str accessor for string methods in Pandas - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of str.upper() on a pandas Series

What is the output of the following code?

import pandas as pd
s = pd.Series(['apple', 'Banana', 'Cherry'])
result = s.str.upper()
print(result)

Pandas

import pandas as pd
s = pd.Series(['apple', 'Banana', 'Cherry'])
result = s.str.upper()
print(result)

A

0    APPLE
1    Banana
2    CHERRY
dtype: object

B

0     APPLE
1    BANANA
2    CHERRY
dtype: object

C

0    apple
1    Banana
2    Cherry
dtype: object

D

0    apple
1    BANANA
2    cherry
dtype: object

💡 Hint

The str.upper() method converts all characters in each string to uppercase.

The str.upper() method applied to a pandas Series converts every string element to uppercase letters.

❓ data_output

intermediate

Count of strings containing 'a' using str.contains()

Given the Series below, what is the count of strings that contain the letter 'a' (case insensitive)?

import pandas as pd
s = pd.Series(['Cat', 'dog', 'Parrot', 'fish', 'ant'])
count = s.str.contains('a', case=False).sum()
print(count)

Pandas

import pandas as pd
s = pd.Series(['Cat', 'dog', 'Parrot', 'fish', 'ant'])
count = s.str.contains('a', case=False).sum()
print(count)

A 3

B 2

C 4

D 1

💡 Hint

Check which strings have 'a' or 'A' in them.

The strings 'Cat', 'Parrot', and 'ant' contain 'a' or 'A', so the count is 3.

🔧 Debug

advanced

Identify the error in using str.replace()

What error does the following code raise?

import pandas as pd
s = pd.Series(['apple', 'banana', 'cherry'])
result = s.str.replace('a', 1)
print(result)

Pandas

import pandas as pd
s = pd.Series(['apple', 'banana', 'cherry'])
result = s.str.replace('a', 1)
print(result)

A AttributeError: 'Series' object has no attribute 'str'

B ValueError: invalid replacement string

C TypeError: expected string or bytes-like object

D No error, output is ['1pple', 'b1n1n1', 'cherry']

💡 Hint

Check the type of the replacement argument in str.replace.

The replacement argument must be a string; passing an integer causes a TypeError.

❓ visualization

advanced

Visualize string length distribution using str.len()

You have a pandas Series of words. Which code snippet correctly creates a histogram of their lengths?

import pandas as pd
import matplotlib.pyplot as plt
s = pd.Series(['dog', 'elephant', 'cat', 'hippopotamus', 'fox'])
# Which code below creates the histogram?

Pandas

import pandas as pd
import matplotlib.pyplot as plt
s = pd.Series(['dog', 'elephant', 'cat', 'hippopotamus', 'fox'])

A

plt.hist(s.str.len())
plt.show()

B

plt.hist(s.len())
plt.show()

C

plt.hist(len(s))
plt.show()

D

plt.hist(s.str.length())
plt.show()

💡 Hint

Use the str accessor to get string lengths.

The str.len() method returns the length of each string element, which can be plotted.

🚀 Application

expert

Filter DataFrame rows where a column's strings start with a vowel

Given the DataFrame below, which code correctly filters rows where the 'Name' column starts with a vowel (a, e, i, o, u), case insensitive?

import pandas as pd
df = pd.DataFrame({'Name': ['Alice', 'bob', 'Eve', 'Oscar', 'Uma', 'Charlie']})

Pandas

import pandas as pd
df = pd.DataFrame({'Name': ['Alice', 'bob', 'Eve', 'Oscar', 'Uma', 'Charlie']})

A

filtered = df[df['Name'].str.match('[aeiou]$', case=False)]
print(filtered)

B

filtered = df[df['Name'].str.startswith('aeiou')]
print(filtered)

C

filtered = df[df['Name'].str.contains('^[aeiou]', case=True)]
print(filtered)

D

filtered = df[df['Name'].str.match('^[aeiou]', case=False)]
print(filtered)

💡 Hint

Use a regex pattern with str.match to check the start of the string.

str.match with regex '^[aeiou]' and case=False matches strings starting with any vowel ignoring case.

Practice

1. What does the str accessor in pandas allow you to do with a DataFrame column?

easy

A. Perform mathematical operations on numeric data

B. Sort the DataFrame rows by index

C. Apply string methods to each element in the column

D. Merge two DataFrames based on a key column

Solution

1. Step 1: Understand the purpose of the str accessor

   The str accessor is designed to apply string methods element-wise on pandas Series or DataFrame columns containing text data.

2. Step 2: Compare options with the str accessor's function

   Performing mathematical operations on numeric data, sorting the DataFrame rows by index, and merging two DataFrames based on a key column are unrelated to string methods. Only applying string methods to each element in the column correctly describes the str accessor's function.

3. Final Answer:

   Apply string methods to each element in the column -> Option C

4. Quick Check:

   str accessor = apply string methods [OK]

Hint: Remember: str accessor works only on text columns [OK]

Common Mistakes:

• Thinking str accessor works on numbers
• Confusing str accessor with DataFrame merge
• Assuming str accessor sorts data

2. Which of the following is the correct syntax to convert all strings in a pandas Series names to lowercase using the str accessor?

easy

A. names.str.lower()

B. names.lower()

C. str.lower(names)

D. names.to_lower()

Solution

1. Step 1: Recall the correct usage of str accessor

   To apply string methods on a pandas Series, use Series.str.method(). Here, lower() converts strings to lowercase.

2. Step 2: Evaluate each option

   names.str.lower() uses names.str.lower(), which is correct. names.lower() misses str. str.lower(names) incorrectly calls str.lower as a function with Series argument. names.to_lower() uses a non-existent method to_lower().

3. Final Answer:

   names.str.lower() -> Option A

4. Quick Check:

   Use Series.str.method() for string methods [OK]

Hint: Always use .str before string methods on Series [OK]

Common Mistakes:

• Calling string methods directly on Series without .str
• Using wrong method names like to_lower()
• Passing Series as argument to str functions

3. Given the pandas Series data = pd.Series(['Apple', 'Banana', 'Cherry']), what is the output of data.str[0]?

medium

A. ['Apple', 'Banana', 'Cherry']

B. ['A', 'B', 'C']

C. ['pple', 'anana', 'herry']

D. Error: cannot index strings this way

Solution

1. Step 1: Understand what data.str[0] does

   The str accessor allows indexing each string element. str[0] extracts the first character of each string in the Series.

2. Step 2: Apply indexing to each element

   For 'Apple', first character is 'A'; for 'Banana', 'B'; for 'Cherry', 'C'. So the result is ['A', 'B', 'C'].

3. Final Answer:

   ['A', 'B', 'C'] -> Option B

4. Quick Check:

   str[0] = first char of each string [OK]

Hint: str[index] extracts character at position index [OK]

Common Mistakes:

• Expecting full strings instead of first chars
• Thinking it slices off first char
• Assuming error occurs on indexing

4. What is wrong with this code snippet?

df = pd.DataFrame({'col': ['a', 'b', 'c']})
df['col'] = df['col'].str.upper

medium

A. Missing parentheses to call the method

B. Using .str.upper on numeric data

C. DataFrame column name is invalid

D. str accessor cannot be used on DataFrame columns

Solution

1. Step 1: Identify the method call syntax error

   The code assigns df['col'].str.upper without parentheses, so it assigns the method itself, not the result of calling it.

2. Step 2: Correct the code

   Adding parentheses df['col'].str.upper() calls the method and returns the uppercase strings.

3. Final Answer:

   Missing parentheses to call the method -> Option A

4. Quick Check:

   Call string methods with () [OK]

Hint: Always add () to call string methods [OK]

Common Mistakes:

• Forgetting parentheses after string methods
• Trying to use str methods on numbers
• Confusing Series and DataFrame usage

5. You have a DataFrame df with a column 'email' containing email addresses. How would you extract the domain part (after '@') for each email using the str accessor?

hard

A. df['email'].str.find('@')

B. df['email'].str.replace('@', '')

C. df['email'].str.upper()

D. df['email'].str.split('@').str[1]

Solution

1. Step 1: Understand the goal to extract domain after '@'

   We want the part of the string after '@' in each email address.

2. Step 2: Use str.split('@') to split each string into two parts

   Splitting by '@' creates a list like ['user', 'domain.com']. Using str[1] selects the domain part.

3. Step 3: Check other options

   df['email'].str.replace('@', '') removes '@' but does not extract domain. df['email'].str.find('@') returns index of '@', not domain. df['email'].str.upper() converts to uppercase, unrelated.

4. Final Answer:

   df['email'].str.split('@').str[1] -> Option D

5. Quick Check:

   Split by '@' then select index 1 [OK]

Hint: Split by '@' then pick second part with str[1] [OK]

Common Mistakes:

• Using replace instead of split
• Trying to find index instead of extracting substring
• Ignoring method chaining with str accessor