np.ix_() for open mesh indexing in NumPy - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create open mesh indices using np.ix_.

NumPy

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

Drag options to blanks, or click blank then click option'

Aix_

Bmeshgrid

Copen_mesh

Dindices

Attempts:

3 left

💡 Hint

Common Mistakes

Using np.meshgrid instead of np.ix_

Trying to use np.indices which creates grid indices for shape, not from arrays

np.ix_() creates open mesh indexing arrays from 1D arrays, perfect for selecting rows and columns.

2fill in blank

medium

Complete the code to select elements from a 2D array using np.ix_.

NumPy

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

Drag options to blanks, or click blank then click option'

Ameshgrid

Bopen_mesh

Cindices

Dix_

Attempts:

3 left

💡 Hint

Common Mistakes

Using np.meshgrid which returns coordinate matrices, not suitable for fancy indexing here

Trying to index directly with lists without np.ix_

np.ix_() creates index arrays that allow selecting the cross product of rows and columns.

3fill in blank

hard

Fix the error in the code to correctly use np.ix_ for indexing.

NumPy

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

Drag options to blanks, or click blank then click option'

Ameshgrid

Bix_

Cindices

Dopen_mesh

Attempts:

3 left

💡 Hint

Common Mistakes

Using np.meshgrid which returns coordinate matrices, not suitable for fancy indexing

Using np.indices which expects shape, not arrays

np.ix_ is the correct function to create open mesh indices for fancy indexing.

4fill in blank

hard

Fill both blanks to create a dictionary of squares for numbers greater than 3 using dictionary comprehension.

NumPy

numbers = [1, 2, 3, 4, 5]
squares = [1]: [2] for n in numbers if n > 3

Drag options to blanks, or click blank then click option'

An

Bn**2

Dn*2

Attempts:

3 left

💡 Hint

Common Mistakes

Using n*2 instead of n**2 for squares

Swapping key and value positions

The dictionary comprehension uses n as key and n squared as value for numbers greater than 3.

5fill in blank

hard

Fill all three blanks to create a dictionary of uppercase keys and values greater than 10.

NumPy

data = {'a': 5, 'b': 15, 'c': 20}
result = [1]: [2] for k, v in data.items() if v [3] 10}}

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Ak.upper()

Bv

C>

Dk

Attempts:

3 left

💡 Hint

Common Mistakes

Using k instead of k.upper() for keys

Using < instead of > in the condition

The dictionary comprehension creates keys as uppercase letters and includes only values greater than 10.

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