python – 递归函数中的计数器

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我是python和编程的新手.我编写了一个函数,它将搜索数组中的相邻元素,并寻找值相互之间0.05的值,就像泛光填充算法一样.唯一的区别是我在计算函数运行的时间时做了一些愚蠢的事情(我想这也会告诉我我发现了多少元素),所以我的计数器值是错误的.代码在找到彼此相差0.05的相邻元素时起作用,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

所以对于测试用例

array = [[5,1,1,3,4],[4,5,6,2,5],[5,8,5,5,9]]
x = 0且y = 0

OUTPUT

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

你可以看到有些东西是可疑的：P
谁能指出我做错了什么？任何帮助,将不胜感激.

解决方法:

除了现在解决的计数问题：

您可以通过每次执行所有递归调用来减少if语句的数量,只需在函数开头检查数组边框,如果x< 0或y <0 0或x> nrows或y> NCOLS.

#akin to a bucket fill in paint, finds the area instead
def floodcount (x,y,array,value,count=0): 
    nrows = len(array)-1          #rows of the image
    ncols = len(array[0])-1       #columns of the image
    if x < 0 or y < 0 or x > nrows or y > ncols:
        return count

    diff = array[x][y] - value
    # the base case, finding a diff more than 0.05 or less than 0 is like finding a boundary
    if (diff < 0.00) or (diff > 0.05): 
        return count

    count = count +1
    print 'count1 ',count

    array[x][y] = -5 # so we do no calculate this pixel again
    #print "[",x,",",y,"]"

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

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

    print 'count2 ',count    
    return count

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