String functions in Spark in Apache Spark - Cheat Sheet & Quick Revision

Recall & Review

beginner

What does the Spark function lower() do to a string column?

It converts all characters in the string column to lowercase letters.

beginner

How does the concat() function work in Spark?

It joins two or more string columns or literals into one single string column.

beginner

What is the purpose of the substring() function in Spark?

It extracts a part of a string from a specified start position for a given length.

beginner

Explain what trim() does in Spark string functions.

It removes leading and trailing spaces from the string values in a column.

intermediate

What does the regexp_replace() function do in Spark?

It replaces all substrings that match a regular expression pattern with a specified replacement string.

Which Spark string function converts all letters in a column to uppercase?

A lower()

B substring()

C trim()

D upper()

The upper() function changes all characters to uppercase.

What does the concat() function do in Spark?

A Joins multiple strings into one

B Extracts a substring

C Removes spaces

D Splits a string into parts

concat() joins two or more strings into a single string.

Which function removes spaces at the start and end of a string in Spark?

A rtrim()

B ltrim()

C trim()

D substring()

trim() removes spaces from both ends of the string.

How do you extract a part of a string starting at position 3 with length 4 in Spark?

A substr(column, 3, 4)

B substr(column, 4, 3)

C substring(column, 4, 3)

D substring(column, 3, 4)

substr(column, start, length) extracts substring; start=3, length=4.

What does regexp_replace() do in Spark?

A Finds a substring

B Replaces substrings matching a pattern

C Splits strings by pattern

D Counts pattern matches

regexp_replace() replaces all matches of a regex pattern with a new string.

Describe how you would clean a string column in Spark to remove extra spaces and convert all text to lowercase.

Think about removing spaces and changing case.

Explain how to combine two string columns into one in Spark and then extract a part of the combined string.

First join strings, then get a substring.

Practice

1. Which Spark string function converts all characters in a column to lowercase?

easy

A. substring()

B. upper()

C. trim()

D. lower()

Solution

1. Step 1: Understand the function purpose

   The lower() function changes all letters in a string to lowercase.

2. Step 2: Compare with other functions

   upper() makes letters uppercase, trim() removes spaces, and substring() extracts part of the string.

3. Final Answer:

   lower() -> Option D

4. Quick Check:

   Lowercase conversion = lower() [OK]

Hint: Lowercase text? Use lower() function [OK]

Common Mistakes:

• Confusing lower() with upper()
• Using trim() to change case
• Thinking substring() changes case

2. Which of the following is the correct syntax to remove leading and trailing spaces from a Spark DataFrame column named name?

easy

A. df.select(trim(col('name')))

B. df.select(trim(name))

C. df.select(trim())

D. df.select(trim('df.name'))

Solution

1. Step 1: Understand Spark column referencing

   In Spark, to refer to a column by name in functions, use col('columnName').

2. Step 2: Apply trim function correctly

   The trim() function expects a Column object, so trim(col('name')) is correct syntax.

3. Final Answer:

   df.select(trim(col('name'))) -> Option A

4. Quick Check:

   Use col() inside trim() for column reference [OK]

Hint: Use col('colName') inside string functions [OK]

Common Mistakes:

• Passing string directly without col()
• Using variable name without quotes or col()
• Using trim() without any argument

3. What will be the output of this Spark code snippet?

from pyspark.sql.functions import substring
df = spark.createDataFrame([('SparkData',)], ['word'])
df.select(substring('word', 1, 5).alias('sub')).collect()

medium

A. [Row(sub='parkD')]

B. [Row(sub='SparkD')]

C. [Row(sub='Spark')]

D. [Row(sub='SparkData')]

Solution

1. Step 1: Understand substring parameters

   substring('word', 1, 5) extracts 5 characters starting at position 1 (1-based index).

2. Step 2: Extract substring from 'SparkData'

   Characters 1 to 5 are 'S', 'p', 'a', 'r', 'k' forming 'Spark'.

3. Final Answer:

   [Row(sub='Spark')] -> Option C

4. Quick Check:

   substring start=1, length=5 extracts 'Spark' [OK]

Hint: substring uses 1-based start index [OK]

Common Mistakes:

• Assuming 0-based index for substring
• Extracting wrong number of characters
• Confusing substring with slice

4. Identify the error in this Spark code that tries to replace all spaces with underscores in column text:

from pyspark.sql.functions import regexp_replace
df.select(regexp_replace(text, ' ', '_')).show()

medium

A. Incorrect number of arguments to regexp_replace

B. Missing column reference using col()

C. Wrong function name, should be replace()

D. Using show() instead of collect()

Solution

1. Step 1: Check argument types for regexp_replace

   The first argument must be a Column object, not a bare identifier.

2. Step 2: Correct usage with col()

   Use regexp_replace(col('text'), ' ', '_') to refer to the column properly.

3. Final Answer:

   Missing column reference using col() -> Option B

4. Quick Check:

   Use col() to reference columns in functions [OK]

Hint: Always wrap column names with col() in functions [OK]

Common Mistakes:

• Passing column name as string without col()
• Confusing regexp_replace with replace
• Ignoring function argument requirements

5. You want to create a new column clean_name in a Spark DataFrame by trimming spaces, converting to lowercase, and replacing all digits with empty strings from column name. Which code snippet correctly does this?

hard

A. df.withColumn('clean_name', regexp_replace(lower(trim(col('name'))), '\\d', ''))

B. df.withColumn('clean_name', lower(trim(regexp_replace(col(name), '\\d', ''))))

C. df.withColumn('clean_name', trim(lower(regexp_replace(col(name), '\\d', ''))))

D. df.withColumn('clean_name', trim(regexp_replace(lower(col(name)), '\\d', '')))

Solution

1. Step 1: Order of operations matters

   First trim spaces, then convert to lowercase, then remove digits.

2. Step 2: Check each option's function nesting

   df.withColumn('clean_name', regexp_replace(lower(trim(col('name'))), '\\d', '')) trims first, then lowers, then replaces digits last, matching the required order.

3. Final Answer:

   df.withColumn('clean_name', regexp_replace(lower(trim(col('name'))), '\\d', '')) -> Option A

4. Quick Check:

   Trim -> Lower -> Replace digits = df.withColumn('clean_name', regexp_replace(lower(trim(col('name'))), '\\d', '')) [OK]

Hint: Chain trim(), lower(), then regexp_replace() in that order [OK]

Common Mistakes:

• Wrong order of function calls
• Missing col() around column name
• Incorrect regex pattern for digits