Workout 7. Functional Programming with Comprehensions

Question 1

Produce a list of squared numbers for each number in an iterable range(5).

Answer 1

[x*x for x in range(5)]

Question 2

Produce a dictionary with keys from an iterable, with key-value pairs of the form (k,2k).

Answer 2

{x : 2*x for x in range(5)}

Question 3

Produce a set of squared numbers for each number in an iterable range(5).

Answer 3

{x*x for x in range(5)}

Question 4

For loop vs List Comprehension:
If your goal is to execute something once for each element in an iterable, throwing away or ignoring any return value, then a _ is the preferred way.
However, if you are interested in transformed output of each element in an iterable, then a _ is the preferred way.

Answer 4

1. for loop
2. list comprehension

Question 5

Write a function (join_numbers) that takes a range of integers.
The function should return those numbers as a string, with commas between the
numbers.
That is, given range(15) as input, the function should return this string:
0,1,2,3,4,5,6,7,8,9,10,11,12,13,14

Question 6

nested list comprehensions:
return list of all visited cities, given the dictionary of (country, [cities]) pairs:

all_places_dict = {'USA': ['Philadelphia', 'New York', 'Cleveland', 'San Jose',
'San Francisco'],
'China': ['Beijing', 'Shanghai', 'Guangzhou'],
'UK': ['London'],
'India': ['Hyderabad']}

Answer 6

def visited_cities(all_places_dict):
  return [city for (_,cities) in all_places_dict.items() for city in cities]

Question 7

Which of the following runs faster?
1. sum([x*x for x in range(100000)])
2. sum((x*x for x in range(100000)))

Answer 7

Second: since it uses generator expression

The generator returns one element at a time, waiting for sum to request the next item in line.
In this way, we’re only consuming one integer’s worth of memory at a time, rather than a huge list of integers’ memory.

Question 8

produce a list of tuples: (x,y) from range(numrange) where x is odd and y is even.
1. with list comprehension
2. with for loop

print(odd_even_tuples(5))

Answer 8

with list comprehension:

def odd_even_tuples(numrange):
  return [(x,y) for x in range(1, numrange) if x%2 for y in range(1, numrange) if not y%2]

with for loop:

def odd_even_tuples_via_loop(numrange):
  output = []
  for x in range(1, numrange):
    if x%2:
      for y in range(1, numrange):
        if not y%2:
          output.append((x,y))
  return output