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认为

list1 = [['a', (1, 1)], ['a', (1, 3)], ['a', (1, 4)], ['b', (2,1)], ['b', (2,2)], ['b',(2, 4)]]

list2 = [[(1, 1), (1, 3), (2, 1), (2, 2), (2, 4)]]

现在我该如何报告 list1 缺少 ['b', (1, 2)] 或 ['b', (2, 3)] 的错误

同样对于 list2 应该有报告错误 (1, 2) 或 (2, 3) 丢失

我的意图是报告错误，例如，如果缺少某些东西，例如 (1,1)，然后是 (1,2)，然后是 (1,3)，如果缺少 (1,2)，则错误

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2 回答 2

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from collections import defaultdict
set1 = set(list1)
set2 = set(list2)
missing = []
dict1 = defaultdict(lambda: defaultdict(list))
dict2 = defaultdict(list)
for key, sublist in set1:
    dict1[key][sublist[0]].append(sublist[1])
for key, value in set2:
    dict2[key].append(value)
for key, subdict in sorted(dict1.iteritems()):
    for subkey, values in sorted(subdict.iteritems()):
        subkey_misses = []
        last_value = None
        for value in values:
            if last_value is not None and last_value + 1 != value:
                subkey_misses.extend(range(last_value + 1, value))
            last_value = value
        if subkey_misses:
            misses.append('%s.%d missing %s' % (key, subkey, subkey_misses))
for key, values in sorted(dict2.iteritems()):
    key_misses = []
    last_value = None
    for value in values:
        if last_value is not None and last_value + 1 != value:
            key_misses.append(range(last_value + 1), value))
        last_value = value
    if key_misses:
        misses.append('%d missing %s' % (key, key_misses))
print misses
于 2012-08-01T20:43:20.410 回答
0

您应该使用字典而不是列表。但这是使用您的结构的解决方案。s1与上一个答案的想法相似，但请注意不必要的长列表理解以获得您拥有的模式list1。而且您需要一个特定的 for 循环来检查而不是设置“ -”运算符。

>>> s1 = [[x, (c, d)] for x in ['a', 'b']
...                   for c in range(1, 3)
...                   for d in range(1, 5)
...                   if x=='a' and c==1 or x=='b' and c==2]
>>> s1
[['a', (1, 1)], ['a', (1, 2)], ['a', (1, 3)], ['a', (1, 4)],
 ['b', (2, 1)], ['b', (2, 2)], ['b', (2, 3)], ['b', (2, 4)]]
>>>
>>> list1 = [['a', (1, 1)], ['a', (1, 3)], ['a', (1, 4)],
...          ['b', (2, 1)], ['b', (2, 2)], ['b', (2, 4)]]
>>> for thing in s1:
...     if thing not in list1:
...         print 'missing: ', thing
...         # or raise an error if you want
...         
missing:  ['a', (1, 2)]
missing:  ['b', (2, 3)]

对 重复相同的操作list2。使用上面的示例创建s2应该更容易s1。

顺便说一句，字典看起来像这样list1：

dict1 = {'a': [(1, 1), (1, 3), (1, 4)], 'b': [(2, 1), (2, 2), (2, 4)]}

然后创建 s1 稍微简化了一点，但是比较循环可能会加长两行。

要回答您的问题以概括，然后是1.先知道字母还是2.知道数字/字母数量？

认识字母：

>>> set_of_letters = ('a', 'b', 'c')
>>> s1 = [[x, (ord(x)-96, d)]
...       for x in set_of_letters
...       for d in range(1, 5)]
>>> s1
[['a', (1, 1)], ['a', (1, 2)], ['a', (1, 3)], ['a', (1, 4)],
 ['b', (2, 1)], ['b', (2, 2)], ['b', (2, 3)], ['b', (2, 4)],
 ['c', (3, 1)], ['c', (3, 2)], ['c', (3, 3)], ['c', (3, 4)]]

认识数字：

>>> number_of_letters = 3
>>> s1 = [[chr(c+96), (c, d)]
...       for c in range(1, number_of_letters + 1)
...       for d in range(1, 5)]
>>> s1
[['a', (1, 1)], ['a', (1, 2)], ['a', (1, 3)], ['a', (1, 4)],
 ['b', (2, 1)], ['b', (2, 2)], ['b', (2, 3)], ['b', (2, 4)],
 ['c', (3, 1)], ['c', (3, 2)], ['c', (3, 3)], ['c', (3, 4)]]
于 2012-08-01T17:04:16.343 回答