python – 函数参数的Dinamically var

发布时间：2020-12-20 11:08:19 所属栏目：Python 来源：网络整理

导读：我正在尝试使用var的名称来调用函数,但我不知道这是否可行,类似于： fight_movies = list() # var how I want to use in function callwin_movies = list() # var how I want to use in function callknowledge_movies = list() # var how I want to use in

我正在尝试使用var的名称来调用函数,但我不知道这是否可行,类似于：

fight_movies = list() # var how I want to use in function call
win_movies = list() # var how I want to use in function call
knowledge_movies = list() # var how I want to use in function call
biography_movies = list() # var how I want to use in function call

for genre in genres:
    .... Ommited #////// Here is where I call the function
    write_jsonl(genre + '_movies',genre,rating,title,genre) #here is the call of the function


def write_jsonl(movie_list,json_name):
    dict = {'title': title,'genre': genre,'rating': rating}
    movie_list.append(dict)
    # print(action_movies)
    with jsonlines.open(json_name+'.jsonl',mode='w') as writer:
        writer.write(movie_list)

我正在尝试动态地将变量名称作为列表传递,但我不确定在python中是否可以使用任何建议？

Error: Traceback (most recent call last):
  File "bucky.py",line 58,in <module>
    web_crawling()
  File "bucky.py",line 34,in web_crawling
    write_jsonl(genre + '_movies',genre)
  File "bucky.py",line 52,in write_jsonl
    movie_list.append(dict)
AttributeError: 'str' object has no attribute 'append'

解决方法

只需使用字典：

genres_dict = {k: [] for k in ('fight','win','knowledge','biography')}

for genre in genres:
    write_jsonl(genres_dict[genre],genre)

可变数量的变量不是推荐的方法.

相关：How do I create a variable number of variables?.

（编辑：李大同）

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