np.mean() for average in NumPy - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of np.mean() on 2D array with axis=0

What is the output of this code snippet using np.mean() on a 2D array with axis=0?

NumPy

import numpy as np
arr = np.array([[1, 2, 3], [4, 5, 6]])
result = np.mean(arr, axis=0)
print(result)

A [2.5 3.5 4.5]

B [2. 3. 4.]

C [3. 4. 5.]

D [1.5 2.5 3.5]

💡 Hint

Remember, axis=0 means averaging down the rows for each column.

The mean along axis=0 calculates the average of each column: (1+4)/2=2.5, (2+5)/2=3.5, (3+6)/2=4.5.

❓ data_output

intermediate

Number of elements in np.mean() output with axis=1

Given a 3x4 numpy array, how many elements does the output have after applying np.mean() with axis=1?

NumPy

import numpy as np
arr = np.arange(12).reshape(3,4)
result = np.mean(arr, axis=1)
print(result)
print(len(result))

A 12

B 3

C 4

D 1

💡 Hint

Axis=1 means averaging across columns for each row.

The array has 3 rows, so averaging across columns (axis=1) produces 3 mean values, one per row.

🔧 Debug

advanced

Identify the error in np.mean() usage

What error does this code raise when trying to compute the mean?

NumPy

import numpy as np
arr = np.array([1, 2, 3])
result = np.mean(arr, axis=1)
print(result)

A TypeError: unsupported operand type(s) for +: 'int' and 'str'

B ValueError: could not broadcast input array from shape (3,) into shape (1,)

C AxisError: axis 1 is out of bounds for array of dimension 1

D No error, prints 2.0

💡 Hint

Check the shape of the array and the axis parameter.

The array is 1D, so axis=1 is invalid and causes an AxisError.

🧠 Conceptual

advanced

Effect of np.mean() with keepdims=True

What is the shape of the output when applying np.mean() on a 2D array with axis=1 and keepdims=True?

NumPy

import numpy as np
arr = np.array([[1, 2, 3], [4, 5, 6]])
result = np.mean(arr, axis=1, keepdims=True)
print(result.shape)

A (2, 1)

B (2,)

C (1, 3)

D (3, 2)

💡 Hint

Keepdims keeps the reduced dimension as size 1.

With keepdims=True, the output keeps the axis dimension with size 1, so shape is (2,1).

🚀 Application

expert

Calculate weighted average using np.mean()

You have values [10, 20, 30] and weights [1, 2, 3]. Which code correctly calculates the weighted average using numpy?

A np.mean([10, 20, 30], weights=[1, 2, 3])

B np.average([10, 20, 30], weights=[1, 2, 3])

C np.mean([10, 20, 30]) * np.mean([1, 2, 3])

D np.sum(np.array([10, 20, 30]) * np.array([1, 2, 3])) / np.sum([1, 2, 3])

💡 Hint

np.mean() does not support weights parameter; use np.average or manual calculation.

np.mean() does not accept weights. np.average supports weights. Option D manually computes weighted average correctly.

Practice

1. What does the function np.mean() do when applied to a list or array?

easy

A. Counts the number of elements in the list or array

B. Finds the maximum value in the list or array

C. Sorts the list or array in ascending order

D. Calculates the average (mean) value of the numbers

Solution

1. Step 1: Understand the purpose of np.mean()

   The function np.mean() is designed to calculate the average value of numbers in a list or array.

2. Step 2: Compare with other options

   Other options describe different functions: maximum value, count, or sorting, which are not what np.mean() does.

3. Final Answer:

   Calculates the average (mean) value of the numbers -> Option D

4. Quick Check:

   np.mean() = average [OK]

Hint: Mean = sum of values ÷ number of values [OK]

Common Mistakes:

• Confusing mean with max or count
• Thinking it sorts the data
• Using sum instead of mean

2. Which of the following is the correct syntax for np.mean() to calculate the mean of a NumPy array named arr?

easy

A. np.mean(arr)

B. mean(arr)

C. arr.mean()

D. np.average(arr)

Solution

1. Step 1: Identify the standard NumPy function for mean

   The correct function to calculate mean in NumPy is np.mean(), which takes the array as an argument.

2. Step 2: Check other options

   mean(arr) is not defined without importing; arr.mean() is valid but not the function asked; np.average(arr) is similar but not the exact function requested.

3. Final Answer:

   np.mean(arr) -> Option A

4. Quick Check:

   Use np.mean(array) for average [OK]

Hint: Use np.mean(array) to get average quickly [OK]

Common Mistakes:

• Calling mean() without np prefix
• Using arr.mean() instead of np.mean(arr)
• Confusing np.mean() with np.average()

3. What is the output of the following code?

import numpy as np
arr = np.array([[1, 2, 3], [4, 5, 6]])
result = np.mean(arr, axis=0)
print(result)

medium

A. [1.5 2.5 3.5]

B. [3. 4. 5.]

C. [2.5 3.5 4.5]

D. [3.5 4.5 5.5]

Solution

1. Step 1: Understand the array and axis parameter

   The array is 2 rows by 3 columns: [[1,2,3],[4,5,6]]. Using axis=0 means calculating the mean down each column.

2. Step 2: Calculate mean for each column

   Column 1: (1+4)/2 = 2.5, Column 2: (2+5)/2 = 3.5, Column 3: (3+6)/2 = 4.5. So the result is [2.5, 3.5, 4.5].

3. Step 3: Check options carefully

   [2.5 3.5 4.5] matches the calculated values exactly.

4. Final Answer:

   [2.5 3.5 4.5] -> Option C

5. Quick Check:

   Mean along axis=0 averages columns [OK]

Hint: Axis=0 averages columns, axis=1 averages rows [OK]

Common Mistakes:

• Mixing axis=0 and axis=1
• Calculating mean of entire array instead of axis
• Misreading array shape

4. The following code throws an error. What is the problem?

import numpy as np
arr = [1, 2, 3, 4]
result = np.mean(arr, axis=1)
print(result)

medium

A. np.mean() cannot be used on lists

B. The list is 1D, so axis=1 is invalid

C. Missing parentheses in np.mean

D. The variable name 'arr' is reserved

Solution

1. Step 1: Check the array shape

   The variable arr is a 1D list with shape (4,). Axis 1 does not exist in 1D data.

2. Step 2: Understand axis parameter

   Axis=1 means operate along columns in 2D or higher arrays. Using axis=1 on 1D data causes an error.

3. Final Answer:

   The list is 1D, so axis=1 is invalid -> Option B

4. Quick Check:

   Axis must exist in array shape [OK]

Hint: Check array dimensions before using axis [OK]

Common Mistakes:

• Using axis parameter on 1D arrays
• Thinking np.mean can't handle lists
• Confusing axis indexing

5. You have a 3D NumPy array representing daily temperatures for 2 cities over 3 days with 4 measurements each day:

temps = np.array([
  [[20, 22, 21, 23], [19, 20, 22, 21]],
  [[21, 23, 22, 24], [20, 21, 23, 22]],
  [[22, 24, 23, 25], [21, 22, 24, 23]]
])

Which code correctly calculates the average temperature for each city over all days and measurements?

hard

A. np.mean(temps, axis=(0,2))

B. np.mean(temps, axis=0)

C. np.mean(temps, axis=1)

D. np.mean(temps, axis=2)

Solution

1. Step 1: Understand the array shape and axes

   The array shape is (3, 2, 4): 3 days, 2 cities, 4 measurements each day.

2. Step 2: Identify axes to average over

   To get average per city over all days and measurements, we average over days (axis 0) and measurements (axis 2), keeping axis 1 (cities).

3. Step 3: Use axis=(0,2) in np.mean

   Using np.mean(temps, axis=(0,2)) averages over days and measurements, resulting in average per city.

4. Final Answer:

   np.mean(temps, axis=(0,2)) -> Option A

5. Quick Check:

   Average over days and measurements = axis 0 and 2 [OK]

Hint: Use tuple axis to average multiple dimensions [OK]

Common Mistakes:

• Averaging over wrong axes
• Using single axis instead of tuple
• Confusing city axis with day or measurement axis