0

如果

list_1 = [(1, 1), (1, 3), (1, 4), (2, 2), (2, 3)]

将元组的一个元素视为 (i, j) 现在我们知道如果我们知道如果 (1, 3) 存在，那么 (1,1) 和 (1,2) 应该存在，如果其中任何一个是漏报错误。错误检测应该首先对 (1,1) 进行，然后对 (1, 2) 进行。类似地，如果我们知道对于另一个序列，如果 (2,3) 存在，那么 (2,1) 和 (2, 2) 将同样存在。

这也是一个示例列表。这个例子也可以是：

list_1 = [(1, 3), (1, 4), (2, 2), (2, 3), (3,1) (3,4)] 现在在这种情况下如果 (3, 4) 存在则(3,1),(3,2) 和 (3,3) 应该存在。这种模式可以由每个“i”的第 j 个元素的最大值来确定我的代码是 gi

for i, e in enumerate(list_1):
              i1 = (1,1)
              if i1 not in list_1:
                 raise ValueError, '%s is missing in %s' %(i1, production)

              if (e[0] == e[1]) and ((e[0],e[1]) not in list_1):
                 #if i1 not in list_1:
                 raise ValueError, '%s is missing in %s' %((e[0], e[1]), production)

              print e, e[0], e[1], (e[0], e[1])
              if i!=len(list_1)-1:


                 if e[0]==list_1[i+1][0] and e[1]!=list_1[i+1][1]-1:

                    raise ValueError, '(%s,%s) is missing in %s ' %(e[0], e[1]+1, production)
4

2 回答 2

0

基于迭代工具的方法

>>> from itertools import groupby, izip, count
>>> from operator import itemgetter
>>> z = [(1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3,1), (3,4)]
>>> z = sorted(z, key=itemgetter(0))
>>> z = groupby(z, key=itemgetter(0))
>>> bad = []
>>> for key, group in z:
...     z1 = izip((y for (x, y) in group), count(1))
...     for (a, b) in z1:
...             if a != b:
...                    bad.append(key) 
... 
>>> print(set(bad))
set([1, 3])
于 2012-08-02T23:41:05.737 回答
0 
>>> list_1 = [(1, 1), (1, 3), (1, 4), (2, 2), (2, 3)]
>>> check_list = zip(list_1, list_1[1:])
>>> check_list
[((1, 1), (1, 3)), ((1, 3), (1, 4)), ((1, 4), (2, 2)), ((2, 2), (2, 3))]
>>> ok = True
>>> for (i1, j1), (i2, j2) in check_list:
...     if i1 == i2 and j1 + 1 != j2:
...         ok = False
...         break
...     elif i1 + 1 != i2: #you may also want to check j2 == 1:
...         ok = False
...         break
...
>>> ok
False
于 2012-08-02T23:26:47.697 回答