Question

可能是这个类似但不知怎的，我无法解决这个问题。我有一个列表列表，我想从中删除每两个元素中的一些字符，并创建另一个列表列表

list = [['A*68:02:01:03', 'A*33:01:01', '1'], 
        ['A*68:02:01:02', 'A*33:01:01', '1'], 
        ['A*68:02:01:01', 'A*33:01:01', '1']]

required output = [['A*68:02', 'A*33:01', '1'], 
                   ['A*68:02', 'A*33:01', '1'], 
                   ['A*68:02', 'A*33:01', '1']]

最后我想只打印独特的元素。与上述情况类似，所有三个元素都相同，因此输出应为：

output = ['A*68:02', 'A*33:01', '1']

感谢您的帮助

Answer 1

>>> lst = [['A*68:02:01:03', 'A*33:01:01', '1'], ['A*68:02:01:02', 'A*33:01:01', '1'], ['A*68:02:01:01', 'A*33:01:01', '1']]
>>> newLst = [tuple(':'.join(data.split(':', 2)[:2]) for data in sublist) for sublist in lst]
>>> set(newLst)
{('A*68:02', 'A*33:01', '1')}

有趣的是':'.join(data.split(':', 2)[:2]。将以冒号分割data，仅取前两部分并再次加入。这样，我们在第二个冒号（包括那个）之后将所有内容都删除。

其余的只是通过嵌套列表的列表理解。我们还需要将内部列表转换为元组，因此当我们在其上调用set()时它们是可以清除的。这样做可以消除所有重复。

Answer 2

我想你想要：

lst = [['A*68:02:01:03', 'A*33:01:01', '1'],
       ['A*68:02:01:02', 'A*33:01:01', '1'],
       ['A*68:02:01:01', 'A*33:01:01', '1']]

output = []

for item in lst:
    processed = [":".join(s.split(":")[:2]) for s in item]
    if processed not in output:
        output.append(processed)

注意：不要调用自己的变量，例如list。

Answer 3

你可以这样做：

def prefix(string):
    """
    Returns the prefix of a string, including all characters 
    up until the second colon. 
    """
    return ":".join(string.split(":", 2)[:2])

def unique(iterable):
    """
    Returns the unique elements in iterable, maintaining the 
    elements' relative order.
    """
    result = []
    seen = set()
    for el in iterable:
        if el not in seen:
            seen.add(el)
            result.append(el)
    return result

L = [
    ['A*68:02:01:03', 'A*33:01:01', '1'], 
    ['A*68:02:01:02', 'A*33:01:01', '1'], 
    ['A*68:02:01:01', 'A*33:01:01', '1'],
]
prefixes = [(prefix(el[0]), prefix(el[1]), el[2]) for el in L]

# The built-in class set accepts an iterable and returns a set, 
# an object with all duplicate elements removed. Since sets are
# unordered, converting the set back to a list will likely 
# produce a list in which the original elements have lost their
# relative order.
# If this is a problem you can use the unique function from above.
uniq = list(set(prefixes))

# If you really need a list of lists, instead of a list of tuples.
uniq = [list(el) for el in uniq]

我已将您的输入列表重命名为L，因为将其命名为list会影响内置函数列表。

我在python中有一个列表列表，我想从每两个元素中删除一些字符

3 个答案: