loc for label-based selection in Pandas - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to select the row with label 'b' using loc.

Pandas

selected_row = df.loc[[1]]

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

A'a'

B'b'

C'c'

D'd'

Attempts:

3 left

💡 Hint

Common Mistakes

Using integer index instead of label.

Using iloc instead of loc.

Using df.loc['b'] selects the row labeled 'b' from the DataFrame.

2fill in blank

medium

Complete the code to select the 'Age' column for the row labeled 'c'.

Pandas

age_value = df.loc[[1], 'Age']

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

A'c'

B'b'

C'a'

D'd'

Attempts:

3 left

💡 Hint

Common Mistakes

Using column name as first argument.

Using integer index instead of label.

df.loc['c', 'Age'] selects the value in the 'Age' column for the row labeled 'c'.

3fill in blank

hard

Fix the error in the code to select rows labeled 'a' to 'c' inclusive.

Pandas

subset = df.loc[[1]]

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

A'a','c'

B['a':'c']

C'a':'c'

D['a', 'b', 'c']

Attempts:

3 left

💡 Hint

Common Mistakes

Using a string slice inside a list.

Using incorrect syntax for slicing.

Using df.loc['a':'c'] selects rows with labels from 'a' to 'c' inclusive.

Note: ['a', 'b', 'c'] also works but is a list of labels, while 'a':'c' is a slice.

4fill in blank

hard

Fill both blanks to select the 'Name' and 'Age' columns for rows labeled from 'b' to 'd' inclusive.

Pandas

subset = df.loc[[1], [2]]

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

A'b':'d'

B['Name', 'Age']

C'Name'

D'Age'

Attempts:

3 left

💡 Hint

Common Mistakes

Using single column names instead of a list for multiple columns.

Using incorrect slicing syntax.

df.loc['b':'d', ['Name', 'Age']] selects rows from 'b' to 'd' and columns 'Name' and 'Age'.

5fill in blank

hard

Fill both blanks to create a dictionary comprehension that maps each word to its length only if the length is greater than 3.

Pandas

lengths = {word: {BLANK_2}} for word in words if {{BLANK_2}}

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

A:

Blen(word)

Clen(word) > 3

Dword

Attempts:

3 left

💡 Hint

Common Mistakes

Putting a colon inside quotes in the first blank.

Using the wrong condition in the if statement.

Confusing key and value positions.

The dictionary comprehension syntax is {key: value for item in iterable if condition}.

Here, key is word, value is len(word), and condition is len(word) > 3.

Practice

1. What does the loc function in pandas primarily do?

easy

A. Deletes rows based on conditions

B. Selects data by row and column positions

C. Sorts the DataFrame by index

D. Selects data by row and column labels

Solution

1. Step 1: Understand the purpose of loc

   loc is designed to select data using labels, not positions.

2. Step 2: Compare with other selection methods

   Unlike iloc which uses positions, loc uses row and column labels.

3. Final Answer:

   Selects data by row and column labels -> Option D

4. Quick Check:

   loc = label-based selection [OK]

Hint: Remember: loc uses labels, iloc uses positions [OK]

Common Mistakes:

• Confusing loc with iloc
• Thinking loc sorts data
• Assuming loc deletes data

2. Which of the following is the correct syntax to select rows with labels 'a' and 'b' and columns 'x' and 'y' using loc?

easy

A. df.loc[a,b][x,y]

B. df.loc[['a', 'b'], ['x', 'y']]

C. df.loc[a:b, x:y]

D. df.loc('a', 'b', 'x', 'y')

Solution

1. Step 1: Identify correct loc syntax

   loc uses square brackets with row labels first, then column labels, both as lists for multiple selections.

2. Step 2: Check each option

   df.loc[['a', 'b'], ['x', 'y']] uses correct list syntax. Options A, B, and C use wrong syntax or parentheses.

3. Final Answer:

   df.loc[['a', 'b'], ['x', 'y']] -> Option B

4. Quick Check:

   Use lists inside loc brackets [OK]

Hint: Use double brackets with lists for rows and columns [OK]

Common Mistakes:

• Using parentheses instead of brackets
• Not using lists for multiple labels
• Mixing label slices without quotes

3. Given the DataFrame:

import pandas as pd
df = pd.DataFrame({
  'A': [10, 20, 30],
  'B': [40, 50, 60]
}, index=['x', 'y', 'z'])

What is the output of df.loc['y', 'B']?

medium

A. 50

B. [50]

C. 40

D. Error

Solution

1. Step 1: Understand the DataFrame structure

   Rows are labeled 'x', 'y', 'z'. Columns are 'A' and 'B'. Value at row 'y' and column 'B' is 50.

2. Step 2: Apply loc selection

   df.loc['y', 'B'] selects the single value at row 'y' and column 'B', which is 50.

3. Final Answer:

   50 -> Option A

4. Quick Check:

   Label 'y' and column 'B' value = 50 [OK]

Hint: Single label loc returns a scalar value [OK]

Common Mistakes:

• Expecting a list or Series instead of scalar
• Confusing row and column labels
• Using iloc syntax instead

4. What is wrong with this code snippet?

df.loc['a':'c', 'x', 'y']

Assuming df has rows labeled 'a' to 'c' and columns 'x' and 'y'.

medium

A. Too many arguments inside loc brackets

B. Row slice labels are invalid

C. Column labels must be integers

D. loc cannot slice rows

Solution

1. Step 1: Check loc argument structure

   loc expects two arguments inside one pair of brackets: rows and columns, separated by a comma.

2. Step 2: Identify error in code

   The code has three arguments inside brackets: 'a':'c', 'x', 'y'. This is invalid syntax.

3. Final Answer:

   Too many arguments inside loc brackets -> Option A

4. Quick Check:

   loc needs exactly two arguments inside brackets [OK]

Hint: Use one comma inside loc brackets for rows and columns [OK]

Common Mistakes:

• Passing multiple column labels as separate arguments
• Using multiple commas inside loc brackets
• Confusing loc with multiple bracket calls

5. You have a DataFrame df with index labels as dates from '2023-01-01' to '2023-01-05' and columns ['temp', 'humidity']. How would you use loc to select all rows from '2023-01-02' to '2023-01-04' and only the 'temp' column as a DataFrame?

hard

A. df.loc[['2023-01-02', '2023-01-04'], 'temp']

B. df.loc['2023-01-02':'2023-01-04', 'temp']

C. df.loc['2023-01-02':'2023-01-04', ['temp']]

D. df.loc['temp', '2023-01-02':'2023-01-04']

Solution

1. Step 1: Understand row and column selection

   Rows are date labels; we want a slice from '2023-01-02' to '2023-01-04'. Columns are selected by label; we want only 'temp'.

2. Step 2: Use loc with slice and list

   Row slice is a string slice, column selection needs a list for one or more columns. So use ['temp'] for columns.

3. Final Answer:

   df.loc['2023-01-02':'2023-01-04', ['temp']] -> Option C

4. Quick Check:

   Slice rows by labels, columns as list [OK]

Hint: Use slice for rows and list for columns in loc [OK]

Common Mistakes:

• Using single string for multiple columns
• Swapping rows and columns in loc
• Using list for row slice instead of slice