如何根据开始和结束元素从列表创建子列表？

Question

正在尝试根据开始和结束元素从列表创建子列表。我无法获取所有出现的开始和结束元素

lst  = ['value0','<!program start>','value1','value2','<!program end>',
        'value3','<!program start>','value4','<!program end>','value5']

预期输出：

[['value0'],['<!program start>','value1','value2','<!program end>'],
 ['value3'],['<!program start>','value4','<!program end>'],['value5']]

代码：

start_idx = lst.index('<!program start>')
end_idx = lst.index('<!program end>')
final_result = lst[:start_idx] + [lst[start_idx:end_idx+1]] + lst[end_idx+1:]
print(final_result)

Answer 1

使用迭代：

lst = ['value0', '<!program start>', 'value1', 'value2', '<!program end>',
       'value3', '<!program start>', 'value4', '<!program end>', 'value5']

res = []
start = False
temp = []

for item in lst:
    if item == '<!program start>':
        start = True
        temp.append(item)

    elif item == '<!program end>':
        start = False
        temp.append(item)
        res.append(temp)
        temp = []

    elif start:
        temp.append(item)
    else:
        res.append([item])

print(res)

输出：

[['value0'], ['<!program start>', 'value1', 'value2', '<!program end>'], ['value3'], ['<!program start>', 'value4', '<!program end>'], ['value5']]

通过start标志，我处理了项目是否在开始和结束标签的中间。

Answer 2

它不像你的单线笔那么酷，但看起来很管用：

def process(input_list, start, end):
    output = []
    while len(input_list) != 0:
        if input_list[0] != start:
            # This isn't a start token, so just add it to the output
            output.append([input_list[0]])
            input_list = input_list[1:]
            continue

        # Looks like we've found a start token, look for the end
        # associated with it and append that. NOTE: You could
        # try/except here if you didn't know that the end token was
        # actually there.
        end_index = input_list.index(end)
        output.append(input_list[:end_index + 1])
        input_list = input_list[end_index + 1:]
    return output

我得到：

[['value0'],
 ['<!program start>', 'value1', 'value2', '<!program end>'],
 ['value3'],
 ['<!program start>', 'value4', '<!program end>'],
 ['value5']]

作为我认为正确的输出

Answer 3

您的代码的问题是索引 returns 第一个看到的索引，而不是所有索引。但可以简单地使用 while 循环来完成。

final_list = []
i = 0
while i < len(lst):
    inner_list = []
    word = lst[i]
    if word == "<!program start>":
        while word != '<!program end>':                  
            word = lst[i]
            inner_list.append(word)
            i += 1    
    else:
        inner_list.append(word)
        i += 1
    final_list.append(inner_list)

print(final_list)

Answer 4

您可以使用相对简单的 FSM (Finite State Machine) 处理数据：

def fsm(lst):
    result = []

    state = 0
    for elem in lst:
        if state == 0:
            result.append([elem])
            state = 1
        elif state == 1:
            if elem == '<!program start>':
                subl = [elem]
                state = 2
            else:
                break  # End of pattern.
        elif state == 2:
            subl.append(elem)
            if elem == '<!program end>':
                result.append(subl)
                state = 0

    return result


lst  = ['value0','<!program start>','value1','value2','<!program end>',
        'value3','<!program start>','value4','<!program end>','value5']

print(fsm(lst))

Answer 5

嵌套 while 循环的类似解决方案。

test_list = ['value0','<!program start>','value1','value2','<!program end>',
        'value3','<!program start>','value4','<!program end>','value5']

answer_list = []
i = 0
while i < len(test_list):
    if test_list[i] == '<!program start>':
        sublist = []
        while test_list[i] != '<!program end>':
            sublist.append(test_list[i])
            i += 1
    elif test_list[i] == '<!program end>':
        sublist.append(test_list[i])
        answer_list.append(sublist)
        i += 1
    else:
        answer_list.append(test_list[i])
        i += 1

print(answer_list)

生产：

['value0', ['<!program start>', 'value1', 'value2', '<!program end>'], 'value3', ['<!program start>', 'value4', '<!program end>'], 'value5']

Answer 6

实际上有一些有趣的综合方法利用基本 str 和 list 处理。

例如，您可以首先根据开始和结束标记的通用子字符串将 lst 分成 chunks：

chunks = [s for s in " ".join(lst).split("<!program ")]

这些块固有地包含区分单个元素和标签之间元素的特征。

list comp 是获得所需输出的一种很好且优雅的方式：

output = [[s.strip('end> ')] if not s.startswith('start>') else ["<!program start>"] + s.strip("start> ").split() + ["<!program end>"] for s in chunks]

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