如何检查列表列表中的两个元组是否不相同

Question

lt1 = [(1, 1), (1, 1), (1, 5), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2)]

How can I report an error if (1,1) in the above list or some other tuple occurs more than once

Similarly for the list

lt22 = [['a', (1,1)], ['a', (1,2)], ['a', (1,2)], ['a', (1,3)], ['b', (2,1)], ['b', (2,2)], ['b', (2,2)]]

How to report an error if ['a', (1,2)] or any other element occurs more than once

Answer 1

Use a set and a loop; the set will let you know if you've seen an element before:

seen = set()
for el in lt1:
    if el in seen:
        raise ValueError, 'More than one %r in your list' % (el,)
    seen.add(el)

Note that for mutable elements such as found in your second list, you want to convert these to non-mutable variants such as tuples first:

seen = set()
for el in lt22:
    el = tuple(el)
    if el in seen:
        raise ValueError, 'More than one %r in your list' % (el,)
    seen.add(el)

Answer 2

If you just want to detect it, not see exactly what element(s) occur more than once, you can just do:

if len(lt1) != len(set(lt1)):
    # Not all unique

In the second example you need to map the lists to tuples before converting to a set:

if len(lt22) != len(set(map(tuple, lt22))):
    # Not all unique

If you want to know how many times each item occurred, use collections.Counter , which was introduced in python 2.7:

>>> from collections import Counter
>>> {k: d for k, d in Counter(lt1).items() if d > 1}
{(1, 1): 2}

Again, you must map lt22 to tuple before using it to have it work.

If you just need a list of which items occur more than once, Martijns solution is probably the most efficient at that particular task.