我有两个t1和t2的Pytorch张量（实际上只是一维列表）。是否可以并行遍历它们，即做类似
for a,b in zip(t1,t2)
？
谢谢。
答案 0 :(得分：3)
对于我来说（Python版本3.7.3和PyTorch版本1.0.0），zip函数可与PyTorch张量正常工作：
>>> import torch
>>> t1 = torch.ones(3)
>>> t2 = torch.zeros(3)
>>> list(zip(t1, t2))
[(tensor(1.), tensor(0.)), (tensor(1.), tensor(0.)), (tensor(1.), tensor(0.))]
只需要list调用即可显示结果。遍历zip正常工作。
答案 1 :(得分：1)
用torch.cat(dim=1) concatenate来使它们更有意义；然后，您可以遍历新的张量。
答案 2 :(得分：0)
您可以尝试：
export class AuthService {
private subscription: Subscription = new Subscription();
user: Observable<firebase.User>;
private currentUserSubject: BehaviorSubject<User | null>;
authState: any = null;
constructor(private firebaseAuth: AngularFireAuth, private router: Router, private userService:UserService) {
this.user = firebaseAuth.authState;
this.subscription.add(this.firebaseAuth.authState.subscribe((auth) => {
this.authState = auth;
//getting the user object if he already signed-in
this.currentUserSubject = new BehaviorSubject<User>(localStorage.getItem('currentUser') | null);
this.currentUser = this.currentUserSubject.asObservable();
}))
signin(email: string, password: string) {
return new Promise<any>((resolve, reject) => {
// --- login with email and password in firebase
this.firebaseAuth.auth.signInWithEmailAndPassword(email, password)
.then(
res => {
// --- Once login is successful, get the user object form the backend
this.subscription =this.userService.findByUsername(res.user.uid).subscribe(data=>{
this.currentUserSubject.next(data);
localStorage.setItem("currentUser", JSON.stringify(data));
resolve(this.firebaseAuth.auth.currentUser);
})
} ,
err => {reject(err); }
).catch(err => {
console.log('Something went wrong:', err.message);
});
});
}
async logout() {
await this.firebaseAuth.auth.signOut().then(()=>{
this.currentUserSubject.next(null);
this.currentUserSubject.complete();
this.subscription.unsubscribe()
localStorage.removeItem('currentUser');
this.router.navigate(['/signin']);
})
}
}
，
有关详细信息，请参见pytoch documentation
答案 3 :(得分：0)
将 PyTorch 中的张量压缩成一个toString()与Date.toString()
示例
torch.stack