1.3.5. Some exercises¶

1.3.5.1. Array manipulations¶

Form the 2-D array (without typing it in explicitly):
```
[[1,  6, 11],
 [2,  7, 12],
 [3,  8, 13],
 [4,  9, 14],
 [5, 10, 15]]
```
and generate a new array containing its 2nd and 4th rows.
Divide each column of the array:
```
>>> import numpy as np
>>> a = np.arange(25).reshape(5, 5)
```
elementwise with the array b = np.array([1., 5, 10, 15, 20]). (Hint: np.newaxis).
Harder one: Generate a 10 x 3 array of random numbers (in range [0,1]). For each row, pick the number closest to 0.5.
- Use abs and argsort to find the column j closest for each row.
- Use fancy indexing to extract the numbers. (Hint: a[i,j] – the array i must contain the row numbers corresponding to stuff in j.)

1.3.5.2. Picture manipulation: Framing a Face¶

Let’s do some manipulations on numpy arrays by starting with an image of a racoon. scipy provides a 2D array of this image with the scipy.misc.face function:

>>> fromscipyimportmisc
>>> face=misc.face(gray=True)# 2D grayscale image

Here are a few images we will be able to obtain with our manipulations: use different colormaps, crop the image, change some parts of the image.

Let’s use the imshow function of pylab to display the image.

>>> importpylabasplt
>>> face=misc.face(gray=True)
>>> plt.imshow(face)
<matplotlib.image.AxesImage object at 0x...>

The face is displayed in false colors. A colormap must be
specified for it to be displayed in grey.
>>> plt.imshow(face, cmap=plt.cm.gray)
<matplotlib.image.AxesImage object at 0x...>
Create an array of the image with a narrower centering : for example,
remove 100 pixels from all the borders of the image. To check the result, display this new array with imshow.
>>> crop_face = face[100:-100, 100:-100]

We will now frame the face with a black locket. For this, we

need to create a mask corresponding to the pixels we want to be black. The center of the face is around (660, 330), so we defined the mask by this condition (y-300)**2 + (x-660)**2

>>> sy,sx=face.shape
>>> y,x=np.ogrid[0:sy,0:sx]# x and y indices of pixels
>>> y.shape,x.shape
((768, 1), (1, 1024))
>>> centerx,centery=(660,300)# center of the image
>>> mask=((y-centery)**2+(x-centerx)**2)>230**2# circle

then we assign the value 0 to the pixels of the image corresponding to the mask. The syntax is extremely simple and intuitive:

>>> face[mask]=0
>>> plt.imshow(face)
<matplotlib.image.AxesImage object at 0x...>

Follow-up: copy all instructions of this exercise in a script called

face_locket.py then execute this script in IPython with %run face_locket.py.

Change the circle to an ellipsoid.

1.3.5.3. Data statistics¶

The data in populations.txt describes the populations of hares and lynxes (and carrots) in northern Canada during 20 years:

>>> data=np.loadtxt('data/populations.txt')
>>> year,hares,lynxes,carrots=data.T# trick: columns to variables
>>> importmatplotlib.pyplotasplt
>>> plt.axes([0.2,0.1,0.5,0.8])
<matplotlib.axes...Axes object at ...>
>>> plt.plot(year,hares,year,lynxes,year,carrots)
[<matplotlib.lines.Line2D object at ...>, ...]
>>> plt.legend(('Hare','Lynx','Carrot'),loc=(1.05,0.5))
<matplotlib.legend.Legend object at ...>

[source code, hires.png, pdf]

Computes and print, based on the data in populations.txt...

The mean and std of the populations of each species for the years in the period.
Which year each species had the largest population.
Which species has the largest population for each year. (Hint: argsort & fancy indexing of np.array(['H', 'L', 'C']))
Which years any of the populations is above 50000. (Hint: comparisons and np.any)
The top 2 years for each species when they had the lowest populations. (Hint: argsort, fancy indexing)
Compare (plot) the change in hare population (see help(np.gradient)) and the number of lynxes. Check correlation (see help(np.corrcoef)).

... all without for-loops.

Solution: Python source file

1.3.5.4. Crude integral approximations¶

Write a function f(a, b, c) that returns $a^b - c$ . Form a 24x12x6 array containing its values in parameter ranges [0,1] x [0,1] x [0,1].

Approximate the 3-d integral

$\int_0^1\int_0^1\int_0^1(a^b-c)da\,db\,dc$

over this volume with the mean. The exact result is: $\ln 2 - \frac{1}{2}\approx0.1931\ldots$ — what is your relative error?

(Hints: use elementwise operations and broadcasting. You can make np.ogrid give a number of points in given range with np.ogrid[0:1:20j].)

Reminder Python functions:

deff(a,b,c):
returnsome_result

Solution: Python source file

1.3.5.5. Mandelbrot set¶

[source code, hires.png, pdf]

Write a script that computes the Mandelbrot fractal. The Mandelbrot iteration:

N_max=50
some_threshold=50
c=x+1j*y
forjinxrange(N_max):
z=z**2+c

Point (x, y) belongs to the Mandelbrot set if $|c|$ < some_threshold.

Do this computation by:

Construct a grid of c = x + 1j*y values in range [-2, 1] x [-1.5, 1.5]
Do the iteration
Form the 2-d boolean mask indicating which points are in the set
Save the result to an image with:

>>> importmatplotlib.pyplotasplt
>>> plt.imshow(mask.T,extent=[-2,1,-1.5,1.5])
<matplotlib.image.AxesImage object at ...>
>>> plt.gray()
>>> plt.savefig('mandelbrot.png')

Solution: Python source file

1.3.5.6. Markov chain¶

Markov chain transition matrix P, and probability distribution on the states p:

0 <= P[i,j] <= 1: probability to go from state i to state j
Transition rule: $p_{new} = P^T p_{old}$
all(sum(P, axis=1) == 1), p.sum() == 1: normalization

Write a script that works with 5 states, and:

Constructs a random matrix, and normalizes each row so that it is a transition matrix.
Starts from a random (normalized) probability distribution p and takes 50 steps => p_50
Computes the stationary distribution: the eigenvector of P.T with eigenvalue 1 (numerically: closest to 1) => p_stationary

Remember to normalize the eigenvector — I didn’t...

Checks if p_50 and p_stationary are equal to tolerance 1e-5

Toolbox: np.random.rand, .dot(), np.linalg.eig, reductions, abs(), argmin, comparisons, all, np.linalg.norm, etc.

Solution: Python source file