python - 递归函数中的计数器

Question

我是 python 和一般编程的新手。我编写了一个函数，它将搜索数组中的相邻元素并查找值彼此相差在 0.05 以内的元素，就像 floodfill 算法一样。唯一的区别是我在计算函数运行次数时做了一些愚蠢的事情(我想这也会告诉我找到了多少元素)，所以我的计数器值是错误的。该代码在查找彼此相距 0.05 以内的相邻元素时有效，只是计数很有趣。

def floodcount (x,y,array,value,count=0):     #akin to a bucket fill in paint, finds the area instead

    nrows = len(array)-1          #rows of the image
    ncols = len(array[0])-1       #columns of the image
    diff = array[x][y] - value
    if (diff < 0.00) or (diff > 0.05): # the base case, finding a diff more than 0.05 or less than 0 is like finding a boundary
        return 0

    count = count +1
    print 'count1 ',count

    array[x][y] = -5 # so we do no calculate this pixel again
    #print "[",x,",",y,"]"
    if x > 0:
        #print '1'# if not the first elemnet then can go back, this makes sure that x is within the array all the time
        floodcount (x-1,y,array,value,count)
    if y > 0:
        #print '2'
        floodcount (x,y-1,array,value,count) 
    if x < nrows:
        #print '3'
        floodcount (x+1,y,array,value,count)
    if y < ncols:
        #print '4'
        floodcount (x,y+1,array,value,count)
    if x > 0 and y > 0:
        #print '5'
        floodcount (x-1,y-1,array,value,count)
    if x < nrows and y < ncols:
        #print '6'
        floodcount (x+1,y+1,array,value,count)
    if x <nrows and y > 0:
        #print '7'
        floodcount (x+1,y-1,array,value,count)
    if x > 0 and y < ncols:
        #print '8'
        floodcount (x-1,y+1,array,value,count)

    print 'count2 ',count    
    return count

所以对于一个测试用例

数组 = [[5,1,1,3,4],[4,5,6,2,5],[5,8,5,5,9]] x=0 和 y=0

输出

count1 1 计数 1 2 计数 1 3 计数 1 4 计数 1 5 计数 2 5 计数 2 4 计数 2 3 计数 1 3 计数 2 3 计数 2 count2 1

如你所见，有些东西是可疑的 :P 谁能指出我做错了什么？任何帮助将不胜感激。

Answer 1

所以 floodcount() 返回新的 count 值。但是你永远不会存储/使用它:)

像这样替换行:

floodcount(x+1, y-1, array, value, count)

与:

count = floodcount(x+1, y-1, array, value, count)