np.ix_() for open mesh indexing in NumPy - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of np.ix_() for 2D mesh indexing

What is the output of the following code snippet using np.ix_() to create an open mesh grid?

NumPy

import numpy as np
rows = np.array([1, 3, 5])
cols = np.array([2, 4])
mesh = np.ix_(rows, cols)
result = mesh[0] + mesh[1]
print(result)

A

[[3 5]
 [5 7]
 [7 9]]

B

[[3 5 7]
 [5 7 9]]

C

[[3 5]
 [4 6]
 [5 7]]

D

[[3 5]
 [4 6]]

💡 Hint

Remember that np.ix_ creates open mesh grids that broadcast arrays for indexing.

np.ix_ creates two arrays: one column vector from rows and one row vector from cols. Adding them broadcasts to a 3x2 array where each element is sum of row and column indices.

❓ data_output

intermediate

Shape of arrays from np.ix_()

Given a = np.array([0, 1, 2]) and b = np.array([10, 20, 30, 40]), what are the shapes of the arrays returned by np.ix_(a, b)?

NumPy

import numpy as np
a = np.array([0, 1, 2])
b = np.array([10, 20, 30, 40])
mesh = np.ix_(a, b)
shapes = (mesh[0].shape, mesh[1].shape)
print(shapes)

A ((4, 1), (3, 1))

B ((3, 1), (1, 4))

C ((1, 3), (4, 1))

D ((3, 4), (3, 4))

💡 Hint

np.ix_ returns arrays shaped to broadcast for indexing: one column vector and one row vector.

np.ix_ returns arrays shaped (len(a),1) and (1,len(b)) to create an open mesh grid for indexing.

🔧 Debug

advanced

Identify the error in using np.ix_() for 3D indexing

What error will the following code raise when using np.ix_() for 3D indexing?

NumPy

import numpy as np
x = np.array([0, 1])
y = np.array([2, 3])
z = np.array([4, 5])
mesh = np.ix_(x, y, z)
arr = np.arange(60).reshape(3,4,5)
result = arr[mesh]
print(result.shape)

A ValueError: cannot reshape array of size 60 into shape (3,4,5)

B TypeError: unhashable type: 'numpy.ndarray'

C IndexError: shape mismatch: indexing arrays could not be broadcast together

D No error, prints (2, 2, 2)

💡 Hint

Check how np.ix_ arrays broadcast with the target array shape.

np.ix_ returns arrays shaped (2,1,1), (1,2,1), (1,1,2). Using these to index a (3,4,5) array causes shape mismatch because the indexing arrays cannot broadcast to the same shape.

🚀 Application

advanced

Using np.ix_() to select submatrix

You have a 5x5 matrix mat. Which option correctly uses np.ix_() to select rows 1, 3 and columns 0, 4 to create a 3x2 submatrix?

NumPy

import numpy as np
mat = np.arange(25).reshape(5,5)

A mat[[0,4]][[1,3]]

B mat[[1,3], [0,4]]

C mat[np.ix_([0,4], [1,3])]

D mat[np.ix_([1,3], [0,4])]

💡 Hint

np.ix_ creates open mesh indexing for selecting multiple rows and columns.

Option D uses np.ix_ with correct row and column indices to select a 2x2 submatrix. Option D selects elements at paired indices, not a mesh. Options C and D select wrong rows/columns or slice incorrectly.

🧠 Conceptual

expert

Why use np.ix_() instead of broadcasting arrays directly?

Why is np.ix_() preferred for open mesh indexing over directly broadcasting arrays for indexing in NumPy?

A np.ix_ ensures the indexing arrays have correct shapes to broadcast without ambiguity, preventing shape mismatch errors.

B np.ix_ compresses the indexing arrays to reduce memory usage.

C np.ix_ converts arrays to lists for compatibility with older NumPy versions.

D np.ix_ automatically sorts the indices to optimize indexing speed.

💡 Hint

Think about how np.ix_ shapes arrays for broadcasting.

np.ix_ reshapes 1D arrays into N-dimensional open mesh grids with singleton dimensions, ensuring they broadcast correctly for indexing without shape conflicts.

Practice

1. What does np.ix_() do in NumPy?

easy

A. Generates random numbers in a given range

B. Creates open mesh index arrays from 1D index lists to select multi-dimensional slices

C. Sorts arrays in ascending order

D. Flattens multi-dimensional arrays into 1D arrays

Solution

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

   It takes multiple 1D index arrays and returns open mesh grids for indexing.

2. Step 2: Recognize its use in multi-dimensional selection

   This allows selecting rows and columns without loops by creating broadcastable index arrays.

3. Final Answer:

   Creates open mesh index arrays from 1D index lists to select multi-dimensional slices -> Option B

4. Quick Check:

   np.ix_() = open mesh indexing [OK]

Hint: np.ix_() makes multi-dim indexing easy without loops [OK]

Common Mistakes:

• Confusing np.ix_() with random number generation
• Thinking it sorts arrays
• Assuming it flattens arrays

2. Which of the following is the correct syntax to use np.ix_() for selecting rows 0 and 2 and columns 1 and 3 from a 2D array arr?

easy

A. arr[np.ix_([0, 1], [2, 3])]

B. arr[np.ix_([1, 3], [0, 2])]

C. arr[np.ix_([0, 2], [1, 3])]

D. arr[np.ix_([2, 0], [3, 1])]

Solution

1. Step 1: Identify the correct row and column indices

   Rows needed are 0 and 2; columns needed are 1 and 3.

2. Step 2: Use np.ix_() with rows first, then columns

   Correct syntax is arr[np.ix_([0, 2], [1, 3])] to select those rows and columns.

3. Final Answer:

   arr[np.ix_([0, 2], [1, 3])] -> Option C

4. Quick Check:

   Rows then columns in np.ix_() [OK]

Hint: np.ix_() takes row indices first, then column indices [OK]

Common Mistakes:

• Swapping row and column indices order
• Using wrong index values
• Confusing np.ix_() with slicing syntax

3. Given the code:

import numpy as np
arr = np.array([[10, 20, 30, 40],
                [50, 60, 70, 80],
                [90, 100, 110, 120]])
rows = [0, 2]
cols = [1, 3]
result = arr[np.ix_(rows, cols)]
print(result)

What is the output?

medium

A. [[ 20 40]
[100 120]]

B. [[ 10 30]
[ 90 110]]

C. [[ 50 80]
[ 90 120]]

D. [[ 20 40 60]
[100 120 140]]

Solution

1. Step 1: Identify rows and columns selected

   Rows 0 and 2 correspond to first and last rows; columns 1 and 3 correspond to second and fourth columns.

2. Step 2: Extract values from arr at these indices

   Row 0: values at cols 1 and 3 are 20 and 40; Row 2: values at cols 1 and 3 are 100 and 120.

3. Final Answer:

   [[ 20 40] [100 120]] -> Option A

4. Quick Check:

   Selected rows and columns match output [OK]

Hint: Match row and column indices to array positions carefully [OK]

Common Mistakes:

• Mixing up row and column indices
• Printing wrong slices
• Confusing np.ix_() output shape

4. What is wrong with this code snippet?

import numpy as np
arr = np.arange(16).reshape(4,4)
rows = [1, 3]
cols = [0, 2]
result = arr[np.ix_(cols, rows)]
print(result)

medium

A. np.ix_() cannot be used with lists

B. The reshape size is incorrect for np.arange(16)

C. The lists rows and cols contain invalid indices

D. The order of indices in np.ix_() is swapped; rows should come first

Solution

1. Step 1: Check the order of indices in np.ix_()

   np.ix_() expects row indices first, then column indices.

2. Step 2: Identify the mistake in the code

   Here, cols are passed first and rows second, which swaps the intended selection.

3. Final Answer:

   The order of indices in np.ix_() is swapped; rows should come first -> Option D

4. Quick Check:

   Rows first, columns second in np.ix_() [OK]

Hint: Always pass rows first, then columns to np.ix_() [OK]

Common Mistakes:

• Passing columns before rows
• Misunderstanding reshape dimensions
• Thinking np.ix_() only accepts arrays, not lists

5. You have a 3D NumPy array data with shape (4, 5, 6). You want to select slices at indices [1, 3] for the first dimension, [0, 2, 4] for the second, and [2, 5] for the third dimension. Which code correctly uses np.ix_() to get this selection?

hard

A. data[np.ix_([1, 3], [0, 2, 4], [2, 5])]

B. data[np.ix_([0, 2, 4], [1, 3], [2, 5])]

C. data[np.ix_([1, 3], [2, 5], [0, 2, 4])]

D. data[np.ix_([2, 5], [1, 3], [0, 2, 4])]

Solution

1. Step 1: Identify the dimension order and indices

   First dimension indices: [1, 3]; second dimension: [0, 2, 4]; third dimension: [2, 5].

2. Step 2: Use np.ix_() with indices in correct dimension order

   np.ix_() takes indices in order of dimensions: first, second, then third.

3. Step 3: Confirm the correct option

   data[np.ix_([1, 3], [0, 2, 4], [2, 5])] matches the correct order and indices.

4. Final Answer:

   data[np.ix_([1, 3], [0, 2, 4], [2, 5])] -> Option A

5. Quick Check:

   Dimension order matches np.ix_() arguments [OK]

Hint: Pass indices in dimension order to np.ix_() for multi-dim arrays [OK]

Common Mistakes:

• Mixing up dimension order in np.ix_()
• Using wrong index lists
• Assuming np.ix_() works differently for 3D arrays