_list)), mode='constant') for sub_list in my_list]

这将生成一个具有相同长度的新列表`padded_list`：

[[1, 2, 3, 0], [4, 5, 0, 0], [6, 7, 8, 9]]

现在，您可以将`padded_list`转换为Tensor，而不会出现`Can't convert non-rectangular Python sequence to Tensor`的错误。

相关问题

Vika and her friends went shopping in a mall, which can be represented as a rectangular grid of rooms with sides of length n and m. Each room has coordinates (a, b), where 1 ≤ a ≤ n, 1 ≤ b ≤ m. Thus we call a hall with coordinates (c, d) a neighbouring for it if ｜a-c｜+|b-d｜=1.Tired of empty fashion talks, Vika decided to sneak away unnoticed. But since she hasn't had a chance to visit one of the shops yet, she doesn't want to leave the mall. After a while, her friends noticed Vika's disappearance and started looking for her.Currently, Vika is in a room with coordinates (z, y), and her k friends are in rooms with coordinates (x1, Y1), (x2, 32),..,(ük) 3k).respectively. The coordinates can coincide. Note that all the girls must move to the neighbouring rooms.Every minute, first Vika moves to one of the adjacent to the side rooms of her choice, and then each friend (seeing Vika's choice) also chooses one of the adjacent rooms to move to.If at the end of the minute (that is, after all the girls have moved on to the neighbouring rooms) at least one friend is in the same room as Vika, she is caught and all the other friends are called.Tell us, can Vika run away from her annoying friends forever, or will she have to continue listening to empty fashion talks after some time?如果能不被抓住输出yes，否则输出no，给出c++代码及中文解释

可以使用深度优先搜索（DFS）来解决这个问题。首先，我们可以将问题抽象为一个图，其中每个房间都是一个节点，相邻房间之间有边相连。

首先，我们需要定义一个辅助函数，用于检查给定的坐标是否在合法的范围内：

bool isValid(int x, int y, int n, int m) {
    return (x >= 1 && x <= n && y >= 1 && y <= m);
}

接下来，我们可以使用DFS来遍历所有可能的移动方式。我们从Vika的起始位置开始，依次向四个方向移动，然后递归地对每个朋友进行相同的操作。如果在递归过程中，任何一个朋友与Vika在同一个房间，则说明Vika会被抓住。

以下是完整的C++代码实现：

#include <iostream>
#include <vector>

using namespace std;

bool isValid(int x, int y, int n, int m) {
    return (x >= 1 && x <= n && y >= 1 && y <= m);
}

bool dfs(int vx, int vy, vector<pair<int, int>>& friends, vector<vector<bool>>& visited, int n, int m) {
    if (vx == 0 && vy == 0) {
        // Vika has successfully escaped
        return true;
    }
    
    visited[vx][vy] = true;
    
    // Possible neighbors' coordinates
    vector<pair<int, int>> neighbors = {{vx-1, vy}, {vx+1, vy}, {vx, vy-1}, {vx, vy+1}};
    
    for (auto neighbor : neighbors) {
        int nx = neighbor.first;
        int ny = neighbor.second;
        
        if (isValid(nx, ny, n, m) && !visited[nx][ny]) {
            bool caught = false;
            
            for (auto friendCoord : friends) {
                int fx = friendCoord.first;
                int fy = friendCoord.second;
                
                if (nx == fx && ny == fy) {
                    caught = true;
                    break;
                }
            }
            
            if (!caught && dfs(nx, ny, friends, visited, n, m)) {
                return true;
            }
        }
    }
    
    return false;
}

string canVikaEscape(int n, int m, int vx, int vy, vector<pair<int, int>>& friends) {
    vector<vector<bool>> visited(n+1, vector<bool>(m+1, false));
    
    if (dfs(vx, vy, friends, visited, n, m)) {
        return "Yes";
    } else {
        return "No";
    }
}

int main() {
    int n, m, vx, vy, k;
    cin >> n >> m >> vx >> vy >> k;
    
    vector<pair<int, int>> friends(k);
    
    for (int i = 0; i < k; i++) {
        cin >> friends[i].first >> friends[i].second;
    }
    
    cout << canVikaEscape(n, m, vx, vy, friends) << endl;
    
    return 0;
}

该代码首先读取输入的n，m，vx，vy和k值。接下来，它读取k个朋友的坐标，并调用canVikaEscape函数来判断Vika是否能逃脱。最后，它输出结果。

希望这可以帮助到你！

def crop(self, box=None): """ Returns a rectangular region from this image. The box is a 4-tuple defining the left, upper, right, and lower pixel coordinate. See :ref:`coordinate-system`. Note: Prior to Pillow 3.4.0, this was a lazy operation. :param box: The crop rectangle, as a (left, upper, right, lower)-tuple. :rtype: :py:class:`~PIL.Image.Image` :returns: An :py:class:`~PIL.Image.Image` object. """ if box is None: return self.copy() if box[2] < box[0]: msg = "Coordinate 'right' is less than 'left'" raise ValueError(msg) elif box[3] < box[1]: msg = "Coordinate 'lower' is less than 'upper'" raise ValueError(msg) self.load() return self._new(self._crop(self.im, box))

根据你提供的代码，错误发生在Image.py文件的crop方法中。具体来说，如果传入的box参数中，右边的坐标值小于左边的坐标值，就会引发ValueError异常并抛出"Coordinate 'right' is less than 'left'"的错误消息。

你需要检查传入crop方法的box参数，确保右边的坐标值大于左边的坐标值。这个错误通常意味着传入的坐标范围不正确，导致无法进行裁剪操作。