我有这个列表对象
[('migrations',), ('users',), ('devices',), ('externals',), ('cloud_securities',), ('operators',), ('promotions',), ('visitors',), ('caches',), ('captures',), ('mirror_settings',), ('wifis',), ('service_plans',), ('auto_provisionings',), ('guest_settings',), ('help_texts',), ('gateways',), ('fb_wifi_tokens',), ('health_check',), ('ubb_settings',), ('templates',), ('ubb_user_settings',), ('captive_portals',), ('languages',)]我试着把它们做成数组列表
我试过了
# table_names = np.asarray(table_names)
我得到了
[['migrations']
['users']
['devices']
['externals']
['cloud_securities']
['operators']
['promotions']
['visitors']
['caches']
['captures']
['mirror_settings']
['wifis']
['service_plans']
['auto_provisionings']
['guest_settings']
['help_texts']
['gateways']
['fb_wifi_tokens']
['health_check']
['ubb_settings']
['templates']
['ubb_user_settings']
['captive_portals']
['languages']]但我想要这样
['migrations', 'users', 'devices', 'externals', 'cloud_securities', 'operators', 'promotions', 'visitors', 'caches', 'captures', 'mirror_settings', 'wifis', 'service_plans', 'auto_provisionings', 'guest_settings', 'help_texts', 'gateways', 'fb_wifi_tokens', 'health_check', 'ubb_settings', 'templates', 'ubb_user_settings', 'captive_portals', 'languages']我怎么能做那条蟒蛇?
我应该使用或调查什么功能?
发布于 2017-02-13 21:44:10
您不需要numpy,您需要简单的python &列表理解:
假设l是您的列表,获取tuple元素的每个第一个&唯一的项,然后从那里重建一个列表:
newl = [x[0] for x in l]如果您的元组包含超过一个元素,那么一个通用的解决方案将涉及itertools.chain来平平列表,但这可能会在这里过度使用:
newl = list(itertools.chain.from_iterable(l))https://stackoverflow.com/questions/42214061
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