#每日一练 根据过滤内容，把值分成两组

def bifurcate_by(lst, fn):
  return [
    [x for x in lst if fn(x)],
    [x for x in lst if not fn(x)]
  ]
bifurcate_by(['beep', 'boop', 'foo', 'bar'], lambda x: x[0] == 'b')
# [ ['beep', 'boop', 'bar'], ['foo'] ]