'tuple' 对象在我的 Python 代码中没有属性 'position'。我该如何解决？
'tuple' object has no attribute 'position' in my Python code. How can I fix it?
我目前正在 python 中实现 A* 寻路。我一直在关注一个来源并对其进行改编，但我遇到了一个我无法解决的问题。错误涉及pythons eq函数，这里是：
Traceback (most recent call last):
File "C:/Users/Patrick/PycharmProjects/untitled/astarmaze.py", line 154, in <module>
main()
File "C:/Users/Patrick/PycharmProjects/untitled/astarmaze.py", line 149, in main
taken_path = a.astarloop()
File "C:/Users/Patrick/PycharmProjects/untitled/astarmaze.py", line 114, in astarloop
if adj == closed_adj:
File "C:/Users/Patrick/PycharmProjects/untitled/astarmaze.py", line 33, in __eq__
return self.position == other.position
File "C:/Users/Patrick/PycharmProjects/untitled/astarmaze.py", line 33, in __eq__
return self.position == other.position
AttributeError: 'list' object has no attribute 'position'
Process finished with exit code 1
这里是有问题的 code/function：
class Cell(object):
def __init__(self, position=None, parent=None):
self.parent = parent
self.position = position
# our g, h, f variables for f = g + h
self.g = 0
self.h = 0
self.f = 0
def __eq__(self, other):
return self.position == other.position
# def __getitem__(self, index):
# return self.position[index]
#Class for our A* pathfinding, which can take the
class Astar(object):
#initialise the pathfinding and all relevant variables by passing in the maze and its start/end
def __init__(self, maze, start, end, mazeX, mazeY):
# Initialising the open and closed lists for the cells
self.list_open = []
self.list_closed = []
self.maze = maze
self.mazeX = mazeX
self.mazeY = mazeY
self.start_cell = Cell(start, None)
self.start_cell.g = 0
self.start_cell.h = 0
self.start_cell.f = 0
self.end_cell = Cell(end, None)
self.end_cell.g = 0
self.start_cell.h = 0
self.start_cell.f = 0
def astarloop(self):
# adds the starting cell to the open list
self.list_open.append(self.start_cell)
# finding the end cell, loops until found
while len(self.list_open) > 0:
# current cell
current_cell = self.list_open[0]
current_i = 0
for i, item in enumerate(self.list_open):
if item.f < current_cell.f:
current_cell = item
current_i = i
# now it takes the current cell off the open list, and adds it to the closed list, since it has been visited
self.list_open.pop(current_i)
self.list_closed.append(current_cell)
# when the end cell is found, return an array that shows the path taken through the maze to reach the end
if current_cell == self.end_cell:
taken_path = []
current = current_cell
while current is not None:
taken_path.append(current.position)
current = current.parent
return taken_path[::-1] # return the taken path but in reverse
# Generate the children from the current cell, this is important for the algorithm to decide where to go
# based on the 'f' value - it should pick the lowest costed cell to get to the end fastest, adjecent cells to current one
adj_cells = []
for new_position in [(0, -1), (0, 1), (-1, 0), (1, 0), (-1, -1), (-1, 1), (1, -1), (1, 1)]: #adjacent squares in 8 directions
# getting the current cell position
cell_position = (current_cell.position[0] + new_position[0], current_cell.position[1] + new_position[1])
# make sure the cell pos is within the range of the maze array
if cell_position[0] > (len(self.maze) - 1) or cell_position[0] < 0 or cell_position[1] > (len(self.maze[len(self.maze)-1]) - 1) or cell_position[1] < 0:
continue
if self.maze[cell_position[0]][cell_position[1]] != 0:
continue
# new child cell to be appended to adjacent cells
new_cell = Cell(current_cell, cell_position)
adj_cells.append(new_cell)
# time to loop through the adjacent cells
for adj in adj_cells:
for closed_adj in self.list_closed:
if adj == closed_adj:
continue
# create f, g and h for the child adjacent cell
adj.g = current_cell.g + 1
adj.h = ((adj.position[0] - self.end_cell.position[0]) ** 2) + ((adj.position[1] - self.end_cell.position[1]) ** 2)
adj.f = adj.g + adj.h
# already on open list
for open_adj in self.list_open:
if adj == open_adj and adj.g > open_adj.g:
continue
# add to open
self.list_open.append(adj)
def main():
maze = [[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0]]
start = (0, 0)
end = (7, 6)
a = Astar(maze, start, end, 0, 0)
taken_path = a.astarloop()
print(taken_path)
if __name__ == "__main__":
main()
如您所见，eq 函数在 Cell class 中似乎没有正常工作，并且没有返回我想要的结果。我试过更改从元组传递到列表的迷宫的起点和终点，但同样的错误仍然存在。我该如何修复我的代码？
这是构造函数
class Cell(object):
def __init__(self, position=None, parent=None):
self.parent = parent
self.position = position
这是你使用它的地方
new_cell = Cell(current_cell, cell_position)
看来你把position传给parent，parent传给position了
稍后，这会导致比较 Cell.eq(self.position, other.position) 时出现问题，因为 self.position 实际上是一个单元格，而 other.position 是一个如您所料的原始元组。所以它再次在 self.position 上调用 eq 试图访问 other.position.position （它不存在，因为它是原始元组而不是 Cell 的位置 属性）
您可以使用
修复它
new_cell = Cell(parent=current_cell, position=cell_position)
或者只是将参数按正确的顺序排列。
我目前正在 python 中实现 A* 寻路。我一直在关注一个来源并对其进行改编，但我遇到了一个我无法解决的问题。错误涉及pythons eq函数，这里是：
Traceback (most recent call last):
File "C:/Users/Patrick/PycharmProjects/untitled/astarmaze.py", line 154, in <module>
main()
File "C:/Users/Patrick/PycharmProjects/untitled/astarmaze.py", line 149, in main
taken_path = a.astarloop()
File "C:/Users/Patrick/PycharmProjects/untitled/astarmaze.py", line 114, in astarloop
if adj == closed_adj:
File "C:/Users/Patrick/PycharmProjects/untitled/astarmaze.py", line 33, in __eq__
return self.position == other.position
File "C:/Users/Patrick/PycharmProjects/untitled/astarmaze.py", line 33, in __eq__
return self.position == other.position
AttributeError: 'list' object has no attribute 'position'
Process finished with exit code 1
这里是有问题的 code/function：
class Cell(object):
def __init__(self, position=None, parent=None):
self.parent = parent
self.position = position
# our g, h, f variables for f = g + h
self.g = 0
self.h = 0
self.f = 0
def __eq__(self, other):
return self.position == other.position
# def __getitem__(self, index):
# return self.position[index]
#Class for our A* pathfinding, which can take the
class Astar(object):
#initialise the pathfinding and all relevant variables by passing in the maze and its start/end
def __init__(self, maze, start, end, mazeX, mazeY):
# Initialising the open and closed lists for the cells
self.list_open = []
self.list_closed = []
self.maze = maze
self.mazeX = mazeX
self.mazeY = mazeY
self.start_cell = Cell(start, None)
self.start_cell.g = 0
self.start_cell.h = 0
self.start_cell.f = 0
self.end_cell = Cell(end, None)
self.end_cell.g = 0
self.start_cell.h = 0
self.start_cell.f = 0
def astarloop(self):
# adds the starting cell to the open list
self.list_open.append(self.start_cell)
# finding the end cell, loops until found
while len(self.list_open) > 0:
# current cell
current_cell = self.list_open[0]
current_i = 0
for i, item in enumerate(self.list_open):
if item.f < current_cell.f:
current_cell = item
current_i = i
# now it takes the current cell off the open list, and adds it to the closed list, since it has been visited
self.list_open.pop(current_i)
self.list_closed.append(current_cell)
# when the end cell is found, return an array that shows the path taken through the maze to reach the end
if current_cell == self.end_cell:
taken_path = []
current = current_cell
while current is not None:
taken_path.append(current.position)
current = current.parent
return taken_path[::-1] # return the taken path but in reverse
# Generate the children from the current cell, this is important for the algorithm to decide where to go
# based on the 'f' value - it should pick the lowest costed cell to get to the end fastest, adjecent cells to current one
adj_cells = []
for new_position in [(0, -1), (0, 1), (-1, 0), (1, 0), (-1, -1), (-1, 1), (1, -1), (1, 1)]: #adjacent squares in 8 directions
# getting the current cell position
cell_position = (current_cell.position[0] + new_position[0], current_cell.position[1] + new_position[1])
# make sure the cell pos is within the range of the maze array
if cell_position[0] > (len(self.maze) - 1) or cell_position[0] < 0 or cell_position[1] > (len(self.maze[len(self.maze)-1]) - 1) or cell_position[1] < 0:
continue
if self.maze[cell_position[0]][cell_position[1]] != 0:
continue
# new child cell to be appended to adjacent cells
new_cell = Cell(current_cell, cell_position)
adj_cells.append(new_cell)
# time to loop through the adjacent cells
for adj in adj_cells:
for closed_adj in self.list_closed:
if adj == closed_adj:
continue
# create f, g and h for the child adjacent cell
adj.g = current_cell.g + 1
adj.h = ((adj.position[0] - self.end_cell.position[0]) ** 2) + ((adj.position[1] - self.end_cell.position[1]) ** 2)
adj.f = adj.g + adj.h
# already on open list
for open_adj in self.list_open:
if adj == open_adj and adj.g > open_adj.g:
continue
# add to open
self.list_open.append(adj)
def main():
maze = [[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0]]
start = (0, 0)
end = (7, 6)
a = Astar(maze, start, end, 0, 0)
taken_path = a.astarloop()
print(taken_path)
if __name__ == "__main__":
main()
如您所见，eq 函数在 Cell class 中似乎没有正常工作，并且没有返回我想要的结果。我试过更改从元组传递到列表的迷宫的起点和终点，但同样的错误仍然存在。我该如何修复我的代码？
这是构造函数
class Cell(object):
def __init__(self, position=None, parent=None):
self.parent = parent
self.position = position
这是你使用它的地方
new_cell = Cell(current_cell, cell_position)
看来你把position传给parent，parent传给position了
稍后，这会导致比较 Cell.eq(self.position, other.position) 时出现问题，因为 self.position 实际上是一个单元格，而 other.position 是一个如您所料的原始元组。所以它再次在 self.position 上调用 eq 试图访问 other.position.position （它不存在，因为它是原始元组而不是 Cell 的位置 属性）
您可以使用
修复它new_cell = Cell(parent=current_cell, position=cell_position)
或者只是将参数按正确的顺序排列。