小编Muh*_*fil的帖子

ValueError                                Traceback (most recent call last)
<ipython-input-30-33821ccddf5f> in <module>
     23         output = model(data)
     24         # calculate the batch loss
---> 25         loss = criterion(output, target)
     26         # backward pass: compute gradient of the loss with respect to model parameters
     27         loss.backward()

C:\Users\mnauf\Anaconda3\envs\federated_learning\lib\site-packages\torch\nn\modules\module.py in __call__(self, *input, **kwargs)
    487             result = self._slow_forward(*input, **kwargs)
    488         else:
--> 489             result = self.forward(*input, **kwargs)
    490         for hook in self._forward_hooks.values():
    491             hook_result = hook(self, input, result)

C:\Users\mnauf\Anaconda3\envs\federated_learning\lib\site-packages\torch\nn\modules\loss.py in forward(self, input, target)
    593                                                   self.weight,
    594                                                   pos_weight=self.pos_weight,
--> …

&代码中的目的是什么&i in list？如果我删除&，则会在 中产生错误largest = i，因为它们的类型不匹配（其中iis&32和iis i32）。但是如何&i转换i为i32？

fn largest(list: &[i32]) -> i32 {
    println!("{:?}", list);
    let mut largest = list[0];
    for &i in list {
        if i > largest {
            largest = i;
        }
    }
    largest
}

fn main() {
    let hey = vec![1, 3, 2, 6, 90, 67, 788, 12, 34, 54, 32];
    println!("The largest number is: {}", largest(&hey)); …

我知道这setTimeout是一个异步运行的 API，我想同步运行它。使用async/await如下所示应该bla bla首先打印，但我bla先打印。

  async function testing(){
    console.log("testing function has been triggered");
      await setTimeout(function () {
        console.log("bla bla")
      }, 4200)
    console.log("bla");
  }

我按照 Diesel 的官方文档开始使用。我还安装了 PostgreSQL。我的数据库用户名是postgres，密码是1schoollife@。

我开始于

$ echo DATABASE_URL=postgres://postgres:1schoollife@@localhost/diesel_demo > .env
$ diesel setup

结果：

Creating migrations directory at: /home/naufil/Desktop/rust/3june/testing/migrations
Creating database: diesel_demo
database "diesel_demo" already exists

我创建了一个迁移：

$ diesel migration generate create_posts
Creating migrations/2019-06-03-182531_create_posts/up.sql
Creating migrations/2019-06-03-182531_create_posts/down.sql

迁移数据库时出现以下错误：

$ diesel migration run
thread 'main' panicked at 'called `Result::unwrap()` on an `Err` value: ConnectionError(BadConnection("could not translate host name \"@localhost\" to address: Name or service not known\n"))', src/libcore/result.rs:997:5
note: Run with `RUST_BACKTRACE=1` environment variable to display a …

当我搜索“ 什么是2 + 2 ” 时，我试图抓取Google结果，但是下面的代码正在返回'NoneType' object has no attribute 'text'。请帮助我实现所需的目标。

text="What is 2+2"
search=text.replace(" ","+")
link="https://www.google.com/search?q="+search
headers={'User-Agent':'Mozilla/5.0 (X11; Linux x86_64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/51.0.2704.103 Safari/537.36'}
source=requests.get(link,headers=headers).text
soup=BeautifulSoup(source,"html.parser")
answer=soup.find('span',id="cwos")

self.respond(answer.text)

唯一的问题是idin soup.find，但是我非常仔细地选择了此id。我不要误会 我也尝试过answer=soup.find('span',class_="cwcot gsrt")，但是都没有用。

Rust文档告诉我们cargo build在编译后创建一个二进制文件，可以使用执行该文件cargo run。cargo run如果在cargo build执行命令后发现任何更改，它将再次编译代码。它还说该cargo build --release命令创建了最终程序，它将运行得更快。

我的问题是，为什么当我这样做时cargo build --release，它会编译代码，这很好。但是，当我执行时cargo run，即使此后我没有更改任何代码，它也会再次编译代码。它与一起正常工作cargo build，然后cargo run与前一个命令一起编译一次。

naufil@naufil-Inspiron-7559:~/Desktop/rust/20April/variables$ cargo build
   Compiling variables v0.1.0 (/home/naufil/Desktop/rust/20April/variables)
    Finished dev [unoptimized + debuginfo] target(s) in 0.35s
naufil@naufil-Inspiron-7559:~/Desktop/rust/20April/variables$ cargo run
    Finished dev [unoptimized + debuginfo] target(s) in 0.02s
     Running `target/debug/variables`
Hello, world! 6
naufil@naufil-Inspiron-7559:~/Desktop/rust/20April/variables$ cargo build --release
   Compiling variables v0.1.0 (/home/naufil/Desktop/rust/20April/variables)
    Finished release [optimized] target(s) in 0.34s
naufil@naufil-Inspiron-7559:~/Desktop/rust/20April/variables$ cargo run …

Epoch: 1    Training Loss: 0.816370     Validation Loss: 0.696534
Validation loss decreased (inf --> 0.696534).  Saving model ...
Epoch: 2    Training Loss: 0.507756     Validation Loss: 0.594713
Validation loss decreased (0.696534 --> 0.594713).  Saving model ...
Epoch: 3    Training Loss: 0.216438     Validation Loss: 1.119294
Epoch: 4    Training Loss: 0.191799     Validation Loss: 0.801231
Epoch: 5    Training Loss: 0.111334     Validation Loss: 1.753786
Epoch: 6    Training Loss: 0.064309     Validation Loss: 1.348847
Epoch: 7    Training Loss: 0.058158     Validation Loss: 1.839139
Epoch: 8    Training Loss: 0.015489     Validation …

Rust文档说默认的整数类型是i32，这意味着变量默认可以保存的最大数字是2147483647ie 2e31 - 1。事实证明，这是千真万确的：如果我试图挽救数量大于2e31 - 1在x变，我得到的错误literal out of range。

码

fn main() {
    let x = 2147483647;
    println!("Maximum signed integer: {}", x);
    let x = 2e100;
    println!("x evalues to: {}", x);
}

但是，如果我2e100在x变量中保存值，为什么我不会出错？它的计算结果肯定大于2e31 - 1。

输出量

fn main() {
    let x = 2147483647;
    println!("Maximum signed integer: {}", x);
    let x = 2e100;
    println!("x evalues to: {}", x);
}

码

fn main() {
    let …

为什么我需要使这个闭包变量可变？闭包不返回任何内容，因此闭包变量中没有存储任何内容。这个闭包只是从环境中捕获一个值并递增它。

fn main() {
    let mut x = 1;
    let mut y = || x = x + 1;
    y();
    println!("{}", x);
}

我正在制作十进制到二进制的转换程序。为什么我无法将字符串解析为数字？binary_vec[i]是一个character。我应用to_string()了将其转换为字符串的方法，因为parse()它不适用于字符，但仍然给我一个错误。

use std::io;

fn main() {
    let mut binary = String::new();
    println!("Enter a decimal: ");
    io::stdin().read_line(&mut binary)
        .ok()
        .expect("Couldn't read line");
    println!("{}",to_decimal(binary));

}

fn to_decimal(mut binary_str:String) -> String {
    let mut binary_no: u32 = binary_str.trim().parse().expect("invalid input");
    if binary_no == 0 {
        format!("{}",binary_no)
                } 
    else {
        let mut bits = String::new();
        let mut binary_vec: Vec<char> = binary_str.chars().collect();
        let mut result = 0;
        let mut i = 0;
        while i <=binary_str.len()-2{
            let mut result …

如何替换Node.js中的单词，例如，如果我想将is字符串下面的单词替换为done：

He is a Pakistani and is a good person

输出仅给出：

He done a pakdonetani and done a good person代替He done a Pakistani and done a good person。我正在使用此代码：

var str = "He is a Pakistani and is a good person";
var res = str.replace(/is/g, "done");
console.log(res);

可以这样说，改为：

var res = str.replace(/ is /g, "done");

但它对于以下字符串将失败：

var str = "He is a Pakistani and a good person he is.";

为了应对这种情况，我们可以：

var res = str.replace(/ …