使用不同的参数调用relativelta时，创建通用代码并避免冗余

Question

有什么方法可以总结以下代码吗？每个if-else条件后面的代码非常相似。唯一的区别是传递给relativedelta的参数。

def change_datetime_objects(dt_objs=[], resolution="year", frequency=1):
new_dt_list = []
if resolution="year":
for time_obj in dt_objs:
new_dt_list.append(time_obj + relativedelta(years=frequency))
elif resolution="month":
for time_obj in dt_objs:
new_dt_list.append(time_obj + relativedelta(months=frequency))
elif resolution="week":
for time_obj in dt_objs:
new_dt_list.append(time_obj + relativedelta(weeks=frequency))
elif resolution="day":
for time_obj in dt_objs:
new_dt_list.append(time_obj + relativedelta(days=frequency))
elif resolution="hour":
for time_obj in dt_objs:
new_dt_list.append(time_obj + relativedelta(hours=frequency))
elif resolution="minute":
for time_obj in dt_objs:
new_dt_list.append(time_obj + relativedelta(minutes=frequency))
elif resolution="second":
for time_obj in dt_objs:
new_dt_list.append(time_obj + relativedelta(seconds=frequency))

附言：我想也许eval()、exec()甚至装饰师可以帮助创建new_dt_list.append(time_obj + relativedelta(<>=frequency))的通用形式，但不知道他们能提供什么帮助。

Answer 1

我在这篇文章的启发下找到了问题的答案。

因此，基本上为了避免所有冗余，参数应该在之前作为字典生成，并使用**作为关键字参数传递给函数。

def change_datetime_objects(dt_objs=[], resolution="year", frequency=1):
args = {resolution: frequency}
new_dt_list = []
for time_obj in dt_objs:
new_dt_list.append(time_obj + relativedelta(**args))