你可以用 regexp_extract 取而代之的是：

import pyspark.sql.functions as F

df1 = df.withColumn(
    'CONV_ID',
    F.regexp_extract('URL', 'conversations/(.*)', 1)
)

df1.show()
+---+--------------------+----------+
| ID|                 URL|   CONV_ID|
+---+--------------------+----------+
|  1|https://app.xyz.c...|2686735685|
|  2|https://app.xyz.c...|2938415796|
|  3|https://app.drift...|2938419189|
+---+--------------------+----------+

或者如果你想用 split ，您不需要指定 .* . 您只需要指定用于拆分的模式。

import pyspark.sql.functions as F

df1 = df.withColumn(
    'CONV_ID',
    F.split('URL', '(?<=conversations/)')[1]    # just using 'conversations/' should also be enough
)

df1.show()
+---+--------------------+----------+
| ID|                 URL|   CONV_ID|
+---+--------------------+----------+
|  1|https://app.xyz.c...|2686735685|
|  2|https://app.xyz.c...|2938415796|
|  3|https://app.drift...|2938419189|
+---+--------------------+----------+

如果你是URL一直都是那种形式，你实际上可以只使用 substring_index 要获取最后一个路径元素：

import pyspark.sql.functions as F

df1 = df.withColumn("CONV_ID", F.substring_index("URL", "/", -1))

df1.show(truncate=False)

# +---+-------------------------------------------------------------+----------+

# |ID |URL                                                          |CONV_ID   |

# +---+-------------------------------------------------------------+----------+

# |1  |https://app.xyz.com/inboxes/136636/conversations/2686735685  |2686735685|

# |2  |https://app.xyz.com/inboxes/136636/conversations/2938415796  |2938415796|

# |3  |https://app.drift.com/inboxes/136636/conversations/2938419189|2938419189|

# +---+-------------------------------------------------------------+----------+