Answer-first summary for fast verification
Answer: df.withColumn('domain', regexp_extract(col('url'), 'https?://([^/]+)', 1))
The correct answer is A because it correctly uses the `regexp_extract` function to extract the domain part of the URL, using a regular expression that captures the domain after the protocol and before any path segments.
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Suppose you have a DataFrame df with a column url containing URLs. How would you extract the domain name from these URLs and create a new column domain using Spark? Provide the code snippet.
A
df.withColumn('domain', regexp_extract(col('url'), 'https?://([^/]+)', 1))
B
df.select(regexp_extract(col('url'), 'https?://([^/]+)', 1).alias('domain'))
C
df.withColumn('domain', split(col('url'), '/')[2])
D
df.select(split(col('url'), '/')[2].alias('domain'))