Views share memory with originals in NumPy - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of modifying a NumPy view

What is the output of the following code?

import numpy as np
arr = np.array([1, 2, 3, 4])
view = arr[1:3]
view[0] = 10
print(arr)

NumPy

import numpy as np
arr = np.array([1, 2, 3, 4])
view = arr[1:3]
view[0] = 10
print(arr)

A [1 2 3 4]

B [1 10 3 4]

C [10 2 3 4]

D [1 10 10 4]

💡 Hint

Remember that views share memory with the original array.

The slice creates a view, so changing view[0] changes arr[1]. The original array updates at index 1.

❓ data_output

intermediate

Memory sharing check with np.shares_memory

Given the code below, what does the print statement output?

import numpy as np
x = np.arange(5)
y = x[2:]
print(np.shares_memory(x, y))

NumPy

import numpy as np
x = np.arange(5)
y = x[2:]
print(np.shares_memory(x, y))

A False

B None

C True

D Raises an error

💡 Hint

Slices usually share memory with the original array.

The slice y is a view of x, so they share memory. np.shares_memory returns True.

🔧 Debug

advanced

Why does modifying a copy not affect the original?

Consider this code:

import numpy as np
arr = np.array([5, 6, 7, 8])
copied = arr.copy()
copied[0] = 100
print(arr)

Why does the original array not change?

NumPy

import numpy as np
arr = np.array([5, 6, 7, 8])
copied = arr.copy()
copied[0] = 100
print(arr)

A Because modifying copied modifies arr internally

B Because copied is a view sharing memory with arr

C Because arr is immutable

D Because copied is a separate copy with its own memory

💡 Hint

Think about what .copy() does to the data.

The .copy() method creates a new array with its own memory, so changes to copied do not affect arr.

🚀 Application

advanced

Detecting if two arrays share memory

You have two arrays:

import numpy as np
base = np.array([1, 2, 3, 4, 5])
arr1 = base[1:4]
arr2 = base.copy()[1:4]

Which code snippet correctly detects which arrays share memory with base?

NumPy

import numpy as np
base = np.array([1, 2, 3, 4, 5])
arr1 = base[1:4]
arr2 = base.copy()[1:4]

A print(np.shares_memory(base, arr1), np.shares_memory(base, arr2)) # True False

B print(np.shares_memory(base, arr1), np.shares_memory(base, arr2)) # False False

C print(np.shares_memory(base, arr1), np.shares_memory(base, arr2)) # False True

D print(np.shares_memory(base, arr1), np.shares_memory(base, arr2)) # True True

💡 Hint

Remember that slicing creates views but copy() creates independent arrays.

arr1 is a view and shares memory with base, arr2 is a slice of a copy and does not share memory.

🧠 Conceptual

expert

Effect of modifying a view on a reshaped array

Given the code below, what is the output?

import numpy as np
original = np.arange(6)
reshaped = original.reshape(2, 3)
view = reshaped[:, 1]
view[0] = 99
print(original)

NumPy

import numpy as np
original = np.arange(6)
reshaped = original.reshape(2, 3)
view = reshaped[:, 1]
view[0] = 99
print(original)

A [0 99 2 3 4 5]

B [0 1 2 3 4 5]

C [0 1 99 3 4 5]

D [99 1 2 3 4 5]

💡 Hint

Think about how reshaping and slicing create views sharing memory.

The view selects the second column of reshaped, which corresponds to original indices 1 and 4. Changing view[0] changes original[1].