python - 在 PySpark 中使用 regexp_extract 提取多个单词

Question

我有一个包含一些单词的列表，我需要从文本行中提取匹配的单词，我找到了这个，但它只提取一个单词。

密钥文件内容

这是一个关键字

part_description 文件内容

32015 这是关键字hello world

代码

import pyspark.sql.functions as F

keywords = sc.textFile('file:///home/description_search/keys') #1
part_description =  sc.textFile('file:///description_search/part_description') #2
keywords = keywords.map(lambda x: x.split(' ')) #3
keywords = keywords.collect()[0] #4
df = part_description.map(lambda r: Row(r)).toDF(['line']) #5
df.withColumn('extracted_word', F.regexp_extract(df['line'],'|'.join(keywords), 0)).show() #6

输出

+--------------------+--------------+
|                line|extracted_word|
+--------------------+--------------+
|32015   this is a...|          this|
+--------------------+--------------+

预期产出

+--------------------+-----------------+
|                line|   extracted_word|
+--------------------+-----------------+
|32015   this is a...|this,is,a,keyword|
+--------------------+-----------------+

我想要

1. 返回所有匹配的关键字及其计数

2. 如果step #4是最有效的方法

可重现的例子：

keywords = ['this','is','a','keyword']
l = [('32015 this is a keyword hello world'      , ),
('keyword this'      ,   ),
('32015 this is a keyword hello world 32015 this is a keyword hello world'      ,   ),
('keyword keyword'      ,   ),
('is a'      , )]

columns = ['line']

df=spark.createDataFrame(l, columns)

Answer 1

在Spark 3.1+ regexp_extract_all中可用：

regexp_extract_all(str, regexp[, idx])- 提取与表达式str匹配regexp并对应于正则表达式组索引的所有字符串。

你原来的问题现在可以这样解决：

re_pattern = '(' + '|'.join([f'\\\\b{k}\\\\b' for k in keywords]) + ')'
df = df.withColumn('matched', F.expr(f"regexp_extract_all(line, '{re_pattern}', 1)"))
df = df.withColumn('count', F.size('matched'))

df.show()
#+--------------------+--------------------+-----+
#|                line|             matched|count|
#+--------------------+--------------------+-----+
#|32015 this is a k...|[this, is, a, key...|    4|
#|        keyword this|     [keyword, this]|    2|
#|32015 this is a k...|[this, is, a, key...|    8|
#|     keyword keyword|  [keyword, keyword]|    2|
#|                is a|             [is, a]|    2|
#+--------------------+--------------------+-----+

Answer 2

我设法通过使用UDF来解决它，如下所示

def build_regex(keywords):
    res = '('
    for key in keywords:
        res += '\\b' + key + '\\b|'
    res = res[0:len(res) - 1] + ')'

    return res


def get_matching_string(line, regex):
    matches = re.findall(regex, line)
    return matches if matches else None


udf_func = udf(lambda line, regex: get_matching_string(line, regex),
               ArrayType(StringType()))

df = df.withColumn('matched', udf_func(df['line'], F.lit(build_regex(keywords)))).withColumn('count', F.size('matched'))

结果

+--------------------+--------------------+-----+
|                line|             matched|count|
+--------------------+--------------------+-----+
|32015    this is ...|[this, is, this, ...|    5|
|12832    Shb is a...|             [is, a]|    2|
|35015    this is ...|          [this, is]|    2|
+--------------------+--------------------+-----+