Python结合了多个类似的函数

Question

我正在使用一个API来使用python收回数据。我的功能很好，但我觉得我在一遍又一遍地重复自己，我可能应该做一些事情来提高效率。

每一个所做的就是获取结果的数量，然后返回api，以返回确切的记录数量。

第一个功能：

def get_categories():
headers = {"Authorization": "Bearer " + access_token} # auth plus token
response = requests.get("https://api.destination.com/categories", headers=headers) # response
data = json.loads(response.text) # load the json data
records = str(data['totalResults']) # get number of results for next call
response = requests.get("https://api.destination.com/categories?$skip=0&$top="+records, headers=headers)
all_data = json.loads(response.text) # load the json data
list_of_dict = all_data['resources'] # get rid of all but lists of dictionaries
df = pd.DataFrame.from_records(list_of_dict) # create dataframe
df['links'] = df['links'].str[0].str['href'] # just grab the links(key) items
return df # return the final dataframe

第二个功能：

def get_groups():
headers = {"Authorization": "Bearer " + access_token} # auth plus token
response = requests.get("https://api.destination.com/groups", headers=headers) # response
data = json.loads(response.text) # load the json data
records = str(data['totalResults']) # get number of results
response = requests.get("https://api.destination.com/groups?$skip=0&$top="+records, headers=headers)
all_data = json.loads(response.text) # load the json data
list_of_dict = all_data['resources']  # get rid of all but lists of dictionaries
df = pd.DataFrame.from_records(list_of_dict) # create dataframe
df['links'] = df['links'].str[0].str['href'] # just grab the links(key) items
return df # return the final dataframe

还有3个类似users的函数做同样的事情。正如您所看到的，它们之间唯一的区别是get(如https://api.destination.com/categories和https://api.destination.com/groups(，并且为每个返回的number of records将不同。有没有一种方法可以将这些结合起来，并称之为某种方式？

Answer 1

看起来您已经知道如何生成函数了，只需将其进一步扩展一步，即可抽象出函数中常见的所有内容。

BASE_URL = "https://api.destination.com/{}"
def make_headers():
headers = {"Authorization": "Bearer " + access_token}
return headers
def make_params(recs):
params = {'$skip': 0, '$top': recs}
return params
def make_df(data):
list_of_dict = data['resources']
df = pd.DataFrame.from_records(list_of_dict)
df['links'] = df['links'].str[0].str['href']
return df
def process(process):
headers = make_headers()
url = BASE_URL.format(process)
resp = requests.get(url, headers=headers)
data = resp.json()
records = data['totalResults']
params = make_params(records)
resp = requests.get(url, headers=headers, params=params)
all_data = resp.json()
return make_df(all_data)

然后你可以这样称呼它：

process('groups')
process('categories')

你可以把它进一步分解，但你明白了。

Answer 2

您只需向该函数添加一个参数即可。例如：

def get_categories():
headers = {"Authorization": "Bearer " + access_token} # auth plus token
response = requests.get("https://api.destination.com/categories", headers=headers) # response
data = json.loads(response.text) # load the json data
records = str(data['totalResults']) # get number of results for next call
response = requests.get("https://api.destination.com/categories?$skip=0&$top="+records, headers=headers)
all_data = json.loads(response.text) # load the json data
list_of_dict = all_data['resources'] # get rid of all but lists of dictionaries
df = pd.DataFrame.from_records(list_of_dict) # create dataframe
df['links'] = df['links'].str[0].str['href'] # just grab the links(key) items
return df # return the final dataframe

你可以重构为：

def get_elements(element):
if element is None:
return 'not found' #defaults to 404 error.
headers = {"Authorization": "Bearer " + access_token} # auth plus token
response = requests.get("https://api.destination.com/{}".format(element), headers=headers) # response
data = json.loads(response.text) # load the json data
records = str(data['totalResults']) # get number of results for next call
response = requests.get("https://api.destination.com/{}?$skip=0&$top={}".format(element,records), headers=headers)
all_data = json.loads(response.text) # load the json data
list_of_dict = all_data['resources'] # get rid of all but lists of dictionaries
df = pd.DataFrame.from_records(list_of_dict) # create dataframe
df['links'] = df['links'].str[0].str['href'] # just grab the links(key) items
return df # return the final dataframe