PySpark查找另一列中是否存在一列中的模式
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我有两个pyspark数据帧。一个包含FullAddress字段(比如col1),另一个数据框包含其中一个列中的city / town / suburb的名称(比如col2)。我想比较col2和col1,如果有匹配则返回col2。
此外,郊区名称可以是郊区名称列表。
包含完整地址的Dataframe1
+--------+--------+----------------------------------------------------------+
|Postcode|District|City/ Town/ Suburb |
+--------+--------+----------------------------------------------------------+
|2000 |Sydney |Dawes Point, Haymarket, Millers Point, Sydney, The Rocks |
|2001 |Sydney |Sydney |
|2113 |Sydney |North Ryde |
+--------+--------+----------------------------------------------------------+
+-----------------------------------------------------------+
|FullAddress |
+-----------------------------------------------------------+
|BADAJOZ ROAD NORTH RYDE 2113, NSW, Australia |
| HAY STREET HAYMARKET 2000, NSW, Australia |
| SMART STREET FAIRFIELD 2165, NSW, Australia |
|CLARENCE STREET SYDNEY 2000, NSW, Australia |
+-----------------------------------------------------------+
我想要这样的东西
+-----------------------------------------------------------++-----------+
|FullAddress |suburb |
+-----------------------------------------------------------++-----------+
|BADAJOZ ROAD NORTH RYDE 2113, NSW, Australia |NORTH RYDE |
| HAY STREET HAYMARKET 2000, NSW, Australia |HAYMARKET |
| SMART STREET FAIRFIELD 2165, NSW, Australia |NULL |
|CLARENCE STREET SYDNEY 2000, NSW, Australia |SYDNEY |
+-----------------------------------------------------------++-----------+
有两个DataFrames
-
DataFrame 1:包含完整地址的DataFrame
。
DataFrame 2:包含基础数据的DataFrame
- Postcode
,District
和City / Town / Suburb
。
问题的目的是从suburb
中为DataFrame 1
提取适当的DataFrame 2
。尽管OP尚未明确指定我们可以加入两个DataFrame的key
,但Postcode
似乎只是合理的选择。
# Importing requisite functions
from pyspark.sql.functions import col,regexp_extract,split,udf
from pyspark.sql.types import StringType
让我们创建DataFrame 1
作为df
。在这个DataFrame
我们需要提取Postcode
。在澳大利亚,所有邮政编码都是4 digit long,因此我们使用regexp_extract()从string
列中提取4位数字。
df = sqlContext.createDataFrame([('BADAJOZ ROAD NORTH RYDE 2113, NSW, Australia ',),
('HAY STREET HAYMARKET 2000, NSW, Australia',),
('SMART STREET FAIRFIELD 2165, NSW, Australia',),
('CLARENCE STREET SYDNEY 2000, NSW, Australia',)],
('FullAddress',))
df = df.withColumn('Postcode', regexp_extract('FullAddress', "(\d{4})" , 1 ))
df.show(truncate=False)
+---------------------------------------------+--------+
|FullAddress |Postcode|
+---------------------------------------------+--------+
|BADAJOZ ROAD NORTH RYDE 2113, NSW, Australia |2113 |
|HAY STREET HAYMARKET 2000, NSW, Australia |2000 |
|SMART STREET FAIRFIELD 2165, NSW, Australia |2165 |
|CLARENCE STREET SYDNEY 2000, NSW, Australia |2000 |
+---------------------------------------------+--------+
现在,我们已经提取了Postcode
,我们创造了key
加入两个DataFrames
。让我们创建DataFrame 2
,我们需要从中提取各自的suburb
。
df_City_Town_Suburb = sqlContext.createDataFrame([(2000,'Sydney','Dawes Point, Haymarket, Millers Point, Sydney, The Rocks'),
(2001,'Sydney','Sydney'),(2113,'Sydney','North Ryde')],
('Postcode','District','City_Town_Suburb'))
df_City_Town_Suburb.show(truncate=False)
+--------+--------+--------------------------------------------------------+
|Postcode|District|City_Town_Suburb |
+--------+--------+--------------------------------------------------------+
|2000 |Sydney |Dawes Point, Haymarket, Millers Point, Sydney, The Rocks|
|2001 |Sydney |Sydney |
|2113 |Sydney |North Ryde |
+--------+--------+--------------------------------------------------------+
加入两个DataFrames
与left
加入 -
df = df.join(df_City_Town_Suburb.select('Postcode','City_Town_Suburb'), ['Postcode'],how='left')
df.show(truncate=False)
+--------+---------------------------------------------+--------------------------------------------------------+
|Postcode|FullAddress |City_Town_Suburb |
+--------+---------------------------------------------+--------------------------------------------------------+
|2113 |BADAJOZ ROAD NORTH RYDE 2113, NSW, Australia |North Ryde |
|2165 |SMART STREET FAIRFIELD 2165, NSW, Australia |null |
|2000 |HAY STREET HAYMARKET 2000, NSW, Australia |Dawes Point, Haymarket, Millers Point, Sydney, The Rocks|
|2000 |CLARENCE STREET SYDNEY 2000, NSW, Australia |Dawes Point, Haymarket, Millers Point, Sydney, The Rocks|
+--------+---------------------------------------------+--------------------------------------------------------+
使用City_Town_Suburb
函数将列split()拆分为数组 -
df = df.select('Postcode','FullAddress',split(col("City_Town_Suburb"), ",s*").alias("City_Town_Suburb"))
df.show(truncate=False)
+--------+---------------------------------------------+----------------------------------------------------------+
|Postcode|FullAddress |City_Town_Suburb |
+--------+---------------------------------------------+----------------------------------------------------------+
|2113 |BADAJOZ ROAD NORTH RYDE 2113, NSW, Australia |[North Ryde] |
|2165 |SMART STREET FAIRFIELD 2165, NSW, Australia |null |
|2000 |HAY STREET HAYMARKET 2000, NSW, Australia |[Dawes Point, Haymarket, Millers Point, Sydney, The Rocks]|
|2000 |CLARENCE STREET SYDNEY 2000, NSW, Australia |[Dawes Point, Haymarket, Millers Point, Sydney, The Rocks]|
+--------+---------------------------------------------+----------------------------------------------------------+
最后创建一个UDF来检查数组City_Town_Suburb
中的每个元素(如果它存在于FullAddress
列中)。如果存在一个,我们立即返回,否则返回None
。
def suburb(FullAddress,City_Town_Suburb):
# Check for the case where there is no Array, otherwise we will get an Error
if City_Town_Suburb == None:
return None
# Checking each and every Array element if it exists in 'FullAddress',
# and if a match is found, it's immediately returned.
for sub in City_Town_Suburb:
if sub.strip().upper() in FullAddress:
return sub.upper()
return None
suburb_udf = udf(suburb,StringType())
应用此UDF
-
df = df.withColumn('suburb', suburb_udf(col('FullAddress'),col('City_Town_Suburb'))).drop('City_Town_Suburb')
df.show(truncate=False)
+--------+---------------------------------------------+----------+
|Postcode|FullAddress |suburb |
+--------+---------------------------------------------+----------+
|2113 |BADAJOZ ROAD NORTH RYDE 2113, NSW, Australia |NORTH RYDE|
|2165 |SMART STREET FAIRFIELD 2165, NSW, Australia |null |
|2000 |HAY STREET HAYMARKET 2000, NSW, Australia |HAYMARKET |
|2000 |CLARENCE STREET SYDNEY 2000, NSW, Australia |SYDNEY |
+--------+---------------------------------------------+----------+
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