Find indices of elements equal to zero in a NumPy array

Sometimes we need to find out the indices of all null elements in the array. Numpy provides many functions to compute indices of all null elements. 

Method 1: Finding indices of null elements using numpy.where()

This function returns the indices of elements in an input array where the given condition is satisfied.

Syntax : 

numpy.where(condition[, x, y])
When True, yield x, otherwise yield y

Python3

# importing Numpy package
import numpy as np

# creating a 1-D Numpy array
n_array = np.array([1, 0, 2, 0, 3, 0, 0, 5,
                    6, 7, 5, 0, 8])

print("Original array:")
print(n_array)

# finding indices of null elements using np.where()
print("\nIndices of elements equal to zero of the \
given 1-D array:")

res = np.where(n_array == 0)[0]
print(res)

Output:

Time complexity: O(n) – where n is the size of the array
Auxiliary space: O(k) – where k is the number of null elements in the array, as we are storing their indices in a separate array.

Method 2: Finding indices of null elements using numpy.argwhere()

This function is used to find the indices of array elements that are non-zero, grouped by element.

Syntax : 

numpy.argwhere(arr)

Python3

# importing Numpy package
import numpy as np

# creating a 3-D Numpy array
n_array = np.array([[0, 2, 3],
                    [4, 1, 0],
                    [0, 0, 2]])

print("Original array:")
print(n_array)

# finding indices of null elements
# using np.argwhere()
print("\nIndices of null elements:")
res = np.argwhere(n_array == 0)

print(res)

Output:

The time complexity of the code is O(m * n) where m and n are the dimensions of the 3-D Numpy array .

The auxiliary space complexity of the code is O(k) where k is the number of null elements in the 3-D Numpy array .

Method 3: Finding the indices of null elements using numpy.nonzero()

This function is used to Compute the indices of the elements that are non-zero. It returns a tuple of arrays, one for each dimension of arr, containing the indices of the non-zero elements in that dimension.

Syntax:

numpy.nonzero(arr)

Python3

# importing Numpy package
import numpy as np

# creating a 1-D Numpy array
n_array = np.array([1, 10, 2, 0, 3, 9, 0,
                    5, 0, 7, 5, 0, 0])

print("Original array:")
print(n_array)

# finding indices of null elements using
# np.nonzero()
print("\nIndices of null elements:")

res = np.nonzero(n_array == 0)
print(res)

Output:

The time complexity  is O(n), where n is the number of elements in the input array. 

The auxiliary space complexity  is O(k), where k is the number of null elements in the input array. 

Method 4: Using numpy.extract() method

Use the numpy.extract() method. This method returns an array of values that satisfy a certain condition. In this case, we can use it to extract the indices of elements that are equal to zero.

Python3

# importing Numpy package
import numpy as np

# creating a 1-D Numpy array
n_array = np.array([1, 0, 2, 0, 3, 0, 0, 5,
                    6, 7, 5, 0, 8])

print("Original array:")
print(n_array)

# finding indices of null elements using np.extract()
print("\nIndices of elements equal to zero of the \
given 1-D array:")

res = np.extract(n_array == 0, np.arange(len(n_array)))
print(res)

Output:

Original array:
[1 0 2 0 3 0 0 5 6 7 5 0 8]
Indices of elements equal to zero of the given 1-D array:
[ 1  3  5  6 11]

Time complexity: O(n), where n is the length of the input array.
Auxiliary space: O(m), where m is the number of elements in the input array that are equal to zero.