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Given the following text description, write Python code to implement the functionality described below step by step
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Description:
 numpy Testing
Miki Tebeka .
Step1:  The Naive Approach
Step2:  This is due to the fact that when we compare two numpy arrays with == we'll get an array of boolean values comparing each element.
Step3:  And the truch value of an array (as the error says) is ambiguous.
Step4:  We need to use np.all to check that all elements are equal.
Step5:  Using np.all
Step6:  This is due to the fact that floating points are not exact.
Step7:  This is not a bug in Python but how floating points are implemented. You'll get the same result in C, Java, Go ...
To overcome this we're going to use np.allclose.
BTW
Step8:  Oh nan, Let Me Count the Ways ...
Step9:  This is due to the fact the nan does not equal itself.
Step10:  To check is a number is nan we need to use np.isnan
Step11:  We have two options to solve this
Step12:  Option 2
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Python Code:
import numpy as np
Explanation: numpy Testing
Miki Tebeka .:. 353solutions .:. Highly effective Python, Scientific Python and Go workshops
We'll explore certain caveats while testing numpy code.
TL;DR
Use np.allclose when comparing numpy arrays. Beware of nan.
End of explanation
def test_mul():
    arr = np.array([0.0, 1.0, 1.1])
    v, expected = 1.1, np.array([0.0, 1.1, 1.21])
    assert arr * v == expected, 'bad multiplication'
    
test_mul()
Explanation: The Naive Approach
End of explanation
np.array([1,2,3]) == np.array([1, 1, 3])
Explanation: This is due to the fact that when we compare two numpy arrays with == we'll get an array of boolean values comparing each element.
End of explanation
bool(np.array([1, 2, 3]))
Explanation: And the truch value of an array (as the error says) is ambiguous.
End of explanation
np.all([True, True, True])
Explanation: We need to use np.all to check that all elements are equal.
End of explanation
def test_mul():
    arr = np.array([0.0, 1.0, 1.1])
    v, expected = 1.1, np.array([0.0, 1.1, 1.21])
    assert np.all(arr * v == expected), 'bad multiplication'
    
test_mul()
Explanation: Using np.all
End of explanation
1.1 * 1.1
Explanation: This is due to the fact that floating points are not exact.
End of explanation
def test_mul():
    arr = np.array([0.0, 1.0, 1.1])
    v, expected = 1.1, np.array([0.0, 1.1, 1.21])
    assert np.allclose(arr * v, expected), 'bad multiplication'
    
test_mul()
Explanation: This is not a bug in Python but how floating points are implemented. You'll get the same result in C, Java, Go ...
To overcome this we're going to use np.allclose.
BTW: If you're really intersted in floating points, read this article.
Using np.allclose
End of explanation
def test_div():
    arr1, arr2 = np.array([1.0, np.inf, 2.0]), np.array([2.0, np.inf, 2.0])
    expected = np.array([0.5, np.nan, 1.0])
    assert np.allclose(arr1 / arr2, expected), 'bad nan'
    
test_div()
Explanation: Oh nan, Let Me Count the Ways ...
End of explanation
np.nan == np.nan
Explanation: This is due to the fact the nan does not equal itself.
End of explanation
np.isnan(np.inf/np.inf)
Explanation: To check is a number is nan we need to use np.isnan
End of explanation
def test_div():
    arr1, arr2 = np.array([1.0, np.inf, 2.0]), np.array([2.0, np.inf, 2.0])
    expected = np.array([0.5, np.nan, 1.0])
    result = arr1 / arr2
    
    result[np.isnan(result)] = 0.0
    expected[np.isnan(expected)] = 0.0
    assert np.allclose(result, expected), 'bad nan'
    
test_div()
Explanation: We have two options to solve this:
Convert all nan to numbers
Use equal_nan argument to np.allclose
Option 1: Convert nan to Numbers
End of explanation
def test_div():
    arr1, arr2 = np.array([1.0, np.inf, 2.0]), np.array([2.0, np.inf, 2.0])
    expected = np.array([0.5, np.nan, 1.0])
    assert np.allclose(arr1 / arr2, expected, equal_nan=True), 'bad nan'
    
test_div()
Explanation: Option 2: Use equal_nan in np.allclose
End of explanation
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