loc for label-based selection in Pandas - Cheat Sheet & Quick Revision

Recall & Review

beginner

What does the loc function in pandas do?

loc is used to select rows and columns by their labels (names) in a DataFrame.

beginner

How do you select a single row with label 'A' using loc?

You use df.loc['A'] to get the row labeled 'A'.

intermediate

How to select multiple rows labeled 'A' and 'B' and columns 'X' and 'Y' using loc?

Use df.loc[['A', 'B'], ['X', 'Y']] to select those rows and columns by label.

intermediate

What happens if you use loc with a label that does not exist?

It raises a KeyError because the label is not found in the DataFrame index or columns.

intermediate

Can loc select rows using a label range like 'A' to 'C'?

Yes, loc supports label slicing, so df.loc['A':'C'] selects rows from 'A' to 'C' inclusive.

What does df.loc['row1'] return?

A The first row by position

B The column named 'row1'

C The row with label 'row1'

D An error

loc selects rows by label, so it returns the row labeled 'row1'.

How do you select columns 'A' and 'B' for rows labeled 'x' and 'y'?

A df.loc['A':'B', 'x':'y']

B df.loc[['x', 'y'], ['A', 'B']]

C df.iloc[['x', 'y'], ['A', 'B']]

D df.iloc[0:2, 0:2]

loc uses labels, so you pass lists of row and column labels.

What error occurs if you use loc with a missing label?

A KeyError

B IndexError

C ValueError

D TypeError

Using a label not in the index or columns raises a KeyError.

Which of these selects rows from label 'a' to 'c' inclusive?

A df.iloc[0:3]

B df.iloc['a':'c']

C df.loc['a','c']

D df.loc['a':'c']

loc supports label slicing and includes the end label.

What does df.loc[:, 'col1'] select?

A All rows for column 'col1'

B All columns for row 'col1'

C Only the first row

D Only the first column

The colon : means all rows, so this selects all rows in column 'col1'.

Explain how to use loc to select specific rows and columns by their labels in a pandas DataFrame.

Think about how you pick items by name in a list or dictionary.

Describe what happens if you try to select a label that does not exist using loc.

What happens when you ask for a key that is not in a dictionary?

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