Question

我试图将蜇伤分成均匀的部分。例如，如果我有['abcdef']，则预期输出应为['ab','cd','ef']。我继续为第一个测试'asdfadsf'/

获得相同的输入

def solution(s):

    lists = []

    for i in s:
        if len(s) % 2 == 0:
            lists.append(s)
            zip(*[iter(lists)]*2)
        return lists

test.describe("Example Tests")

tests = (
    ("asdfadsf", ['as', 'df', 'ad', 'sf']),
    ("asdfads", ['as', 'df', 'ad', 's_']),
    ("", []),
    ("x", ["x_"]),
)

 for inp, exp in tests:
 test.assert_equals(solution(inp), exp)

Answer 1

试试这个，

def solutions(s):
    return [j.ljust(2,'_') for j in (s[i:i+2] for i in  range(0,len(s),2))]

<强>结果

In [38]: tests = (
   ....:     ("asdfadsf", ['as', 'df', 'ad', 'sf']),
   ....:     ("asdfads", ['as', 'df', 'ad', 's_']),
   ....:     ("", []),
   ....:     ("x", ["x_"]),
   ....: )
In [39]: for inp, exp in tests:
   ....:     print solutions(inp) == exp 
   ....:     
True
True
True
True

Answer 2

执行所需操作的最简单方法是使用range。

def chunk(input):

    if len(input) % 2 == 1:
        input += '_'

    output = [input[idx:idx+2] for idx in range(0, len(input), 2]
    return output

列表中的均匀分裂字符串：Python

2 个答案: