我有这个字典:
data = {'name':['Andrea', 'Luca'], 'age':['14', '15']}
我写:
for key, value in data.items():
print (key, value)
在这里,我在(print(key,value))中的结果是:
name ['Andrea', 'Luca']
age ['14', '15']
我想提取这个:
name = Andrea
name = Luca
age = 14
age = 15
我该怎么办?
答案 0 :(得分:3)
只需添加一个内部template<class... Args>
auto uninitialized_construct(Args&&... args){
typename array::element_ptr cur = this->data();
for(size_type n = this->num_elements(); n > 0; --n, ++cur)
alloc_traits::construct(allocator_, std::addressof(*cur), args...);
}catch(...){destroy(cur); throw;}
}
循环:
template<class... Args>
auto uninitialized_construct(Args&&... args){
std::unitialized_fill_n(data_, n, T(args...)); // or should I construct a copy first?
}
答案 1 :(得分:1)
value
是您的列表。遍历并打印每个项目。
for key, value in data.items():
for x in value:
print('{} = {}'.format(key, x))