#每日一练 根据提供函数，返回同时存在的值

def intersection_by(a, b, fn):
  _b = set(map(fn, b))
  return [item for item in a if fn(item) in _b]

from math import floor

intersection_by([2.1, 1.2], [2.3, 3.4], floor) # [2.1]