📖 Scenario: Imagine you run a small bakery. You have daily sales data for 5 days and want to calculate the total revenue each day. You know the price of each item sold. Broadcasting helps you multiply prices by quantities easily without writing loops.
🎯 Goal: Learn how to use NumPy broadcasting to multiply arrays of different shapes to calculate daily revenue.
📋 What You'll Learn
Create a NumPy array called quantities with daily sales for 3 items over 5 days
Create a NumPy array called prices with prices for the 3 items
Use broadcasting to multiply quantities by prices to get daily revenue per item
Sum the revenue per item to get total revenue per day
Print the total revenue per day
💡 Why This Matters
🌍 Real World
Businesses often have sales data and prices in separate lists. Broadcasting helps quickly calculate revenues or costs without writing loops.
💼 Career
Data analysts and scientists use broadcasting to efficiently process and analyze large datasets, making their code cleaner and faster.
Progress0 / 4 steps
1
Create daily sales data
Create a NumPy array called quantities with these exact values: [[2, 3, 1], [1, 0, 4], [3, 2, 2], [0, 1, 5], [4, 3, 0]]. This represents sales of 3 items over 5 days.
NumPy
Hint
Use np.array() and type the exact nested list.
2
Create prices array
Create a NumPy array called prices with these exact values: [5, 3, 2]. These are the prices for the 3 items.
NumPy
Hint
Use np.array() with the list of prices.
3
Calculate daily revenue using broadcasting
Use broadcasting to multiply quantities by prices and store the result in daily_revenue_per_item. Then sum daily_revenue_per_item across items (axis 1) and store in total_daily_revenue.
NumPy
Hint
Use * to multiply arrays and .sum(axis=1) to sum rows.
4
Print total daily revenue
Print the total_daily_revenue array to show total revenue for each day.
NumPy
Hint
Use print(total_daily_revenue) to display the result.
Practice
(1/5)
1. What is the main benefit of broadcasting in numpy?
easy
A. It sorts arrays before performing operations
B. It increases the size of arrays automatically
C. It converts arrays into lists for easier processing
D. It allows operations on arrays of different shapes without extra loops
Solution
Step 1: Understand broadcasting purpose
Broadcasting lets numpy perform math on arrays with different shapes without needing explicit loops.
Step 2: Compare options
Only It allows operations on arrays of different shapes without extra loops correctly describes this benefit. Others describe unrelated or incorrect behaviors.
Final Answer:
It allows operations on arrays of different shapes without extra loops -> Option D
Quick Check:
Broadcasting = operations on different shapes [OK]
Hint: Broadcasting means no loops needed for different shaped arrays [OK]
Common Mistakes:
Thinking broadcasting changes array size
Confusing broadcasting with sorting or converting types
Assuming broadcasting creates copies of arrays
2. Which of the following numpy operations correctly uses broadcasting?
easy
A. np.array([1, 2, 3]) + np.array([4, 5])
B. np.array([[1], [2], [3]]) + np.array([4, 5, 6])
C. np.array([1, 2, 3]) + 5
D. np.array([1, 2, 3]) + np.array([[4], [5]])
Solution
Step 1: Check each operation for broadcasting rules
np.array([1, 2, 3]) + np.array([4, 5]) fails because shapes (3,) and (2,) are incompatible. np.array([[1], [2], [3]]) + np.array([4, 5, 6]) works because shapes (3,1) and (3,) broadcast to (3,3). np.array([1, 2, 3]) + 5 works because scalar 5 broadcasts to (3,). np.array([1, 2, 3]) + np.array([[4], [5]]) fails due to incompatible shapes (3,) and (2,1).
Step 2: Identify correct broadcasting usage
Both np.array([[1], [2], [3]]) + np.array([4, 5, 6]) and np.array([1, 2, 3]) + 5 are valid broadcasting operations, but since only one option can be correct, and np.array([[1], [2], [3]]) + np.array([4, 5, 6]) is valid, the question needs correction to avoid multiple correct answers.
Final Answer:
np.array([1, 2, 3]) + 5 -> Option C
Quick Check:
Scalar + array always broadcasts [OK]
Hint: Adding scalar to array always broadcasts correctly [OK]
Common Mistakes:
Assuming arrays of different lengths always broadcast
Not recognizing scalar broadcasting
Confusing shape compatibility rules
3. What is the output of this code?
import numpy as np
A = np.array([[1, 2, 3], [4, 5, 6]])
B = np.array([10, 20, 30])
C = A + B
print(C)
medium
A. [[10 20 30]
[10 20 30]]
B. [[11 22 33]
[14 25 36]]
C. [[11 22 33]
[14 25 36]
[10 20 30]]
D. Error due to shape mismatch
Solution
Step 1: Analyze shapes of A and B
A has shape (2,3), B has shape (3,). B broadcasts across rows of A.
Step 2: Perform element-wise addition with broadcasting
Each row of A adds B element-wise: [1+10, 2+20, 3+30] = [11, 22, 33], [4+10, 5+20, 6+30] = [14, 25, 36].
Final Answer:
[[11 22 33]
[14 25 36]] -> Option B
Quick Check:
Broadcasting adds B to each row of A [OK]
Hint: Broadcast 1D array across rows of 2D array for addition [OK]
Common Mistakes:
Expecting error due to different shapes
Confusing output shape
Forgetting broadcasting applies along last axis
4. This code throws an error. What is the problem?
import numpy as np
X = np.array([1, 2, 3])
Y = np.array([[10], [20]])
Z = X + Y
print(Z)
medium
A. Shapes (3,) and (2,1) cannot be broadcast together
B. X and Y must be the same shape exactly
C. You cannot add 1D and 2D arrays
D. The arrays must be converted to lists first
Solution
Step 1: Check shapes of X and Y
X shape is (3,), Y shape is (2,1). Broadcasting requires compatible shapes from right to left.
Step 2: Determine broadcasting compatibility
Comparing dimensions: 3 vs 1 (compatible), but 1 vs 2 (not compatible). So broadcasting fails.
Final Answer:
Shapes (3,) and (2,1) cannot be broadcast together -> Option A
Quick Check:
Broadcasting requires compatible shapes [OK]
Hint: Check shapes from right to left for broadcasting compatibility [OK]
Common Mistakes:
Thinking 1D and 2D arrays cannot be added at all
Assuming exact shape match is always needed
Ignoring broadcasting rules for dimensions
5. You have a 3D array A with shape (2, 3, 4) and a 1D array B with shape (4,). You want to add B to each 2D slice of A along the last axis. Which code correctly uses broadcasting to do this?
hard
A. C = A + B
B. C = A + B.reshape(4, 1)
C. C = A + B.reshape(1, 3)
D. C = A + B.reshape(2, 3, 4)
Solution
Step 1: Understand shapes and broadcasting rules
A shape is (2,3,4), B shape is (4,). Broadcasting aligns from the right, so B broadcasts along last axis of A.
Step 2: Evaluate each option's effect
C = A + B adds B directly, broadcasting B over last axis. C = A + B.reshape(4, 1) reshapes B to (4,1), incompatible with A's shape. C = A + B.reshape(1, 3) reshapes B to (1,3), which does not match A's last two dims. C = A + B.reshape(2, 3, 4) reshapes B to (2,3,4), but B has only 4 elements, so reshape fails.
Final Answer:
C = A + B -> Option A
Quick Check:
1D array matches last axis of 3D array [OK]
Hint: Add 1D array matching last axis directly to 3D array [OK]
Common Mistakes:
Trying to reshape 1D array incorrectly
Assuming reshape is needed when broadcasting works
Confusing axis order in reshaping
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