1. Dictionaries. Use a dict comprehension to invert a dictionary. That is, if in the original dict we have a key-value pair k: v, we should now have v: k.
2. Is it possible to have multiple entries in a dict with the same key k? What is the effect of your invert code if there are multiple entries in the original dict with the same value v? Think about it, then try it. 3. Higher-order functions: we want to create a list containing ex ∀x ∈ [0.0,0.1,...,1.0]. Use range, lambda, map and of course math.exp to do this.
4. Exceptions. In the following code, check that the user does not request too large a value of n. If they do, raise ValueError with an informative message such as ValueError: Can't return 7 elements from abcde of length 5. Hint: you could use an f-string to create that string.
def get_last_n_elements(s, n):
return s[-n:]
5. Itertools: a magic square is an n × n grid containing the numbers 1,2,...n2 (used exactly once each) such that each row and column sums to the same value. Here is a 3 × 3 magic square:
(9, 5, 1) (4, 3, 8) (2, 7, 6)
We will generate all magic squares for n = 3. Look up itertools.permutations and use it to generate all permutations of the numbers 1,2,...9. Next, for each permutation p, think of it as a grid, like this:
(p[0], p[1], p[2]) (p[3], p[4], p[5]) (p[6], p[7], p[8])
Check whether the rows and columns sum to the right value, and if so, print it out.
Hint: in Python, you can chain multiple comparisons together, e.g. x == y == z.
1
6. Generators: create a generator (a “function” that uses yield) that yields the squares of all the non-negative integers, starting at 0.
7. Test it by running for s in sq(): print(s). Of course it creates an infinite loop. To exit the loop, we have to interrupt Python:
• In Spyder, type Ctrl-C in the console
• In Terminal or IPython, type Ctrl-C
• In Jupyter Notebook, go to the Kernel menu and select interrupt.
8. Without altering your generator, use it to create a for-loop that prints out all the even squares ≤ 100. This time, your for-loop could use break to avoid the infinite loop.
2
