Question

根据某些限制，我在某些地方会拜访每个地点。例如，我在every day和morning上学校noon，而我不在weekends上学校。还有一个，我morning的时间去过购物中心，所以我今天和以后几天可能都不会再去那里（我不在购物中心工作）。

我的问题是，是否有一种从数据集中选择正确位置的好方法，如：

location         | types
MIT              | 'school'
western union    | 'accounting', 'finance'
Walmart          | 'store', 'home_goods_store'

我知道每个地方的限制，取决于：

the time of the day is it a working day how long I've been there the frequency of visiting have I been there today? this week? this month?

每个位置的约束值都有不同的组合。

现在，我为每个约束使用了很多字典，但对于这么多if statements

来说，这似乎很糟糕

有更好的方法吗？使用Python

编辑：我的程序需要根据所有条件对每个活动进行分类。

freq_dic['low_freq'] =  ['airport','atm','bank','dentist','doctor','hospital','zoo','jewelry_store','spa']

freq_dic['high_freq'] = ['grocery_or_supermarket','gym','home_goods_store','primary_school','school',
                       'secondary_school','shopping_mall','store','supermarket','university']

time_dic['morning'] = ['airport', 'atm','bakery','bank','beauty_salon','bicycle_store','book_store','cafe',
                   'clothing_store','dentist','doctor','drugstore','gym','hair_care','home_goods_store',
                  'hospital','jewelry_store','library','local_government_office','pharmacy','primary_school',
                  'school','secondary_school','shopping_mall','spa','store','supermarket','university','zoo']

time_dic['noon'] = ['atm','bakery','bank','beauty_salon','bicycle_store','book_store','cafe','clothing_store',
               'convenience_store','dentist','department_store','doctor','drugstore','grocery_or_supermarket',
                'gym','hair_care','home_goods_store','hospital','jewelry_store','library','local_government_office',
               'primary_school','restaurant','school','secondary_school','shopping_mall','spa','store',
                'supermarket','university']

time_dic['evening'] = ['bakery','bar','bicycle_store','cafe','convenience_store','department_store','drugstore',
                   'grocery_or_supermarket','gym','home_goods_store','hospital','jewelry_store','movie_theater',
                  'night_club','restaurant','shopping_mall','store','supermarket']

 visited_today_dic = {}
   ...

对于数据集中的每一行，我隔离了时间和频率并将其发送到此函数：

def find_list(c_time, freq):

if  7 <= c_time <= 12:
    if freq <= 6:
        return list(set(time_dic['morning']).intersection(freq_dic['low_freq']))
    elif freq <= 9:
        return list(set(time_dic['morning']).intersection(freq_dic['med_freq']))
    else:
        return list(set(time_dic['morning']).intersection(freq_dic['high_freq']))
elif 12 < c_time <= 17:
    if freq <= 6:
        return list(set(time_dic['noon']).intersection(freq_dic['low_freq']))
    elif freq <= 9:
        return list(set(time_dic['noon']).intersection(freq_dic['med_freq']))
    else:
        return list(set(time_dic['noon']).intersection(freq_dic['high_freq']))
else:
    if freq <= 6:
        return list(set(time_dic['evening']).intersection(freq_dic['low_freq']))
    elif freq <= 9:
        return list(set(time_dic['evening']).intersection(freq_dic['med_freq']))
    else:
        return list(set(time_dic['evening']).intersection(freq_dic['high_freq']))

现在，对于每个活动，我从每个类别（时间，频率，访问，持续时间...）中选择合适的字典，并仅选择类型适合于字典中所有值的位置，即该活动是否具有每天的频率现在是早上，我今天还没去过那里，所以我将查看freq_dic['high_freq']和time_dic['morning']中的所有值，并仅选择常见的地方

TIA

根据许多约束条件选择项目

0 个答案: