我正在制作一个列表,我想在任何方向上连续测试4个或更多的阳性结果(我正在测试B和E(例如:
List = [N, N, B, N, N, E, B, E, N,
N, E, B, N, E, E, E, B, N,
N, N, N, N, N, E, B, E, N,
N, E, B, N, E, E, E, B, N]
它更长,但你明白了,无论如何,我希望它为E
返回为真,因为连续有四个E
但没有B
.我也想测试对角线,但我正在努力弄清楚如何尝试这样做。我没有示例代码,因为我不知道如何解决这个问题。如果我需要以不同的方式解释,请告诉我。
如果我们可以将这些数据表示为网格结构,每个点都有 x、y 坐标,那就更容易了。我们可以使用一个类来以这种方式设置数据。该类需要知道您希望网格有多宽才能执行此操作。要查找项目序列,我们可以遍历行和列,每次找到该项目的匹配项时,我们都会检查每个方向上的相邻方块是否也包含相同的项目。
class Grid(object):
def __init__(self, data, width):
self.data = data
self.width = width
self.height = len(L) / width
def get(self, x, y):
""" Find the item in the grid at the given coordinates """
if 0 <= x < self.width and 0 <= y < self.height:
return self.data[y * self.width + x]
def find_sequence(self, item, times):
""" Check for a sequence of item occuring at least the specified
number of times. Checks right, down, down-right, and down-left. """
for y in range(self.height):
for x in range(self.width):
# if we find the item at x, y...
if self.get(x, y) == item:
# ... then look at adjacent items in each direction
for dx, dy in [(1, 0), (0, 1), (1, 1), (-1, 1)]:
if all(self.get(x + i*dx, y + i*dy) == item
for i in range(1, times)):
return True
return False
N, B, E = "N", "B", "E"
data = [N, N, B, N, N, E, B, E, N,
N, E, B, N, E, E, E, B, N,
N, N, N, N, N, E, B, E, N,
N, E, B, N, E, E, E, B, N]
grid = Grid(data, width=9)
grid.get(3, 3) # N
grid.find_sequence(B, 4) # False
grid.find_sequence(N, 4) # True
grid.find_sequence(E, 4) # True
这是你要找的吗?
List = [N, N, B, N, N, E, B, E, N,
N, E, B, N, E, E, E, B, N,
N, N, N, N, N, E, B, E, N,
N, E, B, N, E, E, E, B, N]
limit = 4
flag = False
oldValue = ''
newValue = ''
counter = 0
for i in range(0, len(List)):
newValue = List[i]
if (newValue == oldValue):
counter += 1
oldValue = newValue
if (counter >= 4):
flag = True
break
print 'Result is' + str(flag)