Get shortest path to a cell in a 2d array python

问题:

I have a 2d array arr where each cell in it has a value 1, 2 or 3 for example arr[0][0] = 3, arr[2][1] = 2 and arr[0][4] = 1. I want to know the shortest path from a given certain cell eg arr[5][5] to the closest cell which has value 2 where the path shouldn't contain any cells that have the value 1. Any advice?

Below is a the script for the BFS but I don't know how I can make it accept a 2d array as graph and starting point as a certain cell location in the array and then go to the nearest 2 from this cell avoiding cells with 1s, so that it looks like bfs(2darray, starting location, 2)?

def bfs(graph, start, end):
    # maintain a queue of paths
    queue = []
    # push the first path into the queue
    queue.append([start])
    while queue:
        # get the first path from the queue
        path = queue.pop(0)
        # get the last node from the path
        node = path[-1]
        # path found
        if node == end:
            return path
        # enumerate all adjacent nodes, construct a new path and push it into the queue
        for adjacent in graph.get(node, []):
            new_path = list(path)
            new_path.append(adjacent)
            queue.append(new_path)

print bfs(graph, '1', '11')

回答1:

You can use a simple breadth first search for this. Basically, each cell in your grid corresponds to a node in the graph, with edges between adjacent cells. Start at the starting position, and keep expanding passable cells until you find a goal cell.

def bfs(grid, start):
    queue = collections.deque([[start]])
    seen = set([start])
    while queue:
        path = queue.popleft()
        x, y = path[-1]
        if grid[y][x] == goal:
            return path
        for x2, y2 in ((x+1,y), (x-1,y), (x,y+1), (x,y-1)):
            if 0 <= x2 < width and 0 <= y2 < height and grid[y2][x2] != wall and (x2, y2) not in seen:
                queue.append(path + [(x2, y2)])
                seen.add((x2, y2))

Grid setup and results: (Note that I'm using symbols instead of numbers, simply for the reason that it's easier to visually parse the grid this way and to very the solution.)

wall, clear, goal = "#", ".", "*"
width, height = 10, 5
grid = ["..........",
        "..*#...##.",
        "..##...#*.",
        ".....###..",
        "......*..."]
path = bfs(grid, (5, 2))
# [(5, 2), (4, 2), (4, 3), (4, 4), (5, 4), (6, 4)]

回答2:

If the list is not too big, the easiest solution I find is using the where function of the numpy library to find the cells which have the value you are looking for. So, you will need to convert your list into a numpy array.

The code below might be simplified to make it shorter and more efficient, but in this way it will be clearer. By the way, you can compute two kind of distances: the typical euclidean and the manhattan.

If there is more than one target cell at the same distance of the origin cell, min_coords corresponds to the first cell found (first by rows, then by columns).

import numpy as np

# The list needs to be transformed into an array in order to use the np.where method
# arr = np.random.randint(5, size=(6, 6))
arr = np.array([[0, 0, 0, 1, 1, 3],
                [0, 0, 2, 1, 1, 0],
                [0, 0, 1, 1, 1, 1],
                [3, 0, 3, 1, 1, 1], ])

# Origin cell to make the search
x0, y0 = (1, 1)
targetValue = 3

# This is the keypoint of the problem: find the positions of the cells containing the searched value
positions = np.where(arr == targetValue)
x, y = positions

dx = abs(x0 - x)  # Horizontal distance
dy = abs(y0 - y)  # Vertical distance

# There are different criteria to compute distances
euclidean_distance = np.sqrt(dx ** 2 + dy ** 2)
manhattan_distance = abs(dx + dy)
my_distance = euclidean_distance  # Criterion choice
min_dist = min(my_distance)
print(min_dist)

min_pos = np.argmin(my_distance)  # This method will only return the first occurrence (!)
min_coords = x[min_pos], y[min_pos]
print(min_coords)