python查看a表中比b表多出来的行

你可以使用pandas来完成这个任务。假设a表和b表都是pandas的DataFrame，以下是一种可能的实现方式：

import pandas as pd

# 读取a表和b表
a = pd.read_csv('a.csv')
b = pd.read_csv('b.csv')

# 找出a表中不在b表中的行
diff = a[~a.isin(b)].dropna()

# 输出结果
print(diff)

这里使用了isin()函数来判断a表中的每一行是否在b表中出现过，然后使用~运算符取反得到不在b表中的行，最后使用dropna()函数删除所有包含NaN值的行。

相关问题

A星算法的多机器人路径规划的python代码

### 回答1：
A星算法是一种用于解决路径规划问题的算法，它可以用于单机器人和多机器人的情况。下面是一段使用 Python 实现多机器人路径规划的代码示例：

import heapq
import numpy as np

def a_star(grid, start, goal, robot_starts, robot_goals):
    def heuristic(position, goal):
        # 计算启发函数（曼哈顿距离）
        return np.abs(position[0] - goal[0]) + np.abs(position[1] - goal[1])

    def get_neighbors(position, grid):
        # 获取当前位置的邻居
        row, col = position
        neighbors = []
        if row > 0 and grid[row-1, col] != 1:
            neighbors.append((row-1, col))
        if row < grid.shape[0]-1 and grid[row+1, col] != 1:
            neighbors.append((row+1, col))
        if col > 0 and grid[row, col-1] != 1:
            neighbors.append((row, col-1))
        if col < grid.shape[1]-1 and grid[row, col+1] != 1:
            neighbors.append((row, col+1))
        return neighbors

    def reconstruct_path(came_from, current):
        # 通过 came_from 数组重建路径
        path = [current]
        while current in came_from:
            current = came_from[current]
            path.append(current)
        path.reverse()
        return path

    robot_num = len(robot_starts)
    closed_set = [set() for i in range(robot_num)]
    came_from = [{} for i in range(robot_num)]
    g_score = [{robot_starts[i]: 0} for i in range(robot_num)]
    f_score = [{robot_starts[i]: heuristic(robot_starts[i], goal)} for i in range(robot_num)]
    heap = [(f_score[i][robot_starts[i]], i, robot_starts[i]) for i in range(robot_num)]
    heapq.heapify(heap)

    while heap:
        f, robot_idx, current = heapq.heappop(heap)
        if current == goal:
            return [reconstruct_path(came_from[i], current) for i in range(robot_num)]

        closed_set[robot_idx].add(current)
         

### 回答2：
A星算法（A*）是一种常用于路径规划的算法，适用于多机器人路径规划问题。下面是一个使用Python实现A*算法来解决多机器人路径规划的示例代码。

import heapq

# 定义节点类
class Node:
def __init__(self, pos, g, h):
self.pos = pos # 当前节点的位置
self.g = g # 从起点到当前节点的实际距离
self.h = h # 从当前节点到终点的估算距离
self.f = self.g + self.h # f = g + h

# 定义节点比较函数，用于优先队列的排序
def __lt__(self, other):
return self.f < other.f

# 定义A*算法函数
def astar(start, goal, obstacles):
open_list = [] # 用优先队列存放待探索节点
closed_list = set() # 存放已探索节点的位置
heapq.heappush(open_list, start) # 将起点加入优先队列

while open_list:
current_node = heapq.heappop(open_list) # 取出f值最小的节点
if current_node.pos == goal.pos: # 到达目标节点，路径规划完成
return True

closed_list.add(current_node.pos) # 将当前节点置为已探索

# 遍历当前节点的相邻节点
for neighbor in get_neighbors(current_node, obstacles):
if neighbor.pos in closed_list: # 忽略已探索节点
continue

new_g = current_node.g + get_distance(current_node, neighbor) # 新的g值
if neighbor in open_list:
if new_g < neighbor.g: # 当前路径更短，更新g值
neighbor.g = new_g
else:
neighbor.g = new_g
heapq.heappush(open_list, neighbor) # 加入优先队列

return False # 未找到路径

# 获取相邻节点
def get_neighbors(node, obstacles):
neighbors = []
dx = [-1, 0, 1, 0] # 定义上下左右四个方向
dy = [0, 1, 0, -1]

for i in range(4):
x = node.pos[0] + dx[i]
y = node.pos[1] + dy[i]

# 判断相邻节点是否合法
if x < 0 or x >= len(obstacles) or y < 0 or y >= len(obstacles[0]) or obstacles[x][y] == 1:
continue

h = get_distance(Node((x, y), 0, 0), goal) # 估算到目标节点的距离
neighbors.append(Node((x, y), node.g + 1, h))

return neighbors

# 获取节点间的距离（此处采用曼哈顿距离）
def get_distance(node1, node2):
return abs(node1.pos[0] - node2.pos[0]) + abs(node1.pos[1] - node2.pos[1])

# 示例使用
start = Node((0, 0), 0, 0) # 起点
goal = Node((5, 5), 0, 0) # 终点
obstacles = [[0, 1, 0, 0, 0, 0],
[0, 1, 0, 1, 1, 0],
[0, 0, 0, 0, 0, 0],
[0, 0, 0, 1, 1, 0],
[0, 1, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0]]

success = astar(start, goal, obstacles) # 运行A*算法

if success:
print("找到路径")
else:
print("未找到路径")

以上代码是一个简单的示例，实现了多机器人路径规划的A*算法。其中，`start`和`goal`分别表示起点和终点，`obstacles`表示障碍物的矩阵。在运行代码后，将输出成功找到路径或未找到路径的提示。  

### 回答3：
A星算法是一种常用的路径规划算法，多机器人路径规划是指多个机器人同时进行路径规划。下面是一个基于Python的多机器人路径规划的A星算法代码：

import heapq

class PriorityQueue:
def __init__(self):
self.elements = []

def empty(self):
return len(self.elements) == 0

def put(self, item, priority):
heapq.heappush(self.elements, (priority, item))

def get(self):
return heapq.heappop(self.elements)[1]

class Node:
def __init__(self, position, g=0, h=0):
self.position = position
self.g = g
self.h = h
self.f = g + h

def heuristic(a, b):
return abs(b[0] - a[0]) + abs(b[1] - a[1])

def a_star(graph, start, goal):
open_set = PriorityQueue()
open_set.put(start, 0)
came_from = {}
g_score = {start: 0}
f_score = {start: heuristic(start, goal)}

while not open_set.empty():
current = open_set.get()

if current == goal:
path = []
while current in came_from:
path.append(current)
current = came_from[current]
path.append(start)
path.reverse()
return path

for neighbor in graph[current]:
temp_g_score = g_score[current] + 1
if neighbor not in g_score or temp_g_score < g_score[neighbor]:
came_from[neighbor] = current
g_score[neighbor] = temp_g_score
f_score[neighbor] = temp_g_score + heuristic(neighbor, goal)
if neighbor not in open_set.elements:
open_set.put(neighbor, f_score[neighbor])

return None

graph = {
(0, 0): [(0, 1)],
(0, 1): [(0, 0), (1, 1)],
(1, 0): [(1, 1)],
(1, 1): [(0, 1), (1, 0)]
}

start = (0, 0)
goal = (1, 1)
path = a_star(graph, start, goal)
if path:
print("Path found:", path)
else:
print("Path not found.")

以上是一个简单的多机器人路径规划的A星算法的Python代码。代码中使用了优先队列实现open_set，并通过一个字典来记录每个节点的g_score和f_score。heuristic函数用来计算启发式值，指示节点到目标节点的估计距离。a_star函数中使用了A星算法来进行路径规划。最后，通过调用a_star函数来执行路径规划，并输出路径结果。

a星算法python

Sure! Here's an example implementation of the A* algorithm in Python:

import heapq

def heuristic(node, goal):
    # Calculate the Euclidean distance between two points
    return ((node[0] - goal[0]) ** 2 + (node[1] - goal[1]) ** 2) ** 0.5

def astar(graph, start, goal):
    open_set = [(0, start)]  # Priority queue for nodes to be explored
    came_from = {}  # Store the parent of each node
    g_score = {start: 0}  # Cost from start along best known path
    f_score = {start: heuristic(start, goal)}  # Estimated total cost from start to goal

    while open_set:
        current = heapq.heappop(open_set)[1]  # Get the node with the lowest f_score
        if current == goal:
            # Reached the goal
            path = []
            while current in came_from:
                path.append(current)
                current = came_from[current]
            path.append(start)
            path.reverse()
            return path

        for neighbor in graph[current]:
            # Calculate the tentative g_score for the neighbor
            new_g_score = g_score[current] + heuristic(current, neighbor)
            if neighbor not in g_score or new_g_score < g_score[neighbor]:
                # Update the g_score and f_score for the neighbor
                g_score[neighbor] = new_g_score
                f_score[neighbor] = new_g_score + heuristic(neighbor, goal)
                came_from[neighbor] = current
                heapq.heappush(open_set, (f_score[neighbor], neighbor))

    return None  # No path found

# Example usage:
graph = {
    'A': {'B': 5, 'C': 10},
    'B': {'D': 7, 'E': 3},
    'C': {'F': 2},
    'D': {'G': 11},
    'E': {'F': 2},
    'F': {'H': 5},
    'G': {},
    'H': {}
}
start_node = 'A'
goal_node = 'H'
path = astar(graph, start_node, goal_node)
print(path)

In this example, the `graph` dictionary represents a weighted graph where the keys are nodes and the values are dictionaries of neighbors with their corresponding edge weights. The `heuristic` function calculates the heuristic (estimated) distance between two nodes, and the `astar` function implements the A* algorithm to find the shortest path from the `start` node to the `goal` node.

Please note that this is a simplified implementation and may not cover all possible scenarios. You may need to modify it based on your specific requirements.