这里有一个非udf解决方案。将字符串分割为你要计数的字符，你想要的值是结果数组的长度减去1：

from pyspark.sql.functions import col, size, split
DF.withColumn('Number_Products_Assigned', size(split(col("assigned_products"), r"\+")) - 1)

你必须转义+，因为它是一个特殊的正则表达式字符。

+--------------------+------------------------+
|   assigned_products|Number_Products_Assigned|
+--------------------+------------------------+
|POWER BI PRO+Powe...|                       3|
+--------------------+------------------------+

from pyspark.sql.functions import col,udf

@udf(returnType='int')
def cnt(s):
    return s.count('+')

DF.withColumn('Number_Products_Assigned', cnt(col("assigned_products")))

Replace将子字符串的出现替换为空字符串。因此，我们可以通过比较替换前后的长度来计算出现次数，如下所示：

使用SparkSQL：

SELECT length(x) - length(replace(x,'+')) as substring_count
FROM  (select 'abc+def+ghi++aaa' as x) -- Sample data

输出：

substring_count
---------------
4

使用PySpark函数：

import pyspark.sql.functions as F

df1 = spark.sql("select 'abc+def+ghi++aaa' as x") # Sample data
df1.withColumn('substring_count', 
                F.length(col('x')) 
               - F.length(F.regexp_replace(col('x'), '\+', '')) 
              ).show()

输出：

+----------------+---------------+
|               x|substring_count|
+----------------+---------------+
|abc+def+ghi++aaa|              4|
+----------------+---------------+

**Spark 3.4+**有regexp_count

F.expr(r"regexp_count(col_name, '\\+')")

完整示例：

from pyspark.sql import functions as F
df = spark.createDataFrame(
    [("POWER BI PRO+Power BI (free)+AUDIO CONFERENCING+OFFICE 365 ENTERPRISE E5 WITHOUT AUDIO CONFERENCING",)],
    ["assigned_products"])

df = df.withColumn('Number_Products_Assigned', F.expr(r"regexp_count(assigned_products, '\\+')"))

df.show()
# +--------------------+------------------------+
# |   assigned_products|Number_Products_Assigned|
# +--------------------+------------------------+
# |POWER BI PRO+Powe...|                       3|
# +--------------------+------------------------+