Pyspark string pattern from columns values and regexp expression

Question

Hi I have dataframe with 2 columns :

+----------------------------------------+----------+
|                  Text                  | Key_word |
+----------------------------------------+----------+
| First random text tree cheese cat      | tree     |
| Second random text apple pie three     | text     |
| Third random text burger food brain    | brain    |
| Fourth random text nothing thing chips | random   |
+----------------------------------------+----------+

I want to generate a 3rd columns with a word appearing before the key_word from the text.

+----------------------------------------+----------+-------------------+--+
|                  Text                  | Key_word | word_bef_key_word |  |
+----------------------------------------+----------+-------------------+--+
| First random text tree cheese cat      | tree     | text              |  |
| Second random text apple pie three     | text     | random            |  |
| Third random text burger food brain    | brain    | food              |  |
| Fourth random text nothing thing chips | random   | Fourth            |  |
+----------------------------------------+----------+-------------------+--+

I tried this but it's not working

df2=df1.withColumn('word_bef_key_word',regexp_extract(df1.Text,('\\w+)'df1.key_word,1))

Here is the code to create a example of the dataframe

df = sqlCtx.createDataFrame(
    [
        ('First random text tree cheese cat' , 'tree'),
        ('Second random text apple pie three', 'text'),
        ('Third random text burger food brain' , 'brain'),
        ('Fourth random text nothing thing chips', 'random')
    ],
    ('Text', 'Key_word') 
)

Answer 1

Update

You can also do this without a udf by using pyspark.sql.functions.expr to pass column values as a parameter to pyspark.sql.functions.regexp_extract:

from pyspark.sql.functions import expr

df = df.withColumn(
    'word_bef_key_word', 
    expr(r"regexp_extract(Text, concat('\\w+(?= ', Key_word, ')'), 0)")
)
df.show(truncate=False)
#+--------------------------------------+--------+-----------------+
#|Text                                  |Key_word|word_bef_key_word|
#+--------------------------------------+--------+-----------------+
#|First random text tree cheese cat     |tree    |text             |
#|Second random text apple pie three    |text    |random           |
#|Third random text burger food brain   |brain   |food             |
#|Fourth random text nothing thing chips|random  |Fourth           |
#+--------------------------------------+--------+-----------------+

---

Original Answer

One way to do this is by using a udf to do the regex:

import re
from pyspark.sql.functions import udf

def get_previous_word(text, key_word):
    matches = re.findall(r'\w+(?= {kw})'.format(kw=key_word), text)
    return matches[0] if matches else None

get_previous_word_udf = udf(
    lambda text, key_word: get_previous_word(text, key_word),
    StringType()
)

df = df.withColumn('word_bef_key_word', get_previous_word_udf('Text', 'Key_word'))
df.show(truncate=False)
#+--------------------------------------+--------+-----------------+
#|Text                                  |Key_word|word_bef_key_word|
#+--------------------------------------+--------+-----------------+
#|First random text tree cheese cat     |tree    |text             |
#|Second random text apple pie three    |text    |random           |
#|Third random text burger food brain   |brain   |food             |
#|Fourth random text nothing thing chips|random  |Fourth           |
#+--------------------------------------+--------+-----------------+

The regex pattern '\w+(?= {kw})'.format(kw=key_word) means match a word followed by a space and the key_word. If there are multiple matches, we will return the first one. If there are no matches, the function returns None.