6.6. Filter

• Select elements from sequence

• Generator (lazy evaluated)

• Built-in

Syntax:

• filter(callable, *iterables)

• required callable - Function

• required iterables - 1 or many sequence or iterator objects

>>> def even(x):
...     return x % 2 == 0
>>>
>>> result = (x for x in range(0,5) if even(x))
>>> result = filter(even, range(0,5))

>>> result = (x for x in range(0,5) if x%2==0)
>>> result = filter(lambda x: x%2==0, range(0,5))

6.6.1. Problem

Plain code:

>>> def even(x):
...     return x % 2 == 0
>>>
>>>
>>> DATA = [1, 2, 3, 4, 5, 6]
>>> result = []
>>>
>>> for x in DATA:
...     if even(x):
...         result.append(x)
>>>
>>> print(result)
[2, 4, 6]

Comprehension:

>>> def even(x):
...     return x % 2 == 0
>>>
>>>
>>> DATA = [1, 2, 3, 4, 5, 6]
>>> result = [x for x in DATA if even(x)]
>>>
>>> print(result)
[2, 4, 6]

6.6.2. Solution

>>> def even(x):
...     return x % 2 == 0
>>>
>>>
>>> DATA = [1, 2, 3, 4, 5, 6]
>>> result = filter(even, DATA)
>>>
>>> list(result)
[2, 4, 6]

6.6.3. Lazy Evaluation

>>> def even(x):
...     return x % 2 == 0
>>>
>>>
>>> DATA = [1, 2, 3, 4, 5, 6]
>>> result = filter(even, DATA)
>>>
>>> next(result)
2
>>> next(result)
4
>>> next(result)
6
>>> next(result)
Traceback (most recent call last):
StopIteration

6.6.4. Rationale

>>> people = [
...     {'age': 21, 'name': 'Pan Twardowski'},
...     {'age': 25, 'name': 'Mark Watney'},
...     {'age': 18, 'name': 'Melissa Lewis'}]
>>>
>>>
>>> def adult(person):
...     return person['age'] >= 21
>>>
>>>
>>> result = filter(adult, people)
>>> list(result)  
[{'age': 21, 'name': 'Pan Twardowski'},
 {'age': 25, 'name': 'Mark Watney'}]

>>> people = [
...     {'is_astronaut': False, 'name': 'Pan Twardowski'},
...     {'is_astronaut': True, 'name': 'Mark Watney'},
...     {'is_astronaut': True, 'name': 'Melissa Lewis'}]
>>>
>>>
>>> def astronaut(person):
...     return person['is_astronaut']
>>>
>>>
>>> result = filter(astronaut, people)
>>> list(result)  
[{'is_astronaut': True, 'name': 'Mark Watney'},
 {'is_astronaut': True, 'name': 'Melissa Lewis'}]

>>> astronauts = ['Mark Watney', 'Melissa Lewis']
>>> people = ['Mark Watney', 'Melissa Lewis', 'Jimenez']
>>>
>>>
>>> def is_astronaut(person):
...     return person in astronauts
>>>
>>>
>>> result = filter(is_astronaut, people)
>>> list(result)
['Mark Watney', 'Melissa Lewis']

6.6.5. Performance

>>> # %%timeit -r 10 -n 100_000
>>> # result = (x for x in range(0,5) if x%2==0)
>>> # 490 ns ± 44 ns per loop (mean ± std. dev. of 10 runs, 100000 loops each)

>>> # %%timeit -r 10 -n 100_000
>>> # result = filter(lambda x: x%2==0, range(0,5))
>>> # 384 ns ± 34.2 ns per loop (mean ± std. dev. of 10 runs, 100000 loops each)

6.6.6. Assignments

Code 6.48. Solution

"""
* Assignment: Idioms Filter Chain
* Complexity: easy
* Lines of code: 5 lines
* Time: 8 min

English:
    1. Use generator expression to create `result`
    2. Use `range()` to get numbers:
       a. from 0 (inclusive)
       b. to 10 (exclusive)
    3. Use `filter()` to get odd numbers from `result`
       (and assign to `result`)
    4. Use `map()` to cube all numbers in `result`
    5. Create `result: float` with arithmetic mean of `result`
    6. Do not use `lambda` expressions
    7. Note, that all the time you are working on one data stream
    8. Run doctests - all must succeed

Polish:
    1. Użyj wyrażenia generatorowego do stworzenia `result`
    2. Użyj `range()` aby otrzymać liczby:
       a. od 0 (włącznie)
       b. do 10 (rozłącznie)
    3. Użyj `filter()` aby otrzymać liczby nieparzyste z `result`
       (i przypisz je do `result`)
    4. Użyj `map()` aby podnieść wszystkie liczby w `result` do sześcianu
    5. Stwórz `result: float` ze średnią arytmetyczną z `result`
    6. Nie używaj wyrażeń lambda
    7. Zwróć uwagę, że cały czas pracujesz na jednym strumieniu danych
    8. Uruchom doctesty - wszystkie muszą się powieść

Hints:
    * type cast to `list()` to expand generator before calculating mean
    * `mean = sum(...) / len(...)`
    * TypeError: object of type 'map' has no len()
    * ZeroDivisionError: division by zero

Tests:
    >>> import sys; sys.tracebacklimit = 0
    >>> from inspect import isfunction

    >>> isfunction(odd)
    True
    >>> isfunction(cube)
    True
    >>> type(result) is float
    True
    >>> result
    245.0
"""

def odd(x):
    return x % 2


def cube(x):
    return x ** 3


# Range numbers from 1 to 10 (exclusive)
# Filter odd numbers
# Cube result
# Calculate mean
# type: float
result = ...