Question

我有一个数据框，其中有两列，地址和街道名称。

from pyspark.sql.functions import *
import pyspark.sql

df = spark.createDataFrame([\
    ['108 badajoz road north ryde 2113, nsw, australia', 'north ryde'],\
    ['25 smart street fairfield 2165, nsw, australia', 'smart street']
  ],\ 
  ['address', 'street_name'])

df.show(2, False)

+------------------------------------------------+---------------+
|address                                         |street_name    |
+------------------------------------------------+---------------+
|108 badajoz road north ryde 2113, nsw, australia|north ryde     |
|25 smart street fairfield 2165, nsw, australia  |smart street   |
+------------------------------------------------+---------------+

我想查找street_name中是否存在address，并在新列中返回布尔值。我可以手动搜索模式，如下所示。

df.withColumn("new col", col("street").rlike('.*north ryde.*')).show(20,False)
----------------------------------------------+---------------+-------+
|address                                         |street_name |new col|
+------------------------------------------------+------------+-------+
|108 badajoz road north ryde 2113, nsw, australia|north ryde  |true   |
|25 smart street fairfield 2165, nsw, australia  |smart street|false  |
+------------------------------------------------+------------+-------+

但是我想用下面的列street_name替换手动值

 df.withColumn("new col", col("street")\
  .rlike(concat(lit('.*'),col('street_name'),col('.*))))\
  .show(20,False)

Answer 1

您只需使用contains函数就可以做到这一点。有关更多详细信息，请参见this：

from pyspark.sql.functions import col, when

df = df.withColumn('new_Col',when(col('address').contains(col('street_name')),True).otherwise(False))
df.show(truncate=False)

+------------------------------------------------+------------+-------+ 
|address                                         |street_name |new_Col|      
+------------------------------------------------+------------+-------+ 
|108 badajoz road north ryde 2113, nsw, australia|north ryde  |true   | 
|25 smart street fairfield 2165, nsw, australia  |smart street|true   | 
+------------------------------------------------+------------+-------+

Answer 2

一个简单的解决方案是定义一个UDF并使用它。例如，

from pyspark.sql.functions import udf

def contains_address(address, street_name):
  return street_name in address

contains_address_udf = udf(contains_address, BooleanType())
df.withColumn("new_col", contains_address_udf("address", "street_name")

在这里，可以简单地使用in，但是如果需要更复杂的功能，只需将其替换为正则表达式即可。

Answer 3

只需使用expr函数：

from pyspark.sql import functions as F

df.select(
    "address", 
    "street_name", 
    F.expr("address like concat('%',street_name,'%')")
).show()

+--------------------+------------+--------------------------------------+
|             address| street_name|address LIKE concat(%, street_name, %)|
+--------------------+------------+--------------------------------------+
|108 badajoz road ...|  north ryde|                                  true|
|25 smart street f...|smart street|                                  true|
+--------------------+------------+--------------------------------------+

在另一列中的一列中有Pyspark查找模式

3 个答案: