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Problem 5: Split (50pts)

Problem

my_split takes in a list l and a function f and splits the list l into two separate (i.e., a pair of) lists, so that for any element e in the first returned list, f(e) evaluates to True; and for any element e in the second returned list, f(e) evaluates to False.

It is guaranteed that

  1. f is a pure function
  2. f always terminates and
  3. f returns either True or False

Such functions are called predicates, a term from the field of logic.

my_split 接受一个列表 l 和一个函数 f，并将列表 l 分割成两个独立的列表（即一个对）， 使得对于第一个返回列表中的任何元素 e，f(e) 求值为 True； 对于第二个返回列表中的任何元素 e，f(e) 求值为 False。

保证以下条件成立：

  1. f 是一个纯函数。
  2. f 总是终止。
  3. f 返回 True 或 False。

这类函数被称为谓词（predicate），这是一个来自逻辑学领域的术语。

def my_split(f, l):
    """Splits the list into a pair of lists according to predicate f.
    >>> my_split(lambda x: True, [1, 2, 3, 4]) # All elements from l conforms to the predicate
    ([1, 2, 3, 4], [])
    >>> my_split(lambda x: x % 2 == 0, [1, 2, 3, 4]) # Split into even and odd numbers
    ([2, 4], [1, 3])
    >>> my_split(lambda x: x < 5, [3, 1, 4, 1, 5, 9, 2])
    ([3, 1, 4, 1, 2], [5, 9])
    >>> my_split(lambda x: not x, [True, False, True, True]) # There might be things other than integers
    ([False], [True, True, True])
    >>> is_ldx = lambda x: not x.startswith('24')
    >>> studentIDs = ['24122', '22122', '502024', '24183']
    >>> ldx, xdx = my_split(is_ldx, studentIDs)
    >>> ldx
    ['22122', '502024']
    >>> studentIDs # You should not mutate the original list
    ['24122', '22122', '502024', '24183']
    """
    "*** YOUR CODE HERE ***"

Hints

Hint: use list comprehensions.

  • 使用列表推导式（list comprehensions）。

Solutions

def my_split(f, l):
    return ([i for i in l if f(i)], [i for i in l if not f(i)])