python - 在Python中迭代列表并连接字母顺序

Question

我试图迭代一个列表，并为列表中的每个元素分配一个按字母顺序排列的字母，如果有重复项，则为其分配字母表中的下一个字母以获得唯一的项目。

sequence = [0, 1, 2, 3, 1, 4, 2, 1]
unique_seq = [0A, 1A, 2A, 3A, 1B, 4A, 2B, 1C]

我尝试生成这样的字母列表:

alpha = list(map(chr, range(65, 91)))

然后我想像这样迭代序列:

for i in sequence:
        unique_seq.append(i) for i in sequence if i not in unique_seq else...

我不知道如何处理剩下的事情......

谢谢，

Answer 1

这是一个解决方案，适用于无限大小和无限重复次数的序列(内存允许)

def increment_item(item = 'A'):
    '''
    Given a character sequence item, produces the next item in the character sequence set

    :type item: str
    :param item: The character sequence item to increment
    :rtype: str
    :return: The next element in the sequence. EX: item='A', return ='B'. item='Z', return ='AA'

    '''
    next_char = [ord(char) for char in item]
    next_char[-1] += 1
    for index in xrange(len(next_char)-1, -1, -1):
            if next_char[index] > ord('Z'):
                    next_char[index] = ord('A')
                    if index > 0:
                            next_char[index-1] += 1
                    else:
                            next_char.append(ord('A'))
    return ''.join((chr(char) for char in next_char))

def char_generator(start = 'A'):
    '''
    A generator which yields the next item in the character sequence every time next() is called

    :type start: str
    :param start: The starting item for the generator sequence

    '''
    current = start
    yield start
    while True:
        current = increment_item(current)
        yield current


def build_unique_sequence(sequence):
    '''
    Given an input sequence, returns the same sequence with characters
    appended such that every element in the returned sequence is unique

    :type sequence: list
    :param sequence: The sequence to make unique
    :rtype: list
    :return: The resultant unique sequence. EX: sequence = [0, 1, 2, 3, 1, 4, 2, 1], return = ['0A', '1A', '2A', '3A', '1B', '4A', '2B', '1C']

    '''
    key_set = dict([item, char_generator()] for item in set(sequence))
    return map(lambda item:'{}{}'.format(item, key_set[item].next()), sequence)

结果是:

>>> build_unique_sequence([0, 1, 2, 3, 1, 4, 2, 1])
['0A', '1A', '2A', '3A', '1B', '4A', '2B', '1C']