如何根据python列表中对象的相对列表位置来比较它们？

我有一些文本数据，我正试图清理。此数据的一个不需要的特性是由特定标记分隔的重复标记。我试图找到一种方法来（1）识别文本中的标记，以及（2）删除其中一个重复的标记

玩具示例：

word_list = ['this','is','a','!!','a','list','I','want','to','clean']

这里有两个重复的“a”标记，由标记“！！”分隔。我正试图找到一种最有效的方法，用下面的代码遍历这个列表

#pseudo
for word in word_list
    if word == "!!":
        if word[at word-1] == word[at word+1]  # compare words either side of the "!!" marker
            del word[at word+1]                # removing the duplicate
            del word                           # removing the "!!" marker


output = ['this','is','a','list','I','want','to','clean']

---

我已经尝试了几种涉及

枚举

函数的方法，但似乎无法使其正常工作。

以下几点可以奏效：

word_list = ['this','is','a','!!','a','list','I','want','to','clean']

i, marker, output = 0, "!!", []

while i < len(word_list):
    x = word_list[i]
    output.append(x)
    if word_list[i+1:i+3] == [marker, x]:
        i += 3
    else:
        i += 1

output
# ['this', 'is', 'a', 'list', 'I', 'want', 'to', 'clean']

result=[]

for i in range(len(word_list)-2):
    if not (word_list[i]=='!!' or (word_list[i+1]=='!!' and  word_list[i+2]==word_list[i])):
       result.append(word_list[i])
if word_list[-2]!='!!':
    result.append(word_list[-2])
if word_list[-1]!='!!':
    result.append(word_list[-1])

print(result)
#['this', 'is', 'a', 'list', 'I', 'want', 'to', 'clean']

---

使用您的逻辑和

枚举

功能：

word_list = ['this','is','a','!!','a','list','I','want','to','clean']

for i, word in enumerate(word_list):
    if word == "!!":
        if word_list[i-1] == word_list[i+1]:
            word_list[i+1] = ""
            word_list[i] = ""

print ([x for x in word_list if x])

输出：

['this', 'is', 'a', 'list', 'I', 'want', 'to', 'clean']

这将有助于：

word_list = ['this', 'is', 'a', '!!', 'a', 'list', 'I', 'want', 'to', 'clean']

for i, word in enumerate(word_list):
    if word == "!!":
        if word_list[i-1] == word_list[i+1]:  # checking if they are duplicates
            del word_list[i+1]  # removing the duplicate
            del word_list[i]  # removing the marker

---

我知道你的分隔符总是“！！”！！“以下各项应起作用：

word_list = ['this','is','a','!!','a','list','I','want','to','clean']

i, marker, output = 0, "!!", []

while i < len(word_list):
    x = word_list[i]
    output.append(x)
    if word_list[i+1:i+3] == [marker, x]:
        i += 3
    else:
        i += 1

output
# ['this', 'is', 'a', 'list', 'I', 'want', 'to', 'clean']

result=[]

for i in range(len(word_list)-2):
    if not (word_list[i]=='!!' or (word_list[i+1]=='!!' and  word_list[i+2]==word_list[i])):
       result.append(word_list[i])
if word_list[-2]!='!!':
    result.append(word_list[-2])
if word_list[-1]!='!!':
    result.append(word_list[-1])

print(result)
#['this', 'is', 'a', 'list', 'I', 'want', 'to', 'clean']

---

请解释一下。