Python 从数组/列表中将字符名及其行添加到新字典

我有一个电影剧本。我的第一项工作是收集字典中每个字符的行

稍后我需要将数据放入一个系列中

现在，我把所有的对话都列在一个列表中，从角色名开始。它的格式如下：

          NAME1
Yo, Yo, good that you're here
man.

          NAME2
     (Laughing)
I don't think that's good!  We were
at the club, smoking, laughing -- doing
stuff.

example = """
                      NAME1
            Yo, Yo, good that you're here
            man.

                      NAME2
                 (Laughing)
            I don't think that's good!  We were
            at the club, smoking, laughing -- doing
            stuff.
"""

lines = example.split('\n')
characters = [line for line in lines if line.startswith(' ' * 22)]
result = {c.strip(): [] for c in characters}
current = ''
for line in lines:
    if line in characters:
        current = line.strip()
    elif current:
        result[current].append(line.strip())

对话[0] 'NAME1\n（16个空格）哟，哟，很高兴你在这里，伙计。'

所有名称都以\n结尾。然后所有的对话都以16个空格开始。我认为这可能是有用的，但我不知道如何利用这一点

我试过很多方法，但几乎没有成功

result={}
对话中的台词：
第一个标记=段落拆分（）[0]
如果第一个_token.endswith（'\n'）：#这将是名称
名称，行=段落拆分（在新行？）
name=name.strip（）
如果结果中没有名称：
结果[名称]=[]
结果[名称]。追加（行）
返回结果

这段代码给了我一大堆错误，所以我认为在这里列出它们是没有用的

理想情况下，我需要将每个字符作为字典中的第一个键，然后将它们的所有行作为数据

大概是这样的：

          NAME1
Yo, Yo, good that you're here
man.

          NAME2
     (Laughing)
I don't think that's good!  We were
at the club, smoking, laughing -- doing
stuff.

example = """
                      NAME1
            Yo, Yo, good that you're here
            man.

                      NAME2
                 (Laughing)
            I don't think that's good!  We were
            at the club, smoking, laughing -- doing
            stuff.
"""

lines = example.split('\n')
characters = [line for line in lines if line.startswith(' ' * 22)]
result = {c.strip(): [] for c in characters}
current = ''
for line in lines:
    if line in characters:
        current = line.strip()
    elif current:
        result[current].append(line.strip())

名称1:[第1行、第2行、第3行…] 名称2:[第1行、第2行、第3行…]

编辑： 有些字符名有两个单词

编辑2： 也许回到原始电影脚本文本文件会更容易

它的格式如下：

          NAME1
Yo, Yo, good that you're here
man.

          NAME2
     (Laughing)
I don't think that's good!  We were
at the club, smoking, laughing -- doing
stuff.

example = """
                      NAME1
            Yo, Yo, good that you're here
            man.

                      NAME2
                 (Laughing)
            I don't think that's good!  We were
            at the club, smoking, laughing -- doing
            stuff.
"""

lines = example.split('\n')
characters = [line for line in lines if line.startswith(' ' * 22)]
result = {c.strip(): [] for c in characters}
current = ''
for line in lines:
    if line in characters:
        current = line.strip()
    elif current:
        result[current].append(line.strip())

---

编辑回答：回到原始文件，如果我们可以假设所有字符名前面都有22个空白字符，我们可以这样做：

          NAME1
Yo, Yo, good that you're here
man.

          NAME2
     (Laughing)
I don't think that's good!  We were
at the club, smoking, laughing -- doing
stuff.

example = """
                      NAME1
            Yo, Yo, good that you're here
            man.

                      NAME2
                 (Laughing)
            I don't think that's good!  We were
            at the club, smoking, laughing -- doing
            stuff.
"""

lines = example.split('\n')
characters = [line for line in lines if line.startswith(' ' * 22)]
result = {c.strip(): [] for c in characters}
current = ''
for line in lines:
    if line in characters:
        current = line.strip()
    elif current:
        result[current].append(line.strip())

现在的结果是：

{'NAME1': ["Yo, Yo, good that you're here", 'man.', ''], 'NAME2': ['(Laughing)', "I don't think that's good!  We were", 'at the club, smoking, laughing -- doing', 'stuff.', '']}

---

这可能需要一些额外的清理

方法1:

data = dict()
for _dialogue in dialogue:
   name, lines = parse(_dialogue)
   data[name] = data.get(name, list()) + lines

由“\n”拆分并剥离。列表的第一个元素是name，剩下的是您的行。str.pop将就地修改您的列表。 如果您的对话有多行文字，此解决方案将不起作用

>>> dialogue
'NAME1\n                abc adbaiuho saidainbw\n                sadi waiudi qoweoq asodhoqndoqndqwdq.\n                qiudwqd aisdiqnd asfiqwofnqofoweqomdomkmq!!'
>>> lines = list(map(str.strip, dialogue.split('\n')))
>>> lines
['NAME1', 'abc adbaiuho saidainbw', 'sadi waiudi qoweoq asodhoqndoqndqwdq.', 'qiudwqd aisdiqnd asfiqwofnqofoweqomdomkmq!!']
>>> name = lines.pop(0)
>>> name
'NAME1'
>>> lines
['abc adbaiuho saidainbw', 'sadi waiudi qoweoq asodhoqndoqndqwdq.', 'qiudwqd aisdiqnd asfiqwofnqofoweqomdomkmq!!']

方法2:

data = dict()
for _dialogue in dialogue:
   name, lines = parse(_dialogue)
   data[name] = data.get(name, list()) + lines

如果您有多行对话，即对话可能包含“\n”字符，则首先按“\n”字符的第一次出现进行拆分。第一个元素是name，下一个元素我们进一步用“16个空格”分割

>>> dialogue
'NAME1\n                abc adbaiuho saidainbw\n                sadi waiudi qoweoq asodhoqndoqndqwdq.\n                qiudwqd aisdiqnd asfiqwofnqofoweqomdomkmq!!'
>>> parse_temp = dialogue.split('\n',1)
>>> name = parse_temp[0]
>>> lines = parse_temp[1].split(" " * 16)[1:]
>>> name
'NAME1'
>>> lines
['abc adbaiuho saidainbw\n', 'sadi waiudi qoweoq asodhoqndoqndqwdq.\n', 'qiudwqd aisdiqnd asfiqwofnqofoweqomdomkmq!!']

作为一种功能

def parse(dialogue):
    parse_temp = dialogue.split('\n',1)
    name = parse_temp[0].strip()
    lines = list(map(str.strip, parse_temp[1].split(" " * 16)[1:]))
    return name, lines

注意：对于第二次拆分，您可以用任何空白模式替换。你甚至可以用正则表达式将其拆分。我在这里使用了简单的16个空格。

请求迭代时添加的代码：

data = dict()
for _dialogue in dialogue:
   name, lines = parse(_dialogue)
   data[name] = data.get(name, list()) + lines

• 拆分文本行
• 为每个参与者创建具有唯一键的dict
• 在dict中添加演员行

编辑：在名称正则表达式中添加空格，删除名称空白

import re
lines = [
    "Dialogue[0] 'NAME1 \n                YO, YO, good that you're here man.'",
    "Dialogue[1] 'NAME 1\n                YO, YO, ",
    "Dialogue[2] 'NAME2\n                YO, YO, good that ",
    "Dialogue[3] 'NAME2\n                YO, YO, good that you're here'",
]

regex = h = re.compile("'([A-Z 0-9]+)\n[ ]{16}(.+)")
lineslist = [re.findall(regex, line) for line in lines]
lineslist = [ match[0] for match in lineslist if len(match)]
keys = [l[0].strip() for l in lineslist]
result = {k:[] for k in set(keys)}
[result[l[0].strip()].append(l[1]) for l in lineslist]
result

输出：

{'NAME 1': ['YO, YO, '],
 'NAME1': ["YO, YO, good that you're here man.'"],
 'NAME2': ['YO, YO, good that ', "YO, YO, good that you're here'"]}

---

---

使用此正则表达式拆分数据：

”（[A-Z0-9]+）\\n[]{16}（+）

在转到原始电影脚本之前，请检查我的回答方法2，因为您似乎有多行对话。既然您已经知道如何在角色对话之间分割，这可能会起作用。在原始文本文件中，名称前面是否总是有几个空格？对话从一行的开头开始？谢谢你的回复。我得到了这个错误：ValueError：太多的值无法解包（预期为2）您能试试这个吗

name，line=item.split（maxslit=1）

给出了相同的错误消息：ValueError:没有足够的值来解包（预期为2，得到1）这不是相同的错误。。。但无论如何，我猜您的数据格式不一致，需要首先清理@fishmanI@fishmanI。我已经进行了编辑以显示原始脚本。你能解释一下我是如何从中得到同样的词典结果的吗？也许这样会更容易给我一个索引错误：indexer错误：列表索引超出范围这已经起作用了，但是有一个小错误，它显示一个字符名，后面有一个空格，作为一个不同的名称起作用了，谢谢！但我刚刚注意到另一个问题，每组行有4个副本。在中，有duplicates@fishman你所说的一组线是什么意思，你能举个例子吗？比如，它能找到所有的线，这很好。但是有四行代码：当运行函数时，它给了我一个错误：AttributeError:“list”对象没有属性“split”，你能把一个对话示例粘贴到pastebin函数中作为输入，并在这里放一个链接吗？啊，你误解了。我让你把它用在原来的问题上。i、 e.对话[0]我假设是一个字符的对话。我以为你已经有了这个逻辑，不是吗？所以我希望你在对话列表中的每一个对话中都使用这个，正如你在原始问题中提到的。我将如何在每一个对话项上运行这个函数？