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',)]
我试图将它们作为数组列表
我试过
# 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']]
但我希望它像这样
['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']
我该怎么做那个python？
我应该使用或研究哪些功能？
答案 0 :(得分：5)
你不希望tuple想要简单的python＆amp;列表理解：
说出newl = [x[0] for x in l]
是你的清单，先把每个放在第一位。仅itertools.chain元素的项目并从那里重建列表：
newl = list(itertools.chain.from_iterable(l))
一般的解决方案，如果你的元组包含多于1个元素会导致#include <string>
#include <sstream>
int add(int a, int b) {
return a + b;
}
int main() {
std::string str = "1 2";
std::stringstream ss(str);
int a, b;
ss >> a;
ss >> b;
printf("Result: %d\n", add(a, b));
}
使列表变平，但这可能是一种过度杀伤：
int a, b;
ss >> a;
ss >> b;
printf("Result: %d\n", add(a, b));