Question

我有一个数组10x10，我需要解决两点之间对角线上的所有点，并检查列表中是否有。我这样做但我不知道它为什么不工作：

让你进入上下文：

a = [i，j] b = [i，j]，i和j =范围（11）

例如，下面的代码适用于a = [5,4] b = [8,7]。

     ...
     elif (b[1] - a[1]) == (b[0] - a[0]):
        #to southeast, code for the other 3 cases are almost the same
        if b[0] > a[0] and b[1] > a[1]:
            n = a[0]
            m = a[1]
            while (n != b[0]) and (m != b[1]):
                n +=1
                m +=1 

                #don't think this part below is relevant
                if board[n][m] in somelist:
                    mov_inv += 1
                else:
                    mov_inv += 0

这是在函数内部，如果mov_inv＆gt;返回False。如果mov_inv = 0，则为1，如果不是那样，则为True。希望你明白我的意思。感谢

Answer 1

def diag(pointa, pointb):
    """ pointa and pointb are tuples
        returns a generator that yields values along the diagonal 
        from pointa to pointb"""

    if abs(pointa[0] - pointb[0]) != abs(pointa[1] - pointb[1]):
        #Sanity check. Diagonal are equal along x and y
        raise ValueError("Points {} and {} are not diagonal".format(pointa, pointb))
    x_dir = 1 if pointa[0] < pointb[0] else -1
    y_dir = 1 if pointa[1] < pointb[1] else -1
    while pointa != pointb:
        pointa = (pointa[0] + x_dir, pointa[1] + y_dir)
        yield pointa

以上不会产生pointb，因为您已经知道了

Answer 2

board=xrange(10*10)
diagonal1=[board[C:(10-C)*10:10+1] for C in xrange(10)]
diagonal2=[board[C:(C*10)+1:10-1] for C in xrange(10)]
print 'diagonal ASC ',diagonal1
print 'diagonal DESC ',diagonal2

然后你只需检查a和b是否在同一对角线

assume a,b= [5,4] , [8,7]
def Square(m,n):
    return 10*(m-1)+n;
m,n=a
A=Square(m,n)
m,n=b
B=Square(m,n)

**print ('ASCENDENT',[diagonal1[x] for x in xrange(10) if A in diagonal[x] and B in diagonal[x]])**

数组对角线python

2 个答案: