Question

我有一个带有4个参数的函数：

   def uploadDatabase(user,filter1,filter2, name):
        """uploads playlists to database

        if a playlist satisfying both filters is not long enough,
        we pick a playlist based on one single filter"""

        playlist_filter1 = [x['track'] for x in filter1]
        playlist_filter2 = [x['track'] for x in filter2]
        playlist_both_filters = [x['track'] for x in filter2 if x['track'] in [y['track'] for y in filter1]]

       #create an empty dict for user as key
       double_filter = {str(user):{}}
       single_filter1 = {str(user):{}}
       single_filter2 = {str(user):{}}
       #set initial playback number for each track
       count = 1
       #set counts to tracks
       double_filter[str(user)] = dict(zip(playlist_both_filters, [count for i in playlist_both_filters]))
       db_double = double_filter 
       #set counts to tracks
       single_filter1[str(user)] = dict(zip(playlist_filter1, [count for i in playlist_filter1]))
       db_single1 = single_filter1 
       (...)

但是，有时候filter2将不会传递：

uploadDatabase(user,energy_playlist, None, 'my playlist')

我如何修复uploadDatabase()并告诉它忽略None Type被分配给变量playlist_filter2和double_filter，因为两者都需要有效lists ？

Answer 1

您可以通过执行以下操作将None类型转换为两个过滤器的有效空列表：

playlist_filter1 = [x['track'] for x in filter1] if filter1 is not None else []
playlist_filter2 = [x['track'] for x in filter2] if filter2 is not None else []

而不是：

playlist_filter1 = [x['track'] for x in filter1]
playlist_filter2 = [x['track'] for x in filter2]

Answer 2

如果我理解正确，你想检查是否传递参数 filter2 。

这样做很简单，你实际上给它runserver的默认值，然后在代码中检查参数是否为None。

示例代码

None

当你使用这个函数时，你可以传递filter2，它会覆盖默认的def uploadDatabase(user,filter1,filter2=None, name): if not filter2: # filter2 argument in not passed else: # filter2 argument is there.值，或者只是不传递它，它将是None

Python - 没有参数

2 个答案:

示例代码