Question

在检索各自的业务ID后，我试图提取点击Yelp API的餐馆列表的营业时间：

我最初定义的函数是：

def is_clocked(business_id):
    #import pdb; pdb.set_trace()
    try:
        clocked_ind = get_business(API_KEY, business_id)
        clocked_ind1 = clocked_ind['hours']
    except:
        clocked_ind1 = 'None'
    return clocked_ind1
clocked_ind = is_clocked(b_id)
print(clocked_ind)

但是，此函数向我返回的是长数据而不是宽数据格式：

bad_format：

Querying https://api.yelp.com/v3/businesses/9GzjKeifGJ6KzWkaPftYHg ...
[{'open': [{'is_overnight': False, 'start': '1100', 'end': '2200', 'day': 0}, {'is_overnight': False, 'start': '1100', 'end': '2200', 'day': 1}, {'is_overnight': False, 'start': '1100', 'end': '2200', 'day': 2}, {'is_overnight': False, 'start': '1100', 'end': '2200', 'day': 3}, {'is_overnight': False, 'start': '1100', 'end': '2200', 'day': 4}, {'is_overnight': False, 'start': '1100', 'end': '2200', 'day': 5}, {'is_overnight': False, 'start': '1100', 'end': '2100', 'day': 6}], 'hours_type': 'REGULAR', 'is_open_now': True}]

我希望我的最终输出在csv中看起来像这样：

（输入）

day = [0, 1,2,3,4,5,6]
start = [1100, 1100, 1100, 1100, 1100, 1100, 1100]
end = [2200, 2200, 2200, 2200, 2200, 2200, 2100]
day1 = []
for i in day:
    day1.append("start"+str(i))

for i in range(len(day1)):
        merge_HOO[day1[i]]=start[i]

pd.DataFrame(merge_HOO, index=[0])

#Desired Output[115]: 
   item  day start   end  end0   ...    start2  start3  start4  start5  start6
0     0    0  1100  2200  2200   ...      1100    1100    1100    1100    1100

但是您要注意：我为一次特定业务手动编码了输入。我想创建一个循环，使其在csv中返回每个business_id的所需输出。我也在下面编写了代码，但是我觉得必须有一种更好的方法来进行此循环。下面的代码必须是一个函数：

day = day_open(b_id)
start = day_start(b_id)
end = day_end(b_id)
day1 = []
for i in day:
    day1.append("start"+str(i))
dict1 = {}
for i in range(len(day1)):
        dict1[day1[i]]=start[i]
start_df = pd.DataFrame(dict1, index=[0])


day2 = []
for i in day:
    day2.append("end"+str(i))
dict2 = {}
for i in range(len(day2)):
        dict2[day2[i]]=end[i]
end_df = pd.DataFrame(dict2, index=[0])
start_df['end0']=end_df['end0']
start_df['end0']=end_df['end0']
start_df['end0']=end_df['end0']
start_df['end0']=end_df['end0']
start_df['end0']=end_df['end0']
start_df['end0']=end_df['end0']

我想像下面这样使它适应循环：

def id_loop(a):
    empty = []
    for i in input_range:
        review_count_ind = is_review_count(a[i])
        empty.append(review_count_ind)
    return empty

c = id_loop(a)

Answer 1

您的预期输出与您的数据不一致。 item和is_overnight是一样的东西吗？您有end和end0，这是什么意思？因此，我假设item是is_overnight，并且所有列名都应附加日期。因为您的结果是一个列表，所以我还假设其长度可以更改。

import pandas as pd

clocked_ind = [{'open': [{'is_overnight': False, 'start': '1100', 'end': '2200', 'day': 0}, {'is_overnight': False, 'start': '1100', 'end': '2200', 'day': 1}, {'is_overnight': False, 'start': '1100', 'end': '2200', 'day': 2}, {'is_overnight': False, 'start': '1100', 'end': '2200', 'day': 3}, {'is_overnight': False, 'start': '1100', 'end': '2200', 'day': 4}, {'is_overnight': False, 'start': '1100', 'end': '2200', 'day': 5}, {'is_overnight': False, 'start': '1100', 'end': '2100', 'day': 6}], 'hours_type': 'REGULAR', 'is_open_now': True}]

new_list = [[] for _ in range(len(clocked_ind)+1)]
for i in range(len(clocked_ind)):
    new_list[i+1] = []
    for day_dict in clocked_ind[i]['open']:
        suffix = str(day_dict['day'])
        for k, v in day_dict.items():
            key = k if k != 'is_overnight' else 'item'
            column_name = key + suffix
            if column_name not in new_list[0]:
                new_list[0].append(column_name)
            new_list[i+1].append(v)  # if you want item to store int instead of bool replace with append(int(v))

df  = pd.DataFrame(new_list[1:], columns=new_list[0])
print(df)

输出

   item0  day0  end0 start0  item1  day1  end1 start1  item2  day2  ...    \
0  False     0  2200   1100  False     1  2200   1100  False     2  ...     

   end4 start4  item5  day5  end5 start5  item6  day6  end6 start6  
0  2200   1100  False     5  2200   1100  False     6  2100   1100  

[1 rows x 28 columns]

如何在python中循环此功能？

1 个答案: