loc for label-based selection in Pandas - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of loc with single label selection

What is the output of the following code snippet?

Pandas

import pandas as pd

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

result = df.loc['y']
print(result)

A

   A   B
x  10  40
y  20  50
z  30  60

B

A    20
B    50
Name: y, dtype: int64

C

A    10
B    40
Name: x, dtype: int64

D KeyError: 'y'

💡 Hint

Remember loc selects rows by label. Here 'y' is a row label.

Using df.loc['y'] selects the row with label 'y'. It returns a Series with values from columns A and B for that row.

❓ data_output

intermediate

Selecting multiple rows and columns with loc

What is the output of this code?

Pandas

import pandas as pd

df = pd.DataFrame({
    'X': [1, 2, 3, 4],
    'Y': [5, 6, 7, 8],
    'Z': [9, 10, 11, 12]
}, index=['a', 'b', 'c', 'd'])

result = df.loc[['b', 'd'], ['Y', 'Z']]
print(result)

A KeyError: "['Y', 'Z'] not in index"

B

   X   Y
b  2   6
d  4   8

C

   Y   Z
b  5   9
d  7  11

D

   Y   Z
b  6  10
d  8  12

💡 Hint

loc uses labels for both rows and columns. Check the selected labels carefully.

df.loc[['b', 'd'], ['Y', 'Z']] selects rows 'b' and 'd' and columns 'Y' and 'Z'. The values correspond to those positions.

🔧 Debug

advanced

Error when using loc with missing label

What error does this code raise?

Pandas

import pandas as pd

df = pd.DataFrame({'A': [1, 2, 3]}, index=['p', 'q', 'r'])

result = df.loc['s']
print(result)

A KeyError: 's'

B IndexError: single positional indexer is out-of-bounds

C ValueError: cannot index with vector containing NA / NaN values

D TypeError: unhashable type: 'list'

💡 Hint

Check if the label 's' exists in the DataFrame index.

Using loc with a label not in the index raises a KeyError.

🚀 Application

advanced

Using loc to update values in a DataFrame

Given the DataFrame below, which option correctly updates the value in row 'b' and column 'Y' to 100 using loc?

Pandas

import pandas as pd

df = pd.DataFrame({
    'X': [10, 20, 30],
    'Y': [40, 50, 60]
}, index=['a', 'b', 'c'])

A df.loc[['b'], 'Y'] = 100

B df.loc['b', ['Y']] = 100

C df.loc['b', 'Y'] = 100

D df.loc['b']['Y'] = 100

💡 Hint

Use loc with row and column labels directly for assignment.

Option C correctly assigns 100 to the cell at row 'b' and column 'Y'. Option C returns a copy and does not update the original DataFrame. Options A and C assign to slices and may cause unexpected behavior.

🧠 Conceptual

expert

Understanding loc with slice labels

Consider this DataFrame:

import pandas as pd

df = pd.DataFrame({'Val': [10, 20, 30, 40]}, index=['a', 'b', 'c', 'd'])

result = df.loc['b':'d']
print(result)

What is the output?

A

   Val
b   20
c   30
d   40

B

   Val
b   20
c   30

C

   Val
c   30
d   40

D KeyError: 'b':'d'

💡 Hint

loc slice includes both start and end labels.

When slicing with loc, both the start and end labels are included in the result.

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