np.in1d() for membership testing in NumPy - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to check which elements of array1 are in array2 using np.in1d().

NumPy

import numpy as np
array1 = np.array([1, 2, 3, 4])
array2 = np.array([3, 4, 5, 6])
result = np.in1d(array1, [1])
print(result)

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Aarray2

Barray1

C[3, 4]

Dnp.array([1, 2])

Attempts:

3 left

💡 Hint

Common Mistakes

Using the same array for both arguments.

Passing a list instead of a numpy array.

The second argument of np.in1d() is the array to check membership against. Here, array2 is correct.

2fill in blank

medium

Complete the code to get a boolean array showing membership of elements in arr1 within arr2.

NumPy

import numpy as np
arr1 = np.array(['apple', 'banana', 'cherry'])
arr2 = np.array(['banana', 'date', 'fig'])
membership = np.in1d(arr1, [1])
print(membership)

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Aarr2

B['banana', 'date']

Carr1

D['apple', 'fig']

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping the arrays in the function call.

Passing a list instead of a numpy array.

The second argument is the array to check membership against. Here, arr2 is correct.

3fill in blank

hard

Fix the error in the code to correctly test membership using np.in1d().

NumPy

import numpy as np
x = np.array([10, 20, 30])
y = np.array([20, 40, 60])
result = np.in1d([1], y)
print(result)

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Ay

Bnp.array([20, 40])

C[10, 20]

Dx

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping the order of arrays in np.in1d.

Passing a list instead of a numpy array.

The first argument should be the array whose elements you want to test for membership in the second array. Here, x is correct.

4fill in blank

hard

Fill both blanks to create a dictionary showing elements of words and whether their length is greater than 5 and if they are in the list check_list using np.in1d().

NumPy

words = ['apple', 'banana', 'cherry', 'date']
check_list = ['banana', 'date', 'fig']
result = {word: (len(word) [1] 5 and np.in1d([word], [2])[0]) for word in words}
print(result)

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A>

B<

Ccheck_list

Dwords

Attempts:

3 left

💡 Hint

Common Mistakes

Using the wrong comparison operator.

Checking membership against the wrong list.

The condition checks if the length of the word is greater than 5 and if the word is in check_list. So, > and check_list are correct.

5fill in blank

hard

Fill all three blanks to create a dictionary comprehension that maps uppercase words to their membership boolean in check_words if their length is less than 6.

NumPy

words = ['apple', 'banana', 'cherry', 'date']
check_words = ['APPLE', 'DATE', 'FIG']
result = {word[1]: np.in1d([word[2]], [3])[0] for word in words if len(word) < 6}
print(result)

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A.upper()

B.lower()

Ccheck_words

Dwords

Attempts:

3 left

💡 Hint

Common Mistakes

Using .lower() instead of .upper().

Checking membership against the wrong list.

The keys are uppercase words, so .upper() is used twice. Membership is checked against check_words.