itertools

Question 1

count()

Answer 1

an infinite interator

counter = itertools.count()

for num in counter:

if num > 10:

break

print(num)

Question 2

To use the itertools methods what is required?

Answer 2

import itertools

Question 3

What does this return, if the preceding counter was 12?

print(next(counter)

Answer 3

13 or the next value in the counter.

Question 4

What does this return?

data = [100, 200, 300, 400]

daily_data = list(zip(itertools.count(), data))

daily_data

Answer 4

[(0, 100), (1, 200), (2, 300), (3, 400)]

Question 5

What does this return?

data = [100, 200, 300, 400]

daily_data = list(enumerate(data))

daily_data

Answer 5

the same result as itertools.count()

[(0, 100), (1, 200), (2, 300), (3, 400)]

Question 6

Provide the values returned If given

counter = itertools.count(start=5, step=5)

print(next(counter))

print(next(counter))

print(next(counter))

print(next(counter))

Answer 6

5

10

15

20

Question 7

Why won’t this zip method return 10 values?

data = [100, 200, 300, 400]

daily_data = list(zip(range(10), data))

daily_data

[(0, 100), (1, 200), (2, 300), (3, 400)]

Answer 7

Because the zip() is exhausted by the data, only four values being inputed.

Question 8

What would be the outcome of this if itertool.zip_longest() was utilized instead of zip()?

data = [100, 200, 300, 400]

daily_data = list(zip(range(10), data))

daily_data

[(0, 100), (1, 200), (2, 300), (3, 400)]

Answer 8

Full range would be outputed with None values for those that do not have a subsequent match

[(0, 100), (1, 200), (2, 300), (3, 400), (4, None), (5, None), (6, None), (7, None), (8, None), (9, None)]

Question 9

cycle()

Answer 9

An Infinite iterator

counter = itertools.cycle([1,2,3])

print(next(counter)) #1

print(next(counter)) #2

print(next(counter)) #3

print(next(counter)) #1

print(next(counter)) #2

Question 10

What is this cycle() mimicing?

counter = itertools.cycle((‘On’, ‘Off’))

Answer 10

an on off toggle switch, value is either on or off.

Question 11

repeat()

Answer 11

Infinite iterator

counter = itertools.repeat(2, times= 3)

print(next(counter)) #2

print(next(counter)) #2

print(next(counter)) #2

print(next(counter)) Traceback (most recent call last) StopIteration

Question 12

What is the output?

squares = map(pow, range(10), itertools.repeat(2))

print(list(squares))

Answer 12

0-9 squares

[0, 1, 4, 9, 16, 25, 36, 49, 64, 81]

Question 13

What is the ouput?

squares = itertools.starmap(pow, [(2,5), (3,2), (10,3)])

Answer 13

2 to the 5th = 32

3 squared = 9

10 cubed = 1000

Question 14

combinations()

Answer 14

Combinatoric iterator

order does not matter

letters = [‘a’, ‘b’, ‘c’, ‘d’]

result = itertools.combinations(letters, 2)

for item in result:

print(item)

Question 15

permutations()

Answer 15

Combinatoric iterator

good for racing events where order matters

letters = [‘a’, ‘b’, ‘c’, ‘d’]

for item in itertools.permutations(letters, 2):

print(item)

Question 16

product()

Answer 16

Combinatoric iterator

equivalent to a nested for-loop, similiar to combination_with_replacement. good for password cracking type app

for item in itertools.product(numbers, repeat=4):

print(item)

Question 17

What does this return?

numbers = [0, 1, 2, 3]

for item in itertools.combinations_with_replacement(numbers, 4):

print(item)

Answer 17

(0, 0, 0, 0) (0, 0, 0, 1) (0, 0, 0, 2) (0, 0, 0, 3) (0, 0, 1, 1) (0, 0, 1, 2) (0, 0, 1, 3) (0, 0, 2, 2) (0, 0, 2, 3) (0, 0, 3, 3) (0, 1, 1, 1) (0, 1, 1, 2) (0, 1, 1, 3) (0, 1, 2, 2)

Question 18

chain()

Answer 18

Iterator terminating on the shortest input sequence

combines lists and then iterates over them in order

for items in itertools.chain(letters, numbers, names):

print(items)

Question 19

islice()

Answer 19

Iterator terminating on the shortest input sequence

elements from seq[start:stop:step]

range of 10 cut by half

for item in itertools.islice(range(10), 5):

print(item)

0
1
2
3
4

Question 20

compress()

Answer 20

Iterator terminating on the shortest input sequence

data, selectors

selectors = [True, True, False, True]
letters = ['a', 'b', 'c', 'd']

for item in itertools.compress(letters, selectors):

print(item)

a
b
c

Question 21

filterfalse()

Answer 21

Iterator terminating on the shortest input sequence

this values are NOT less than 2
# returns values that are false in the list 
# that was passed (numbers)

for item in itertools.filterfalse(lt_2, numbers):

print(item)

Question 22

dropwhile()

Answer 22

Iterator terminating on the shortest input sequence

any values less than 2 will be dropped
# once a True conditon is met it will return 
# rest of the values in the list True or not.

numbers = [0,1,2,3,2,1,0]

for item in itertools.dropwhile(lt_2, numbers):

print(item)

Question 23

takewhile()

Answer 23

Iterator terminating on the shortest input sequence

similiar to dropwhile()
# first two values are True 
# breaks from the iterable once a False value is met

for item in itertools.takewhile(lt_2, numbers):

print(item)

Question 24

accumulate()

Answer 24

Iterator terminating on the shortest input sequence

running total of the values in the list

print(numbers)

for item in itertools.accumulate(numbers):

print(item)

Question 25

accumulate()

Answer 25

Iterator terminating on the shortest input sequence

import operator

running product mul_tiplication

num = [1,2,3,2,1,0]

for item in itertools.accumulate(num, operator.mul):

print(item)

Question 26

tee()

Answer 26

Iterator terminating on the shortest input sequence

copies an iterator
# don't use original once copies are made!

copy1, copy2 = itertools.tee(person_group)

Question 27

groupby()

Answer 27

Iterator terminating on the shortest input sequence

people, state in dict object

for key, group in itertools.groupby(people, get_state):

print(key)

for person in group:

print(person)

print()