时间戳（“.71910”）中的小数秒有五位数。Spark要求小数秒（毫秒）最多有三位数。超过三位数可能会导致解析错误。
下面是修改后的代码。

import sys
from pyspark import SparkContext, SQLContext
from pyspark.sql import functions as F
import dateutil.parser
sc = SparkContext('local')
sqlContext = SQLContext(sc)
### This is very important setting if you want legacy behaviour
sqlContext.setConf("spark.sql.legacy.timeParserPolicy", "LEGACY")
data1 = [["2023-11-17T08:28:40.71910 +01:00"]]
df1Columns = ["ts"]
df1 = sqlContext.createDataFrame(data=data1, schema=df1Columns)
new_df = df1.withColumn("adjusted_ts", F.regexp_replace("ts", "(\\.[0-9]{3})[0-9]*", "$1"))
new_df = new_df.withColumn("converted_ts", F.to_timestamp(F.col("adjusted_ts"), "yyyy-MM-dd'T'HH:mm:ss.SSSSS XXX"))
new_df.show(truncate=False)

字符串
输出量：

+--------------------------------+------------------------------+-----------------------+
|ts                              |adjusted_ts                   |converted_ts           |
+--------------------------------+------------------------------+-----------------------+
|2023-11-17T08:28:40.71910 +01:00|2023-11-17T08:28:40.719 +01:00|2023-11-17 12:58:40.719|
+--------------------------------+------------------------------+-----------------------+

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