#!/usr/bin/env python
# coding: utf-8

# # numpy Testing
# ### Miki Tebeka .:. [353solutions](http://353solutions.com) .:. Highly effective Python, Scientific Python and Go workshops
# 
# We'll explore certain caveats while testing [numpy](http://docs.scipy.org/doc/numpy/reference/) code.
# 
# #### TL;DR
# Use [np.allclose](http://docs.scipy.org/doc/numpy/reference/generated/numpy.allclose.html) when comparing numpy arrays. Beware of `nan`.

# In[1]:


import numpy as np


# ## The Naive Approach

# In[2]:


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()


# 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.

# In[3]:


np.array([1,2,3]) == np.array([1, 1, 3])


# And the truch value of an array (as the error says) is ambiguous.

# In[4]:


bool(np.array([1, 2, 3]))


# We need to use [np.all](http://docs.scipy.org/doc/numpy/reference/generated/numpy.all.html) to check that all elements are equal.

# In[5]:


np.all([True, True, True])


# ## Using np.all

# In[6]:


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()


# This is due to the fact that floating points are not exact.

# In[7]:


1.1 * 1.1


# 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](http://docs.scipy.org/doc/numpy/reference/generated/numpy.allclose.html).
# 
# BTW: If you're really intersted in floating points, read [this article](http://docs.oracle.com/cd/E19957-01/806-3568/ncg_goldberg.html).

# ## Using np.allclose

# In[8]:


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()


# ## Oh nan, Let Me Count the Ways ...

# In[9]:


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()


# This is due to the fact the `nan` does not equal itself.

# In[10]:


np.nan == np.nan


# To check is a number is `nan` we need to use [np.isnan](http://docs.scipy.org/doc/numpy/reference/generated/numpy.isnan.html)

# In[11]:


np.isnan(np.inf/np.inf)


# We have two options to solve this:
# 
# 1. Convert all `nan` to numbers
# 2. Use `equal_nan` argument to `np.allclose`

# ## Option 1: Convert `nan` to Numbers

# In[12]:


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()


# ## Option 2: Use `equal_nan` in `np.allclose`

# In[13]:


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()


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