Question

我们说我有一个这样的列表：

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

我想把它映射到这个：

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

有谁知道一个聪明的解决方案？我能够想出来把它转换成这个：

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

使用：

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

Answer 1

<强>代码：

def convert_quote_list(input_list):
    quotes = [x if x else 1 for x in input_list]

    counter = 0
    ans = []
    for each in quotes:
        if each == 1:
            counter += 1
        else:
            if counter:
                ans.append(counter)
            ans.append(each)
            counter = 0
    return ans

convert_quote_list(['','','','','monty','','','','python',])
convert_quote_list(['monty','','','python',])

<强>输出：

[4, 'monty', 3, 'python']
['monty', 2, 'python']

Answer 2

py = ['','','','','monty','','','','python']
py2 = ['','','','','monty','','','','python', '', '', '']

def conv(lst):
    out = []
    count = 0
    for s in lst:
        if s == '': count += 1
        else:
            if count > 0:
                out.append(count)
            count = 0
            out.append(s)
    if count > 0: out.append(count)

    return out

print(conv(py))
print(conv(py2))

输出

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

Answer 3

我认为Rob's solution是最好的：最易读且最容易理解。我修复了它的情况，最后一个元素是空白的，并改为工作为生成器（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]

Answer 4

您可以使用python的itertools.groupby创建按列表中的值分组的迭代器列表。然后使用列表推导和短路评估，获取迭代器列表的长度或字符串值

import itertools

py = ['','','','','monty','','','','python',]
out = [(k == 1 and len(list(v))) or k for k, v in (grp for grp in itertools.groupby(py, lambda x: x or 1))]
print out
[4, 'monty', 3, 'python']

编辑：经过反思，代码可能难以阅读，因此您可以这样做：

import itertools

out = []

for k, v in itertools.groupby(py, lambda x: x or 1):
    if k == 1:
        out.append(len(list(v)))
    else:
        out.append(k)

print out

用count替换空字符串

4 个答案: