python - 用计数替换空白字符串

Question

假设我有一个这样的列表:

py = ['','','','','monty','','','','python',]

我想将其映射到此:

[4,'monty',3,'python']

有人知道一个聪明的解决方案吗？我能够弄清楚将其转换为:

[1,1,1,1,'monty',1,1,1,'python',]

使用:

quotes = [x if x else 1 for x in quotes]

Answer 1

我认为Rob's solution是最好的:最具可读性且最容易理解。我修复了它，因为最后一个元素为空，并更改为生成器(它比使用 append 更有效):

def blank_to_count(iterable):
    counter = 0
    for val in iterable:
        if val == '':
            counter += 1
        else:
            if counter > 0: yield counter  # yield count of blank elements
            counter = 0
            yield val  # yield current non-blank element
    if counter > 0: yield counter  # in case last element was blank

py = ['','','','','monty','','','','python',]
print(list(blank_to_count(py)))  # [4, 'monty', 3, 'python']

py = ['monty','','','','python']
print(list(blank_to_count(py)))  # ['monty', 3, 'python']

py = ['','','','','monty','','','','python','','']
print(list(blank_to_count(py)))  # [4, 'monty', 3, 'python', 2]