Consolidation - `numpy` riddles

Solve the following numpy riddles using numpy and its documentation.

If possible, work in the pair programming paradigm: work in pairs, with one person taking the role of the driver (writing the code) and one taking the role of the navigator (reading and understanding the documentation). Alternate the roles. Try to find solutions that are short (i.e. few line sof code) but easy to understand.

Diagonal Sum: - Riddle: Write a function that takes a square 2D NumPy array as input and returns the sum of the elements along the main diagonal. - Example: diagonal_sum([[1, 2, 3], [4, 5, 6], [7, 8, 9]]) -> 15

Unique Elements Count: - Riddle: Write a function that takes a 1D NumPy array as input and returns the count of unique elements in the array. - Example: unique_count([1, 2, 3, 2, 4, 1, 5]) -> 5

Random sample ands cumulative sum: - Riddle: A fair coin is tossed 20 times, and we win 1£ for every head and lose 1£ for every tail. Assuming that we start with no money at the beginning, and that the seed of teh default random number generator is seed=1234, how much money do we have at every succesive step?

Rolling Window: - Riddle: Write a function that takes a 1D NumPy array and a window size as input, and returns a 2D array where each row is a sliding window of the input array of a given size. - For example, a 1d array with a rolling window of size 3: rolling_window([1, 2, 3, 4, 5], 3) -> [[1, 2, 3], [2, 3, 4], [3, 4, 5]]

Hint: you can use list comprehension and convert the final list to an array.

Product of elements: - Riddle: The geometric mean of a number of observations $x_1, x_2,\dots, x_n$ is defined as $M = \sqrt{x_1\times x_2\times \dots x_n }$. Define a custom function to calculate the geometric mean. - Example: geometric_mean([1, 2, 3, 4, 5]) -> 10.954451150103322

Vectorised calculations and visualisation: - Riddle: Draw 100 thousand points uniformly distributed inside a circle of radius 1 centered at (0,0). Plot them using scatplotter() from matplotlibacoording to their radial coordinate: - use the hexadecimal colour "#76d6ff" for points at a distance below 0.5 from the origin$. - use the hexadecimal colour "ffe701" for points furtehr away.

Hint1: disk point picking is not trivial: https://mathworld.wolfram.com/DiskPointPicking.html

Hint2: For matplotlib’s plot, use the pixel style ',', and remmber to set the axis to be in the same units ("equal")

--- title: Consolidation - `numpy` riddles jupyter: python3 --- Solve the following `numpy` riddles using `numpy` and its documentation. If possible, work in the **pair programming** paradigm: work in pairs, with one person taking the role of the **driver** (writing the code) and one taking the role of the **navigator** (reading and understanding the documentation). Alternate the roles. Try to find solutions that are **short** (i.e. few line sof code) but **easy to understand**. ```{pyodide} #| caption: "▶ Ctrl/Cmd+Enter | ⇥ Ctrl/Cmd+] | ⇤ Ctrl/Cmd+[" import numpy as np ``` **Diagonal Sum**: - Riddle: Write a function that takes a square 2D NumPy array as input and returns the sum of the elements along the main diagonal. - Example: `diagonal_sum([[1, 2, 3], [4, 5, 6], [7, 8, 9]]) -> 15` ```{pyodide} #| caption: "▶ Ctrl/Cmd+Enter | ⇥ Ctrl/Cmd+] | ⇤ Ctrl/Cmd+[" ## YOUR CODE HERE def diagonal_sum(seq): arr = np.asarray(seq) return np.diagonal(arr).sum() diagonal_sum([[1, 2, 3], [4, 5, 6], [7, 8, 9]]) ``` **Unique Elements Count**: - Riddle: Write a function that takes a 1D NumPy array as input and returns the count of unique elements in the array. - Example: `unique_count([1, 2, 3, 2, 4, 1, 5]) -> 5` ```{pyodide} #| caption: "▶ Ctrl/Cmd+Enter | ⇥ Ctrl/Cmd+] | ⇤ Ctrl/Cmd+[" ## YOUR CODE HERE def unique_count(sequence): return np.unique(sequence).size unique_count([1, 2, 3, 2, 4, 1, 5]) ``` **Random sample ands cumulative sum**: - Riddle: A fair coin is tossed 20 times, and we win 1£ for every head and lose 1£ for every tail. Assuming that we start with no money at the beginning, and that the seed of teh default random number generator is `seed=1234`, how much money do we have at every succesive step? ```{pyodide} #| caption: "▶ Ctrl/Cmd+Enter | ⇥ Ctrl/Cmd+] | ⇤ Ctrl/Cmd+[" ## YOUR CODE HERE n_trials = 20 rng = np.random.default_rng(1234) print("The amount we have after every coin tossing is", rng.choice([-1,1],n_trials).cumsum()) ``` **Rolling Window**: - Riddle: Write a function that takes a 1D NumPy array and a window size as input, and returns a 2D array where each row is a sliding window of the input array of a given size. - For example, a 1d array with a rolling window of size 3: `rolling_window([1, 2, 3, 4, 5], 3) -> [[1, 2, 3], [2, 3, 4], [3, 4, 5]]` Hint: you can use **list comprehension** and convert the final list to an array. ```{pyodide} #| caption: "▶ Ctrl/Cmd+Enter | ⇥ Ctrl/Cmd+] | ⇤ Ctrl/Cmd+[" ## YOUR CODE HERE def rolling_window(seq,window): arr = np.asarray(seq) num_rows = len(seq)- window + 1 return np.array([arr[i:i+window] for i in range(num_rows)]) rolling_window([1, 2, 3, 4, 5], 3) ``` **Product of elements**: - Riddle: The geometric mean of a number of observations $x_1, x_2,\dots, x_n$ is defined as $M = \sqrt{x_1\times x_2\times \dots x_n }$. Define a custom function to calculate the geometric mean. - Example: `geometric_mean([1, 2, 3, 4, 5]) -> 10.954451150103322` ```{pyodide} #| caption: "▶ Ctrl/Cmd+Enter | ⇥ Ctrl/Cmd+] | ⇤ Ctrl/Cmd+[" ## YOUR CODE HERE def geometric_mean(seq): return np.sqrt(np.prod(seq)) geometric_mean([1, 2, 3, 4, 5]) ``` **Vectorised calculations and visualisation**: - Riddle: Draw 100 thousand points uniformly distributed inside a circle of radius 1 centered at (0,0). Plot them using `scatplotter()` from `matplotlib`acoording to their radial coordinate: - use the hexadecimal colour `"#76d6ff"` for points at a distance below 0.5 from the origin$. - use the hexadecimal colour `"ffe701"` for points furtehr away. Hint1: disk point picking is not trivial: https://mathworld.wolfram.com/DiskPointPicking.html Hint2: For matplotlib's `plot`, use the pixel style `','`, and remmber to set the axis to be in the same units (`"equal"`) ```{pyodide} #| caption: "▶ Ctrl/Cmd+Enter | ⇥ Ctrl/Cmd+] | ⇤ Ctrl/Cmd+[" ## YOUR CODE HERE npoints = 100_000 theta = np.random.uniform(0, 2*np.pi, npoints) r = np.sqrt(np.random.uniform(0, 1, npoints)) x = r * np.cos(theta) y = r * np.sin(theta) inside = r<0.5 import matplotlib.pyplot as plt plt.plot(x[inside],y[inside],',',color="#76d6ff") plt.plot(x[~inside],y[~inside],',',color="#ffe701") plt.axis("equal") ```