Question

在我的数据框中：。

df = pd.DataFrame(zip(datetimes, from_, message), columns=['timestamp', 'sender', 'message'])
df['timestamp'] = pd.to_datetime(df.timestamp, format='%d/%m/%Y, %I:%M %p')

有一些有问题的值，由清晰的模式定义：

    timestamp                              sender                               message
    113381 2020-06-04 11:59:24              Jose                               bom te ver feliz\r\n
    113382 2020-06-04 11:59:29              Jose                                              ❤\r\n
    113383 2020-06-04 11:59:40              Maria                Estar bem com você me faz feliz\r\n
    113384 2020-06-04 12:00:57              Maria   Estava falando com uma amiga de infância aque...
    113385 2020-06-04 12:01:14              Maria           Ela teve uma briga feia com o marido\r\n
    113386 2020-06-04 12:01:24   Maria: ‎<attached        00113509-PHOTO-2020-06-04-12-01-25.jpg>\r\n
    113387 2020-06-04 12:02:54              Maria                       e assim leva-se a vida, um\n
    113388 2020-06-04 12:03:21              Maria                  Pelo menos ela riu isso ajuda\r\n
    113389 2020-06-04 13:06:39    Jose: ‎<attached        00113512-PHOTO-2020-06-04-13-06-40.jpg>\r\n

名称始终会有所不同，并且很可能是

John
John: <attached
Mary
Mary: <attached

但是: <attached将永远在那里。

我如何执行字符串替换，从而独立于字符串而纠正，最后以：

timestamp                              sender                               message
113381 2020-06-04 11:59:24              Jose                               bom te ver feliz\r\n
113382 2020-06-04 11:59:29              Jose                                              ❤\r\n
113383 2020-06-04 11:59:40              Maria                Estar bem com você me faz feliz\r\n
113384 2020-06-04 12:00:57              Maria   Estava falando com uma amiga de infância aque...
113385 2020-06-04 12:01:14              Maria           Ela teve uma briga feia com o marido\r\n
113386 2020-06-04 12:01:24              Maria        00113509-PHOTO-2020-06-04-12-01-25.jpg>\r\n
113387 2020-06-04 12:02:54              Maria                       e assim leva-se a vida, um\n
113388 2020-06-04 12:03:21              Maria                  Pelo menos ela riu isso ajuda\r\n
113389 2020-06-04 13:06:39              Jose       00113512-PHOTO-2020-06-04-13-06-40.jpg>\r\n

Answer 1

这应该可行；

df['sender'] = df['sender'].str.replace(u': \u200e<attached', '')

Answer 2

数据

df = pd.DataFrame({'sender': ['Jose','Jose','Maria','Maria','Maria','Maria: <attached','Maria','Maria','Jose: <attached']})

解决方案

df.sender = df.sender.str.split(': <attached').str[0]

   sender
0   Jose
1   Jose
2   Maria
3   Maria
4   Maria
5   Maria
6   Maria
7   Maria
8   Jose

Answer 3

8位Borges，您的数据中可能包含\u200e字符。我遇到了类似的问题，因为像这样的奇怪字符，split什么也不做。这是我的解决方案：

a = df['sender'].to_dict()

然后，我看到了将其发送到字典时的实际值。该值为: \u200e<attached。然后，我只是做了：

df['sender'] = df['sender'].str.split(': \u200e<attached').str[0]

此处有关\u200e的更多信息：decoding \u200e to string

熊猫-用特定模式替换值

3 个答案: