Selecting rows (loc, iloc) in Data Analysis Python - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to select the row with label 2 using loc.

Data Analysis Python

selected_row = df.loc[[1]]

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

A2

Biloc[2]

Cdf[2]

D'2'

Attempts:

3 left

💡 Hint

Common Mistakes

Using quotes around the number when the index is integer.

Using iloc instead of loc.

Using loc with the label 2 selects the row with index label 2.

2fill in blank

medium

Complete the code to select the third row using iloc.

Data Analysis Python

selected_row = df.iloc[[1]]

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

A3

B2

C'2'

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Using 3 instead of 2 for the third row.

Using loc instead of iloc.

iloc uses zero-based integer positions, so the third row is at position 2.

3fill in blank

hard

Fix the code to select rows 1 to 3 (inclusive) using loc.

Data Analysis Python

subset = df.loc[[1]]

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

A1:3

B[1:3]

C1:4

D[1,3]

Attempts:

3 left

💡 Hint

Common Mistakes

Using 1:4, confusing loc (inclusive end) with iloc (exclusive end).

Using list syntax instead of slice notation.

loc slicing with labels is inclusive of both start and end, so 1:3 selects rows with index labels 1, 2, and 3.

4fill in blank

hard

Fill both blanks to select rows with iloc from position 0 to 2 and columns 'A' and 'B'.

Data Analysis Python

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

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

A0:3

B['A', 'B']

C[0, 1]

Dslice(0, 3)

Attempts:

3 left

💡 Hint

Common Mistakes

Using column labels instead of positions with iloc.

Using list syntax for rows instead of slice.

iloc uses integer slicing for rows (0:3 selects positions 0,1,2) and list of integer positions for columns ([0,1] assuming 'A' and 'B' are the first two columns).

5fill in blank

hard

Fill all three blanks to create a dictionary of row labels and values from column 'score' for rows where score > 50 using loc.

Data Analysis Python

result = {index: row[[1]] for index, row in df.loc[df[[2]] [3] 50].iterrows()}

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

A'score'

Bscore

C>

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using score without quotes as a string key.

Using < instead of > for the condition.

We use 'score' as the column label in both loc filtering and row access. The condition uses > to filter scores greater than 50.

Practice

1. Which of the following statements correctly describes the difference between loc and iloc in pandas?

easy

A. loc selects rows by label, iloc selects rows by position.

B. loc selects columns by position, iloc selects rows by label.

C. loc selects rows by position, iloc selects columns by label.

D. loc and iloc both select rows by position.

Solution

1. Step 1: Understand loc usage

   loc selects rows using the row labels or names, not their numeric position.

2. Step 2: Understand iloc usage

   iloc selects rows based on their integer position, starting at zero.

3. Final Answer:

   loc selects rows by label, iloc selects rows by position. -> Option A

4. Quick Check:

   loc = label, iloc = position [OK]

Hint: Remember: loc = label, iloc = integer position [OK]

Common Mistakes:

• Confusing label with position
• Using iloc with labels
• Assuming loc uses integer positions

2. Which of the following is the correct syntax to select the row with label '2023-01-01' from a DataFrame df using loc?

easy

A. df.loc[2023-01-01]

B. df.iloc[2023-01-01]

C. df.iloc['2023-01-01']

D. df.loc['2023-01-01']

Solution

1. Step 1: Identify correct method for label selection

   To select by label, use loc with the label as a string if it is a string label.

2. Step 2: Check syntax for string label

   The label '2023-01-01' must be in quotes inside loc brackets.

3. Final Answer:

   df.loc['2023-01-01'] -> Option D

4. Quick Check:

   Use quotes for string labels with loc [OK]

Hint: Use quotes around string labels inside loc brackets [OK]

Common Mistakes:

• Omitting quotes around string labels
• Using iloc with labels
• Using loc without brackets

3. Given the DataFrame df below:

import pandas as pd
data = {'Name': ['Anna', 'Bob', 'Cara', 'Dave'], 'Age': [23, 35, 45, 29]}
df = pd.DataFrame(data, index=['a', 'b', 'c', 'd'])

What is the output of df.loc['b']?

medium

A. Name Bob
Age 35
Name: 1, dtype: object

B. Name Cara
Age 45
Name: c, dtype: object

C. Name Bob
Age 35
Name: b, dtype: object

D. Error: KeyError

Solution

1. Step 1: Understand DataFrame index labels

   The DataFrame has index labels 'a', 'b', 'c', 'd'. The row with label 'b' corresponds to Bob, age 35.

2. Step 2: Use loc to select row by label

   df.loc['b'] returns the row with label 'b' as a Series with Name and Age values.

3. Final Answer:

   Name Bob Age 35 Name: b, dtype: object -> Option C

4. Quick Check:

   loc selects row 'b' with Bob's data [OK]

Hint: loc uses index labels exactly as given [OK]

Common Mistakes:

• Using iloc instead of loc
• Confusing label 'b' with position 1
• Expecting a DataFrame instead of Series

4. Consider this code snippet:

import pandas as pd
data = {'Value': [10, 20, 30]}
df = pd.DataFrame(data, index=[1, 2, 3])
row = df.loc[2]

What is wrong with this code if the intention was to select the second row by position?

medium

A. It selects the row with label 2, not the second row by position.

B. It raises a KeyError because 2 is not a label.

C. It correctly selects the second row by position.

D. It raises a TypeError because loc needs a string.

Solution

1. Step 1: Check DataFrame index labels

   The DataFrame index labels are [1, 2, 3]. Using loc[2] selects the row with label 2, not the second row by position.

2. Step 2: Understand difference between label and position

   The second row by position is at index 1 (zero-based), so iloc[1] should be used to select by position.

3. Final Answer:

   It selects the row with label 2, not the second row by position. -> Option A

4. Quick Check:

   loc uses labels, iloc uses positions [OK]

Hint: Use iloc for position, loc for label [OK]

Common Mistakes:

• Using loc to select by position
• Confusing index labels with positions
• Expecting loc to be zero-based

5. You have a DataFrame df with a non-numeric index:

import pandas as pd
data = {'Score': [88, 92, 95, 70]}
labels = ['a', 'b', 'c', 'd']
df = pd.DataFrame(data, index=labels)

How would you select the first two rows by position and the rows labeled 'b' and 'c' by label, then combine these selections into one DataFrame?

hard

A. df.loc[:2] + df.loc[['b', 'c']]

B. pd.concat([df.iloc[:2], df.loc[['b', 'c']]])

C. df.iloc[['a', 'b']] + df.loc[1:2]

D. df.loc[['a', 'b']] + df.iloc[1:2]

Solution

1. Step 1: Select first two rows by position

   Use df.iloc[:2] to select rows at positions 0 and 1.

2. Step 2: Select rows labeled 'b' and 'c'

   Use df.loc[['b', 'c']] to select rows by their labels.

3. Step 3: Combine selections into one DataFrame

   Use pd.concat to join the two selections vertically.

4. Final Answer:

   pd.concat([df.iloc[:2], df.loc[['b', 'c']]]) -> Option B

5. Quick Check:

   Combine iloc and loc selections with concat [OK]

Hint: Use pd.concat to join iloc and loc selections [OK]

Common Mistakes:

• Trying to add DataFrames with + operator
• Mixing labels and positions incorrectly
• Using loc with positions or iloc with labels