熊猫解析文本列

Question

我有一个 csv 表，其中有一列包含聊天日志中的文本。每个文本行都遵循相同的格式：消息的人员姓名和时间（带有额外的前后空格填充），后跟消息内容。文本列的单行示例：

'  Siri (3:15pm)  Hello how can I help you?  John Wayne (3:17pm)  what day of the week is today  Siri (3:18pm)  it is Monday.'

我想将这个单个字符串列转换为多列（列数取决于消息数量），每条消息对应一列，如下所示：

• Siri （下午 3:15）您好，需要什么帮助吗
• John Wayne (下午 3:17) 今天是星期几
• Siri (下午 3:18) 今天是星期一 Siri (下午 3:18) 今天是星期一代码>

我如何解析这段文本pandas dataframe 列将聊天日志分成单独的消息列？

Answer 1

如果您有此数据框：

                                                                                                                     Messages
0  Siri (3:15pm)  Hello how can I help you?  John Wayne (3:17pm)  what day of the week is today  Siri (3:18pm)  it is Monday.

那么您可以执行以下操作：

x = df["Messages"].str.split(r"\s{2,}").explode()

out = (x[::2] + " " + x[1::2]).to_frame()
print(out)

打印：

                                            Messages
0            Siri (3:15pm) Hello how can I help you?
0  John Wayne (3:17pm) what day of the week is today
0                        Siri (3:18pm) it is Monday.

注意：仅当名称和文本之间有 2 个以上空格时才有效。

Answer 2

这就是我的做法，花了我一段时间，但我们做到了！

s = pd.Series(['  Siri (3:15pm)  Hello how can I help you?  John Wayne (3:17pm)  what day of the week is today  Siri (3:18pm)  it is Monday.'])
s = s.str.split(r"  ", expand=True)
s = s.drop(labels=[0], axis=1)
s = s.transpose()

for i in s.index:
    list_1 = list(s[0])

odd_i = []
even_i = []
for i in range(0, len(list_1)):
    if i % 2:
        even_i.append(list_1[i])
    else :
        odd_i.append(list_1[i])

d = {'Name': odd_i, 'Message': even_i}
df = pd.DataFrame(data=d)
df

Output:
                   Name                               Message
0         Siri (3:15pm)             Hello how can I help you?
1   John Wayne (3:17pm)         what day of the week is today
2         Siri (3:18pm)                         it is Monday.