我有以下内容:
# Possible neighbors as tuples
UP = (-1,0)
DOWN = (1,0)
LEFT = (0,-1)
RIGHT = (0,1)
POTENTIAL_NEIGHBORS = (UP, DOWN, LEFT, RIGHT)
def LocateNeighbors(self):
print '\nWorking on: {}:{}'.format(self.index,self.value)
for n in POTENTIAL_NEIGHBORS:
try:
if theBoard[self.index[0]+n[0]][self.index[1]+n[1]]:
print "Adding neighbor: " + theBoard[self.index[0]+n[0]][self.index[1]+n[1]]
self.neighbors.append(n)
except IndexError:
print 'failed on: {}'.format(n)
..其中self
是游戏板中的一个单元格,例如:
A B C D E F
1 | {66} | {76} | {28} | {66} | {11} | {09}
-------------------------------------------
2 | {31} | {39} | {50} | {08} | {33} | {14}
-------------------------------------------
3 | {80} | {76} | {39} | {59} | {02} | {48}
-------------------------------------------
4 | {50} | {73} | {43} | {03} | {13} | {03}
-------------------------------------------
5 | {99} | {45} | {72} | {87} | {49} | {04}
-------------------------------------------
6 | {80} | {63} | {92} | {28} | {61} | {53}
-------------------------------------------
但是,这是我正在检查的第一个单元格的输出(0,0
):
Working on: (0, 0):66
Adding neighbor: 80
Adding neighbor: 31
Adding neighbor: 09
Adding neighbor: 76
基本上,如果索引(我认为是)无效,它就会回滚到列表的末尾。有没有办法防止或至少检测到这种行为?或者,或许更优雅的方式来做到这一点?
答案 0 :(得分:2)
您可以使用范围检查功能:
def check(n):
if n<0:
raise IndexError(n)
return n
def LocateNeighbors(self):
print '\nWorking on: {}:{}'.format(self.index,self.value)
for n in POTENTIAL_NEIGHBORS:
try:
if theBoard[check(self.index[0]+n[0])][check(self.index[1]+n[1])]:
print "Adding neighbor: " + theBoard[self.index[0]+n[0]][self.index[1]+n[1]]
self.neighbors.append(n)
except IndexError:
print 'failed on: {}'.format(n)
答案 1 :(得分:2)
按照以下行添加支票:
if (0 <= (self.index[0] + n[0]) < num_cols) and
(0 <= (self.index[1] + n[1]) < num_rows):
# Do stuff here
当你在董事会的对面边缘时,你应该避免IndexError
。
编辑:
丹尼尔的回答更像是Pythonic,我想。它遵循EAFP原则。