## Numpy: Boolean Indexing

import numpy as np A = np.array([4, 7, 3, 4, 2, 8]) print(A == 4)

[ True False False True False False]

Every element of the Array A is tested, if it is equal to 4. The results of these tests are the Boolean elements of the result array.

Of course, it is also possible to check on "<", "<=", ">" and ">=".

print(A < 5)

[ True False True True True False]

It works also for higher dimensions:

B = np.array([[42,56,89,65], [99,88,42,12], [55,42,17,18]]) print(B>=42)

[[ True True True True] [ True True True False] [ True True False False]]

It is a convenient way to threshold images.

import numpy as np A = np.array([ [12, 13, 14, 12, 16, 14, 11, 10, 9], [11, 14, 12, 15, 15, 16, 10, 12, 11], [10, 12, 12, 15, 14, 16, 10, 12, 12], [ 9, 11, 16, 15, 14, 16, 15, 12, 10], [12, 11, 16, 14, 10, 12, 16, 12, 13], [10, 15, 16, 14, 14, 14, 16, 15, 12], [13, 17, 14, 10, 14, 11, 14, 15, 10], [10, 16, 12, 14, 11, 12, 14, 18, 11], [10, 19, 12, 14, 11, 12, 14, 18, 10], [14, 22, 17, 19, 16, 17, 18, 17, 13], [10, 16, 12, 14, 11, 12, 14, 18, 11], [10, 16, 12, 14, 11, 12, 14, 18, 11], [10, 19, 12, 14, 11, 12, 14, 18, 10], [14, 22, 12, 14, 11, 12, 14, 17, 13], [10, 16, 12, 14, 11, 12, 14, 18, 11]]) B = A < 15 B.astype(np.int)The above code returned the following:

array([[1, 1, 1, 1, 0, 1, 1, 1, 1], [1, 1, 1, 0, 0, 0, 1, 1, 1], [1, 1, 1, 0, 1, 0, 1, 1, 1], [1, 1, 0, 0, 1, 0, 0, 1, 1], [1, 1, 0, 1, 1, 1, 0, 1, 1], [1, 0, 0, 1, 1, 1, 0, 0, 1], [1, 0, 1, 1, 1, 1, 1, 0, 1], [1, 0, 1, 1, 1, 1, 1, 0, 1], [1, 0, 1, 1, 1, 1, 1, 0, 1], [1, 0, 0, 0, 0, 0, 0, 0, 1], [1, 0, 1, 1, 1, 1, 1, 0, 1], [1, 0, 1, 1, 1, 1, 1, 0, 1], [1, 0, 1, 1, 1, 1, 1, 0, 1], [1, 0, 1, 1, 1, 1, 1, 0, 1], [1, 0, 1, 1, 1, 1, 1, 0, 1]])

If you have a close look at the previous output, you will see, that it the upper case 'A' is hidden in the array B.

### Fancy Indexing

We will index an array C in the following example by using a Boolean mask. It is called fancy indexing, if arrays are indexed by using boolean or integer arrays (masks). The result will be a copy and not a view.

In our next example, we will use the Boolean mask of one array to select the corresponding elements of another array. The new array R contains all the elements of C where the corresponding value of (A<=5) is True.

C = np.array([123,188,190,99,77,88,100]) A = np.array([4,7,2,8,6,9,5]) R = C[A<=5] print(R)

[123 190 100]

### Indexing with an Integer Array

In the following example, we will index with an integer array:

C[[0, 2, 3, 1, 4, 1]]The above Python code returned the following:

array([123, 190, 99, 188, 77, 188])

Indices can appear in every order and multiple times!

#### Exercises

Extract from the array np.array([3,4,6,10,24,89,45,43,46,99,100]) with Boolean masking all the number

which are not divisible by 3

which are divisible by 5

which are divisible by 3 and 5

which are divisible by 3 and set them to 42

#### Solutions

import numpy as np A = np.array([3,4,6,10,24,89,45,43,46,99,100]) div3 = A[A%3!=0] print("Elements of A not divisible by 3:") print(div3) div5 = A[A%5==0] print("Elements of A divisible by 5:") print(div5) print("Elements of A, which are divisible by 3 and 5:") print(A[(A%3==0) & (A%5==0)]) print("------------------") # A[A%3==0] = 42 print("""New values of A after setting the elements of A, which are divisible by 3, to 42:""") print(A)

Elements of A not divisible by 3: [ 4 10 89 43 46 100] Elements of A divisible by 5: [ 10 45 100] Elements of A, which are divisible by 3 and 5: [45] ------------------ New values of A after setting the elements of A, which are divisible by 3, to 42: [ 42 4 42 10 42 89 42 43 46 42 100]

### nonzero and where

There is an ndarray method called nonzero and a numpy method with this name. The two functions are equivalent.

For an ndarray a both numpy.nonzero(a) and a.nonzero() return the indices of the elements of a that are non-zero. The indices are returned as a tuple of arrays, one for each dimension of 'a'. The corresponding non-zero values can be obtained with:

`a[numpy.nonzero(a)]`

import numpy as np a = np.array([[0, 2, 3, 0, 1], [1, 0, 0, 7, 0], [5, 0, 0, 1, 0]]) print(a.nonzero())

(array([0, 0, 0, 1, 1, 2, 2]), array([1, 2, 4, 0, 3, 0, 3]))

If you want to group the indices by element, you can use transpose:

`transpose(nonzero(a))`

A two-dimensional array is returned. Every row corresponds to a non-zero element.

np.transpose(a.nonzero())The previous Python code returned the following output:

array([[0, 1], [0, 2], [0, 4], [1, 0], [1, 3], [2, 0], [2, 3]])

The corresponding non-zero values can be retrieved with:

a[a.nonzero()]The previous Python code returned the following output:

array([2, 3, 1, 1, 7, 5, 1])

The function 'nonzero' can be used to obtain the indices of an array, where a condition is True. In the following script, we create the Boolean array B >= 42:

B = np.array([[42,56,89,65], [99,88,42,12], [55,42,17,18]]) print(B >= 42)

[[ True True True True] [ True True True False] [ True True False False]]

np.nonzero(B >= 42) yields the indices of the B where the condition is true:

#### Exercise

Calculate the prime numbers between 0 and 100 by using a Boolean array.

Solution:

import numpy as np is_prime = np.ones((100,), dtype=bool) # Cross out 0 and 1 which are not primes: is_prime[:2] = 0 # cross out its higher multiples (sieve of Eratosthenes): nmax = int(np.sqrt(len(is_prime))) for i in range(2, nmax): is_prime[2*i::i] = False print(np.nonzero(is_prime))

(array([ 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97]),)

#### Flatnonzero and count_nonzero

similar functions:

flatnonzero :

`Return indices that are non-zero in the flattened version of the input array.`

count_nonzero :

`Counts the number of non-zero elements in the input array.`