计算矩阵中邻居数量的最佳方法？

Question

I need to write a function def amountofNeighbours(row, column) that prints the amount of neighbours there are to a certain element in the matrix. For example, given the matrix [[2, 3, 4], [5, 6, 7], [8, 9, 10]] , there are three neighbours to the element 2 at position [0][0], while there are eight neighbours to the element 6 at position [1][1]. I'm not sure what is the best way to handle a problem like this. I went through all the possibilities, and this gave me the following:

def amountofNeighbours(row, column):
    neighbours = 0
    for i in range(row):
        for j in range(column):
            if i == 0 and j == 0 or i == 0 and j == column - 1:
                neighbours = 3
            elif i == row - 1 and j == 0 or i == row-1 and j == column - 1:
                neighbours = 3

            elif i > 0 and j == 0 or i == 0 and j > 0:
                neighbours = 5

            elif i == row - 1 and j > 0:
                neighbours = 5

            elif j == column - 1 and i > 0:
                neighbours = 5

            else:
                neighbours = 8

    return neighbours

When I called this with amountofNeighbours(1, 1) it gave me the correct answer, namely 3, but if I called it with amountofNeighbours(2,2) the answer should be 8 while it gave me 3. Anyone has an idea for improvement?

Answer 1

A straight forward way to do it is to say, "If the cell is at corner, it has three neighbors, otherwise if it is on an edge, it has five, otherwise it has 8."

def numberOfNeighbors(rows, columns, row, column):
    topBottom = row in (0, rows-1)
    leftRight = column in (0, columns-1)
    if topBottom and leftRight:
       return 3
    if topBottom or leftRight:
       return 5
    return 8

Answer 2

Your function as it is designed now does not do what you specified. It takes in the number of rows and columns. Then it loops through all elements of your matrix and calculates the number of neighbours. Then it returns the last value calculated , so the bottom right element of your matrix, which has 3 neighbours indeed.

You should get rid of the loops to get it to do what you want. To clarify:

def amountofNeighbours(row, column, n_rows, n_cols):
    neighbours = 0
    if row == 0 and column == 0 or row == 0 and column == n_cols - 1:
        neighbours = 3
    elif row == n_rows - 1 and column == 0 or row == n_rows-1 and column == n_cols - 1:
        neighbours = 3

    elif row > 0 and column == 0 or row == 0 and column > 0:
        neighbours = 5

    elif row == n_rows - 1 and column > 0:
        neighbours = 5

    elif column == n_cols - 1 and row > 0:
        neighbours = 5

    else:
        neighbours = 8

    return neighbours

Answer 3

One solution to avoid many IFs is to start from the element and compute an "enlarged" box around it (a 3x3 square) possibly placed partially outside the matrix.

Then you clamp the result and return the number of elements minus one:

def neighbors(row, col, rows, cols):
    ra, rb = row-1, row+2
    ca, cb = col-1, col+2
    dx = min(cb, cols) - max(0, ca)
    dy = min(rb, rows) - max(0, ra)
    return dx*dy - 1

The image shows the selected element and the enlarged box around it. The min/max operation will cut off the extra squares leaving a 2x3 box resulting in 2*3-1=5 neighbors count.