我只想用 string functions ，没有理由使用regex。

from pyspark.sql import functions as F

testdf\
      .withColumn("Food1", F.expr("""substring(Food,1,instr(Food,',')-1)"""))\
      .withColumn("Food2", F.expr("""substring(Food,instr(Food,',')+1,length(Food))""")).show()

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

# |Animal|           Food|Food1|     Food2|

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

# |   Dog|meat,bread,milk| meat|bread,milk|

# |   Cat|     mouse,fish|mouse|      fish|

# +------+---------------+-----+----------+*

稍微不同的方法是使用切片和修剪：

from pyspark.sql.functions import expr, split

df.withColumn("food_ar", split("food", ",")) \
  .select( \
         df.Animal,
         df.Food,
         expr("food_ar[0]").alias("Food1"),
         expr("trim('[]', string(slice(food_ar, 2, size(food_ar) - 1)))").alias("Food2"))

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

# |Animal|           Food|Food1|     Food2|

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

# |   Dog|meat,bread,milk| meat|bread,milk|

# |   Cat|     mouse,fish|mouse|      fish|

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

首次使用 split 首先生成数组。接下来，我们使用singlesparksql访问器访问这些项 a[0] 填充头部和 slice 一起 trim 对于数组的尾部。

一种使用正则表达式从列表中只拆分第一个匹配项的方法

testdf.withColumn('Food1',f.split('Food',"(?<=^[^,]*)\\,")[0]).\
       withColumn('Food2',f.split('Food',"(?<=^[^,]*)\\,")[1]).show()

+------+---------------+-----+----------+
|Animal|           Food|Food1|     Food2|
+------+---------------+-----+----------+
|   Dog|meat,bread,milk| meat|bread,milk|
|   Cat|     mouse,fish|mouse|      fish|
+------+---------------+-----+----------+