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Basic numpy exercises

Simple arrays

Create an array with variable name a and the following contents (shape (3, 4)):

2  7 12  0
3  9  3  4
4  0  1  3

>>> #- create array "a" with values
>>> #-     2  7 12  0
>>> #-     3  9  3  4
>>> #-     4  0  1  3
>>> import numpy as np
>>> a = np.array([[2, 7, 12, 0], [3, 9, 3, 4], [4, 0, 1,  3]])
>>> a
array([[ 2,  7, 12,  0],
       [ 3,  9,  3,  4],
       [ 4,  0,  1,  3]])

What is the array shape?

>>> #- Array shape?
>>> a.shape
(3, 4)

What is the array ndim?

>>> #- Array ndim?
>>> a.ndim
2

How about the len of the array?

>>> #- Array length
>>> len(a)
3

Can you get the ndim and len from the shape?

>>> #- Get ndim and length from the shape
>>> len(a.shape) == a.ndim
True
>>> a.shape[0] == len(a)
True

Creating arrays using functions

1. Create a 1D array from 2 through 5 inclusive.

>>> #- 1D array 2 through 5
>>> np.arange(2, 6)
array([2, 3, 4, 5])

2. Make an array with 10 equally spaced elements between 2 and 5 inclusive.

>>> #- 10 equally spaced elementd between 2 and 5
>>> np.linspace(2, 5, 10)
array([ 2.      ,  2.333333,  2.666667,  3.      ,  3.333333,  3.666667,
        4.      ,  4.333333,  4.666667,  5.      ])

3. Make an all-ones array shape (4, 4).

>>> #- Shape 4,4 array of 1
>>> np.ones((4, 4))
array([[ 1.,  1.,  1.,  1.],
       [ 1.,  1.,  1.,  1.],
       [ 1.,  1.,  1.,  1.],
       [ 1.,  1.,  1.,  1.]])

4. Make an identity array shape (6, 6).

>>> #- Identity array shape 6, 6
>>> np.eye(6)
array([[ 1.,  0.,  0.,  0.,  0.,  0.],
       [ 0.,  1.,  0.,  0.,  0.,  0.],
       [ 0.,  0.,  1.,  0.,  0.,  0.],
       [ 0.,  0.,  0.,  1.,  0.,  0.],
       [ 0.,  0.,  0.,  0.,  1.,  0.],
       [ 0.,  0.,  0.,  0.,  0.,  1.]])

5. Make this array with a single Python / numpy command:

   1  0  0
   0  2  0
   0  0  3
   

>>> #- Array with top left value == 1 etc
>>> np.diag([1, 2, 3])
array([[1, 0, 0],
       [0, 2, 0],
       [0, 0, 3]])

Look at the docstring for np.random.randn. Make a shape (3, 5) array with random numbers from a standard normal distribution (a normal distribution with mean 0 and variance 1).

>>> #- Array of random numbers shape 3, 5
>>> rand_arr = np.random.rand(3, 5)
>>> rand_arr.shape
(3, 5)

Simple visualizations

1. Make an array x with 100 evenly spaced values between 0 and 2 * pi;

>>> #- x is an array with 100 evenly spaced numbers 0 - 2 pi
>>> x = np.linspace(0, 2 * np.pi, 100)
>>> x.shape
(100,)

2. Make an array y which contains the cosine of the corresponding value in x - so y[i] = cos(x[i]) (hint: np.lookfor('cosine')).

>>> #- y has cosines of values in x
>>> y = np.cos(x)
>>> y.shape
(100,)

3. Plot x against y;

>>> #- plot x against y
>>> import matplotlib.pyplot as plt
>>> plt.plot(x, y)
[...]

4. Make a 10 by 20 array of mean 0 variance 1 normal random numbers;

>>> #- Shape 10, 20 array of random numbers
>>> rand_arr = np.random.randn(10, 20)
>>> rand_arr.shape
(10, 20)

5. Display this array as an image;

>>> #- Display as image
>>> plt.imshow(rand_arr)
<...matplotlib.image.AxesImage object at ...>

6. Investigate plt.cm. See if you can work out how to make the displayed image be grayscale instead of color.

>>> #- Grayscale image of array
>>> plt.imshow(rand_arr, cmap=plt.cm.gray)
<...matplotlib.image.AxesImage object at ...>

Indexing and slicing, array creation

See discussion at Index ordering and reshape in NumPy and MATLAB.

1. Create the following array, call this a (you did this before):

   2  7 12  0
   3  9  3  4
   4  0  1  3
   

>>> #- Create array "a"
>>> a = np.array([[2, 7, 12, 0], [3, 9, 3, 4], [4, 0, 1,  3]])

2. Get the 2nd row of a ([ 3 9 3 4]);

>>> #- 2nd row of a
>>> a[1]
array([3, 9, 3, 4])

3. Get the 3rd column of a ([12 3 1]);

>>> #- 3rd column of a
>>> a[:, 2]
array([12,  3,  1])

4. Create the following arrays (with correct data types):

   [[1, 1, 1, 1],
    [1, 1, 1, 1],
    [1, 1, 1, 2],
    [1, 6, 1, 1]]
   
   [[0., 0., 0., 0., 0.],
    [2., 0., 0., 0., 0.],
    [0., 3., 0., 0., 0.],
    [0., 0., 4., 0., 0.],
    [0., 0., 0., 5., 0.],
    [0., 0., 0., 0., 6.]]
   

   Par on course: 3 statements for each

   Hint: Individual array elements can be accessed similarly to a list, e.g. a[1] or a[1, 2].

   Hint: Examine the docstring for diag.

>>> #- Build given arrays
>>> arr1 = np.ones((4, 4), dtype=np.int64)  # Would be float by default
>>> arr1[3, 1] = 6
>>> arr1[2, 3] = 2
>>> arr1
array([[1, 1, 1, 1],
       [1, 1, 1, 1],
       [1, 1, 1, 2],
       [1, 6, 1, 1]])

>>> arr2 = np.diag([2., 3., 4, 5, 6], -1)  # Need a float input to diag for float output
>>> arr2[:, :-1]
array([[ 0.,  0.,  0.,  0.,  0.],
       [ 2.,  0.,  0.,  0.,  0.],
       [ 0.,  3.,  0.,  0.,  0.],
       [ 0.,  0.,  4.,  0.,  0.],
       [ 0.,  0.,  0.,  5.,  0.],
       [ 0.,  0.,  0.,  0.,  6.]])

5. Skim through the documentation for np.tile, and use this function to construct the array:

   [[4, 3, 4, 3, 4, 3],
    [2, 1, 2, 1, 2, 1],
    [4, 3, 4, 3, 4, 3],
    [2, 1, 2, 1, 2, 1]]
   

>>> #- Use np.tile to construct array
>>> np.tile([[4, 3], [2, 1]], (2, 3))
array([[4, 3, 4, 3, 4, 3],
       [2, 1, 2, 1, 2, 1],
       [4, 3, 4, 3, 4, 3],
       [2, 1, 2, 1, 2, 1]])

Fancy indexing using boolean arrays

1. Create the following array a (same as before):

   2  7 12  0
   3  9  3  4
   4  0  1  3
   

>>> #- Create array a
>>> a = np.array([[2, 7, 12, 0], [3, 9, 3, 4], [4, 0, 1,  3]])

2. Use > to make a mask that is true where the elements are greater than 5, like this:

   False True  True  False
   False True  False False
   False False False False
   

>>> #- Make mask for values greater than 5
>>> mask = a > 5
>>> mask
array([[False,  True,  True, False],
       [False,  True, False, False],
       [False, False, False, False]], dtype=bool)

3. Return all the elements in a that are greater than 5.

>>> #- Return all values in a that are greater than 5
>>> a[mask]
array([ 7, 12,  9])

4. Set all the elements greater than 5 to be equal to 5, to get this:

   2  5  5  0
   3  5  3  4
   4  0  1  3
   

>>> #- Set all elements greater than 5 to equal 5
>>> a[mask] = 5
>>> a
array([[2, 5, 5, 0],
       [3, 5, 3, 4],
       [4, 0, 1, 3]])

Elementwise operations

Remember our array a:

2  7 12  0
3  9  3  4
4  0  1  3

1. Use array slicing to get a new array composed of the even columns (0, 2) of a. Now get array that contains the odd columns (1, 3) of a. Add these two arrays.

>>> #- Add even and odd columns of a
>>> a = np.array([[2, 7, 12, 0], [3, 9, 3, 4], [4, 0, 1,  3]])
>>> even_columns = a[:, ::2]
>>> odd_columns = a[:, 1::2]
>>> even_columns + odd_columns
array([[ 9, 12],
       [12,  7],
       [ 4,  4]])

2. Generate this array:

   [2**0, 2**1, 2**2, 2**3, 2**4]
   

>>> #- Generate array of powers of 2
>>> 2 ** np.arange(5)
array([ 1,  2,  4,  8, 16])

3. Generate an array length 10 such that this is true of the elements (where x[i] is the element of x at index i):

   x[i] = 2 ** (3 * i) - i
   

>>> #- Generate array
>>> inds = np.arange(10)
>>> x = 2 ** (3 * inds) - inds
>>> x
array([        1,         7,        62,       509,      4092,     32763,
          262138,   2097145,  16777208, 134217719])

Summary functions

Remember our array a:

2  7 12  0
3  9  3  4
4  0  1  3

>>> a = np.array([[2, 7, 12, 0], [3, 9, 3, 4], [4, 0, 1,  3]])

What are the:

• sum of all the values?

>>> #- Sum of values in a
>>> a.sum()
48

• sum of the columns?

>>> #- Sum of the values of the columns in a
>>> a.sum(axis=0)  # Sum over the first axis, leaving the second
array([ 9, 16, 16,  7])

• sum of the rows?

>>> #- Sum of the values of the rows in a
>>> a.sum(axis=1)  # Sum over the second axis, leaving the first
array([21, 19,  8])

• mean?

>>> #- Mean of all the values in a
>>> a.mean()
4.0

• min?

>>> #- Minimum of all the values in a
>>> a.min()
0

• max?

>>> #- Maximum of all the values in a
>>> a.max()
12