python - 迭代列表时检查每 3 个连续元素是否遵循某种模式

Question

我有一个元素列表，其中每 3 个连续元素应遵循某种模式，但是某些数据丢失，我想用 "" 填充丢失的数据。

例如，对于下面的列表，我想检查从第一个元素开始的每个第三个元素是否包含单词 “start”，并且从第二个元素开始的每个第三个元素包含 ”/" 最后，从第三个元素开始，每三个元素都包含 ":"

所以这个:

list = ["start1", "mid/1", "end:1", "start2", "end:2", "mid/3", "end:3"]

会变成这样:

list = ["start1", "mid/1", "end:1", "start2", "", "end:2", "", "mid/3", "end: 3"]

我已设法使用以下代码打印出每个第三个元素和步骤号，但我在实际插入 ""

data = ["start1", "mid/1", "end:1", "start2", "end:2", "mid/3", "end:3", "start4"，“mid/4”，“end:4”，“start5”，“end:5”]

    for step, init_info in enumerate(data[:3:], 1):
        for elem, info in enumerate(data[step-1::3], 1):
            print(step, info)

这将返回:

(1, '开始1')
(1, '开始2')
(1，'结束:3')
(1，'结束:4')
(2，'中/1')
(2, '结束:2')
(2, '开始4')
(2, '开始5')
(3, '结束:1')
(3，'中/3')
(3，'中/4')
(3, '结束:5')


Answer 1

您可以使用源列表上的简单循环和一些检查来实现这一点:

l = ["start1", "mid/1", "end:1", "start2", "end:2", "mid/3", "end:3"]

# defines which position must hold what thing in it to be not substituted
d = {0:"start", 1:"/", 2:":"}

# target list    
k = [] 

# counts the insertions, to adjust dict lookup
inserted = 0

for idx,e in enumerate(l):
    must_be_in_else_substitute = d.get( (idx+inserted) % 3)
    if must_be_in_else_substitute in e:
        k.append(e)
    else:
        inserted+=1
        k.extend([" ",e])
print(k)

输出:

['start1', 'mid/1', 'end:1', 'start2', ' ', 'end:2', ' ', 'mid/3', 'end:3']

<小时/>

您可以按照 @timgeb 的建议使用生成器，但不知道该生成器的可重用性如何，因为大多数“规则”都是已设置的。这与上面的代码大致相同，只是更懒:

l = ["start1", "mid/1", "end:1", "start2", "end:2", "mid/3", "end:3"]    

def insert_something_if_at_pos_not_in(rules,data,tbi=" "):
    """No idea how to docstring this in a sensible way - lets say 
    it does what the OP described :o)"""
    inserted=0
    rule_max = max(rules.keys())+1
    for idx,e in enumerate(data):
        must_be_in = rules.get( (idx+inserted) % rule_max)
        if must_be_in in e:
            yield e
        else:
            yield " "
            yield e

print(    list(insert_something_if_at_pos_not_in({0:"start", 1:"/", 2:":"},l) ))