遍历列表并在Python中串联字母顺序

Question

我试图遍历一个列表，并为列表中的每个元素分配一个字母字母，如果有重复，则向其分配下一个字母以获取唯一项。

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

我试图生成这样的字母列表：

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

然后我想遍历这样的序列：

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

我不确定如何进行其余工作...

谢谢，

Answer 1

这是一个解决方案，可以处理无限大小的序列和无限次重复（允许内存）

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']

Answer 2

>>> import string, collections

>>> c = collections.Counter()
>>> for x in [0, 1, 2, 3, 1, 4, 2, 1]:
        letter = string.letters[c[x]]
        c[x] += 1
        print '%s%s' % (x, letter)


0A
1A
2A
3A
1B
4A
2B
1C

Answer 3

我将其作为发电机。 我正在使用集合中的Counter数据结构。 如果键尚不存在，则返回0，而不是键错误。 因此，我们可以将其用作alpha列表的索引。

添加了使excel列样式字母的功能。 测试了多达一百万个相同版本的版本。

import collections

# Uncomment for Python 2.X
#from __future__ import division, print_function

sequence = [0, 1, 2, 3, 1, 4, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]

def get_excel_column(number):
    dividend = number
    column_name = []
    while dividend > 0:
        modulo = (dividend - 1) % 26
        column_name.insert(0, chr(modulo + 65))
        dividend = (dividend - modulo) // 26
    return ''.join(column_name)

def get_unique(seq):
    cnt = collections.Counter()
    for item in seq:
        cnt[item] += 1
        yield '%s%s' % ( str(item), get_excel_column(cnt[item]) )

for uniq in get_unique(sequence):
    print(uniq)

输出：

0A
1A
2A
3A
1B
...
1AC
1AD
1AE

注意：Python 3语法。 根据需要更改打印和划分。