Question

我正在运行for来检查元组列表。

中的内容

for i in range of b:
   actual=i
   temp1=(actual[0]+1,actual[1])
   temp2=(actual[0],actual[1]-1)
   temp3=(actual[0],actual[1]+1)
   temp4=(actual[0]-1,actual[1])

我想确保temp永远不会在之前的循环中验证元组的值。关于如何做到这一点的任何想法？

Answer 1

首先，您的代码似乎存在问题。 range接受整数输入，因此如果b是整数，for i in range(b)将在列表中为您提供整数[0, 1, 2, .. , b-1 ]。您无法使用i索引[]，就像接下来的两行一样。

如果b不是整数，而是集合，那么你应该使用类似的东西：

# Assuming b is a collection
for i in range(len(b)):
   actual=b[i]
   temp1=(actual[0]+1,actual[1])
   temp2=(actual[0],actual[1]-1)
   temp3=(actual[0],actual[1]+1)
   temp4=(actual[0]-1,actual[1])

   # Check if this is the first one.  If it is, previous won't exist.
   if i == 0:
       continue

   previous = b[i-1]
   if previous in [ temp1, temp2, temp3, temp4 ]:
       # This is what you want not to happen.  Deal with it somehow.
       pass

Answer 2

这是我的两分钱。请注意，如果匹配，这将使temp（1-4）无。

# assuming b is a collection
for i in range(len(b)):
    actual=b[i]
    if i!=0:
        prev = b[i-1]
    if i==0:
        prev = [[['something']],[['ridiculous']]] #this is so that the rest of the code works even if index is 0
    if (actual[0]+1,actual[1]) != prev: #if it is not the previous item
        temp1=(actual[0]+1,actual[1]) #assign temp1
    else:
        temp1 = None  #temp1 would otherwise automatically take on the value of (b[i-1][0]+1,b[i-1][1])
    if (actual[0],actual[1]-1) != prev:
        temp2=(actual[0],actual[1]-1)
    else:
        temp2 = None
    if (actual[0],actual[1]+1) != prev:
        temp3=(actual[0],actual[1]+1)
    else:
        temp3 = None
    if (actual[0]-1,actual[1]) != prev:
        temp4=(actual[0]-1,actual[1])
    else:
        temp4 = None

适用于涉及元组和最后元组的范围

2 个答案: