replace() for value substitution in Pandas - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to replace all occurrences of 0 with 100 in the DataFrame column 'score'.

Pandas

df['score'] = df['score'].[1](0, 100)

Drag options to blanks, or click blank then click option'

Areplace

Bsubstitute

Cswap

Dchange

Attempts:

3 left

💡 Hint

Common Mistakes

Using methods like 'substitute' or 'swap' which do not exist in pandas.

Trying to assign values directly without using a method.

The replace() method is used to substitute values in a pandas Series or DataFrame.

2fill in blank

medium

Complete the code to replace the string 'low' with 'L' in the 'level' column of the DataFrame.

Pandas

df['level'] = df['level'].[1]('low', 'L')

Drag options to blanks, or click blank then click option'

Areplace

Bswap

Cchange

Dupdate

Attempts:

3 left

💡 Hint

Common Mistakes

Using non-existent methods like 'swap' or 'update'.

Trying to replace values without calling a method.

The replace() method works for string substitutions in pandas columns.

3fill in blank

hard

Fix the error in the code to replace 1 with 10 in the 'count' column.

Pandas

df['count'] = df['count'].[1](1, 10)

Drag options to blanks, or click blank then click option'

Aswitch

Breplace

Csubstitute

Dchange

Attempts:

3 left

💡 Hint

Common Mistakes

Using methods that do not exist in pandas like 'substitute' or 'switch'.

Misspelling the method name.

The correct pandas method to replace values is replace(). Other options cause errors.

4fill in blank

hard

Fill both blanks to create a dictionary that replaces 'A' with 'Alpha' in the 'grade' column.

Pandas

df['grade'] = df['grade'].replace([1]: [2])

Drag options to blanks, or click blank then click option'

A'A'

B'Alpha'

C'B'

D'Beta'

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping keys and values in the dictionary.

Not using quotes around string keys and values.

Use a dictionary inside replace() to map old values to new ones. The keys are old values, the values are new.

5fill in blank

hard

Fill all three blanks to replace 0 with 'Zero' and 1 with 'One' in the 'number' column using a dictionary.

Pandas

df['number'] = df['number'].replace([1]: [2], [3]: 'One'})

Drag options to blanks, or click blank then click option'

A0

B'Zero'

C1

D'One'

Attempts:

3 left

💡 Hint

Common Mistakes

Putting quotes around keys which are numbers.

Not closing the dictionary properly.

Use a dictionary with keys as original values and values as replacements. Here, 0 maps to 'Zero', 1 maps to 'One'.

Practice

1. What does the replace() method do in pandas?

easy

A. It sorts the DataFrame based on column values.

B. It deletes rows containing specific values.

C. It adds new columns to the DataFrame.

D. It swaps specified values in a DataFrame or Series with new values.

Solution

1. Step 1: Understand the purpose of replace()

   The replace() method is used to change specific values in pandas data structures.

2. Step 2: Compare with other options

   The other options describe sorting the DataFrame, deleting rows, and adding columns, which are unrelated to value substitution.

3. Final Answer:

   It swaps specified values in a DataFrame or Series with new values. -> Option D

4. Quick Check:

   replace() = value substitution [OK]

Hint: Remember: replace() swaps old values with new ones [OK]

Common Mistakes:

• Confusing replace() with drop() or sort_values()
• Thinking replace() adds or removes rows
• Assuming replace() changes column names

2. Which of the following is the correct syntax to replace all occurrences of 0 with 100 in a pandas Series s?

easy

A. s.replace([100], [0])

B. s.replace(100, 0)

C. s.replace(0, 100)

D. s.replace({100: 0})

Solution

1. Step 1: Identify the correct replace() syntax

   The syntax s.replace(old_value, new_value) replaces all old_value with new_value.

2. Step 2: Check each option

   s.replace(0, 100) correctly replaces 0 with 100. The other options reverse the values: s.replace(100, 0), s.replace([100], [0]), s.replace({100: 0}).

3. Final Answer:

   s.replace(0, 100) -> Option C

4. Quick Check:

   replace(old, new) = s.replace(0, 100) [OK]

Hint: Use s.replace(old_value, new_value) for single value swap [OK]

Common Mistakes:

• Swapping the old and new values by mistake
• Using dict with reversed keys and values
• Confusing list syntax with dict syntax

3. Given the DataFrame:

import pandas as pd
df = pd.DataFrame({'A': [1, 2, 3, 2], 'B': [4, 5, 2, 4]})
df_new = df.replace(2, 99)

What is the content of df_new?

medium

A. {'A': [1, 99, 3, 99], 'B': [4, 5, 99, 4]}

B. {'A': [1, 2, 3, 2], 'B': [4, 5, 2, 4]}

C. {'A': [99, 99, 99, 99], 'B': [99, 99, 99, 99]}

D. {'A': [1, 2, 3, 2], 'B': [4, 5, 99, 4]}

Solution

1. Step 1: Understand replace(2, 99) on DataFrame

   All occurrences of 2 in the DataFrame are replaced by 99.

2. Step 2: Apply replacement to each column

   Column 'A': 2 replaced by 99 at positions 1 and 3.
   Column 'B': 2 replaced by 99 at position 2.

3. Final Answer:

   {'A': [1, 99, 3, 99], 'B': [4, 5, 99, 4]} -> Option A

4. Quick Check:

   All 2s -> 99 in df_new [OK]

Hint: replace() changes all matching values in all columns [OK]

Common Mistakes:

• Assuming replace only affects one column
• Thinking original df changes without assignment
• Replacing only first occurrence instead of all

4. What is wrong with this code snippet?

df = pd.DataFrame({'X': [1, 2, 3]})
df.replace(2, 99)
print(df)

It is intended to replace 2 with 99 in column 'X'.

medium

A. The replace() method does not modify df unless inplace=True or reassigned.

B. The replace() method syntax is incorrect; it needs a dictionary.

C. The print statement should be before replace().

D. replace() cannot be used on DataFrames.

Solution

1. Step 1: Understand replace() behavior

   By default, replace() returns a new DataFrame and does not change the original unless inplace=True or reassigned.

2. Step 2: Analyze the code snippet

   The code calls replace() but does not assign the result back to df or use inplace=True, so df remains unchanged.

3. Final Answer:

   The replace() method does not modify df unless inplace=True or reassigned. -> Option A

4. Quick Check:

   replace() returns new object unless inplace=True [OK]

Hint: Always assign replace() result or use inplace=True [OK]

Common Mistakes:

• Expecting replace() to change original without assignment
• Using wrong syntax for replace()
• Thinking replace() works only on Series

5. You have a DataFrame:

df = pd.DataFrame({'Grade': ['A', 'B', 'C', 'D', 'F']})

You want to replace grades 'A' and 'B' with 'Pass' and 'C', 'D', 'F' with 'Fail' in one call. Which code does this correctly?

hard

A. df.replace(['A', 'B'], ['Pass', 'Pass']).replace(['C', 'D', 'F'], ['Fail', 'Fail', 'Fail'])

B. df.replace({'A': 'Pass', 'B': 'Pass', 'C': 'Fail', 'D': 'Fail', 'F': 'Fail'})

C. df.replace({'Pass': ['A', 'B'], 'Fail': ['C', 'D', 'F']})

D. df.replace({'A': 'Pass', 'B': 'Fail', 'C': 'Pass', 'D': 'Fail', 'F': 'Pass'})

Solution

1. Step 1: Understand replace() with dictionary mapping

   Using a dictionary with keys as old values and values as new values allows multiple replacements in one call.

2. Step 2: Check each option

   df.replace({'A': 'Pass', 'B': 'Pass', 'C': 'Fail', 'D': 'Fail', 'F': 'Fail'}) correctly maps each grade to 'Pass' or 'Fail'. df.replace(['A', 'B'], ['Pass', 'Pass']).replace(['C', 'D', 'F'], ['Fail', 'Fail', 'Fail']) uses two replace calls, which works but is two steps, not one call. df.replace({'Pass': ['A', 'B'], 'Fail': ['C', 'D', 'F']}) reverses keys and values, which is invalid. df.replace({'A': 'Pass', 'B': 'Fail', 'C': 'Pass', 'D': 'Fail', 'F': 'Pass'}) mixes mappings incorrectly.

3. Final Answer:

   df.replace({'A': 'Pass', 'B': 'Pass', 'C': 'Fail', 'D': 'Fail', 'F': 'Fail'}) -> Option B

4. Quick Check:

   Use dict in replace() for multiple value swaps [OK]

Hint: Use dict with old:new pairs for multiple replacements [OK]

Common Mistakes:

• Using lists instead of dict for multiple replacements
• Reversing keys and values in dict
• Chaining replace() unnecessarily