Answer 1

您想在分开列表之后附加单词 -

imie = []
dlugosc = []
third_list = []
for line in plik:
    for word1, word2, *_ in line.split():
        imie.append(word1)
        dlugosc.append(word2)
        third_list.append(_)

Answer 2

存储库中的工件名称是由artifactid生成的，没有可能更改它。如果您愿意或需要更改，则您必须更改artifactid。

对于在特定平台上运行的，我将创建一种安装软件包，其中包含正确命名的工件。

Answer 3

'${fetched_string}'.replaceAll(".", ".\n");

这应该用完整的停止和新线路替换所有全部。请注意，它将替换所有全部。因此，请确保您在派生之间不会得到全部

Answer 4

＆lt; v-expansion-panels＆gt;是始终 ＆lt; v-expansion-panel＆gt;。如果没有＆lt; v-expansion-panel＆gt;没有＆lt; v-expansion-panels＆gt; parent（ demo 1 ）。

如果要嵌套多个＆lt; v-expansion-panel＆gt; s，则必须在自己的自己的 ＆lt; v-expansion-panels＆gt; gt; parent：

<v-expansion-panels>
  <v-expansion-panel>
    <v-expansion-panel-content>

      <!-- NESTED PANELS -->
      <v-expansion-panels>
        <v-expansion-panel>
          <v-expansion-panel-content>I'm a nested panel</v-expansion-panel-content>
        </v-expansion-panel>
      </v-expansion-panels>

    </v-expansion-panel-content>
  </v-expansion-panel>
</v-expansion-panels>

Answer 5

我在测试文件中添加了一些行，以使其完整。

findstr REGEX非常有限（残废）。您必须围绕这些限制进行工作，例如搜索三个，四位和五位数字（因为没有{3,5}）。

^用于“线的开始”，必须逃脱*（使用\），因为它具有特殊的含义（零或以前的更多字符），您必须使用/c来照顾空间

扩展测试文件：

d:\temp>type t.txt
*P1 this line should NOT match - less than three numbers
*C12 this line should NOT match - less than three numbers
*T123456 this line should NOT match - more than five numbers
SampleInput.txt:
This line should not match - no asterisk
 * This line should not match because the first character is not an asterisk
*P123 This line should match because it starts with P123 (followed by a space)
*P123This line should not match because there is no space after the P123
*C1234 This line should match ... just like P123 above
*C1234This line should not match ... just like P123This above
*T12345 This line should match ... just like P123 above
*T12345This line should not match ... just like P123This above
*This line should not match because the string *T12345 is not at the beginning of the line
*TX This line should not match because the T is not followed by a number
*CX This line should not match because the C is not followed by a number
*PX This line should not match because the P is not followed by a number
* This line should NOT match because the Asterisk is followed by only one Space
*  This line should match because it starts with Asterisk Space Space
*   This line should match because it starts with Asterisk Space Space Space
*    This line should match because it starts with Asterisk Space Space Space Space
*     This line should match because it starts with Asterisk Space Space Space Space Space
*      This line should match because it starts with Asterisk Space Space Space Space Space Space
*C456... This line should not match because of the three dots (ie there is no space after the C456)
*T999 Bottom line ... 9 of the lines in this file should match (including this line)

匹配行：

d:\temp>type t.txt |  findstr /rc:"^\*[PCT][0-9][0-9][0-9] " /c:"^\*[PCT][0-9][0-9][0-9][0-9] " /c:"^\*[PCT][0-9][0-9][0-9][0-9][0-9] "  /c:"^\*  "
*P123 This line should match because it starts with P123 (followed by a space)
*C1234 This line should match ... just like P123 above
*T12345 This line should match ... just like P123 above
*  This line should match because it starts with Asterisk Space Space
*   This line should match because it starts with Asterisk Space Space Space
*    This line should match because it starts with Asterisk Space Space Space Space
*     This line should match because it starts with Asterisk Space Space Space Space Space
*      This line should match because it starts with Asterisk Space Space Space Space Space Space
*T999 Bottom line ... 9 of the lines in this file should match (including this line)

非匹配行（为了完整性）

d:\temp>type t.txt |  findstr /rvc:"^\*[PCT][0-9][0-9][0-9] " /c:"^\*[PCT][0-9][0-9][0-9][0-9] " /c:"^\*[PCT][0-9][0-9][0-9][0-9][0-9] "  /c:"^\*  "
*P1 this line should NOT match - less than three numbers
*C12 this line should NOT match - less than three numbers
*T123456 this line should NOT match - more than five numbers
SampleInput.txt:
This line should not match - no asterisk
 * This line should not match because the first character is not an asterisk
*P123This line should not match because there is no space after the P123
*C1234This line should not match ... just like P123This above
*T12345This line should not match ... just like P123This above
*This line should not match because the string *T12345 is not at the beginning of the line
*TX This line should not match because the T is not followed by a number
*CX This line should not match because the C is not followed by a number
*PX This line should not match because the P is not followed by a number
* This line should NOT match because the Asterisk is followed by only one Space
*C456... This line should not match because of the three dots (ie there is no space after the C456)

（注意：i使用的懒惰[0-9]可以给出错误的阳性（2 /代码>而是）

Answer 6

在整个数组中循环循环后，您可以抛出异常（这意味着您没有找到任何员工），如果您找到了一个，则该功能将返回。

private static List<Employee> employee = new ArrayList<>();

public Employee findEmployee(String Name) throws emplyeeNotFoundException {
  for(Employee value : employee) {
    if (value.getName().equals(Name)) {
      return value;
    }
  }
  throw new emplyeeNotFoundException(Name);
}

Answer 7

You can try it :

$tmpIds = [];
foreach($name as $n){
    foreach($response as $r){
        if ($n->id != $r->cid) {
            if(!in_array($n->id, $tmpIds)){
                $tmpIds[] = $n->id;
                echo $n->id;
            }
        }
    }
}

Example

Answer 8

https://stackoverflow.com/a/72472483/9842697 的答案。

（此注释是对答案的回应）

我已经通过Pyenv安装了Python 3.10.6，Pyenv通过Homebrew安装了Pyenv。
（使用-Devel不起作用的卸载）

• 与Pyenv一起安装TCL-TK并用Pyenv重新安装Python 3.10.6，使Python安装使用TCL-TK的自制版本，此后闲置效果很好。

% brew install tcl-tk
% pyenv install 3.10.6
% python
>>> import idlelib.idle

Answer 9

您的问题似乎归结为：如何将两个张量交织在一起。给定X和y两个张量。您可以组合 a>和 reshape 。

>>> torch.stack((x,y),1).transpose(1,2).reshape(2,-1)
tensor([[ 1.1547e+01,  0.0000e+00,  1.3786e+00, -8.1970e-01, -3.2118e-02,
         -2.3900e-02, -3.2898e-01, -3.4610e-01, -1.7916e-01,  1.2308e+00,
         -5.4203e-01,  1.2580e-01,  8.5273e-01,  8.9980e-01, -2.7096e+00,
         -3.8060e-01,  3.0016e-01, -4.5240e-01, -7.7809e-02,  4.5630e-01,
         -4.5805e-03,  0.0000e+00],
        [ 1.1106e+01,  0.0000e+00,  1.3362e-01,  1.3830e-01, -7.4233e-01,
          7.7570e-01, -9.9461e-01,  1.0834e+00,  1.6952e+00,  5.2920e-01,
         -1.1884e+00, -2.5970e-01, -8.7958e-01,  4.3180e-01, -9.3039e-01,
          8.8130e-01, -1.0048e+00,  1.2823e+00,  2.0595e-01, -6.5170e-01,
          1.7209e+00,  0.0000e+00]])

Answer 10

这是您可以从一个人文件中填充数据的方式：

var people = new DataTable();
people.Columns.Add("Name");//string
people.Columns.Add("Age", typeof(int));
people.Columns.Add("HireDate", typeof(DateTime));
people.Columns.Add("IsManager", typeof(bool));

foreach(var line in File.ReadLines("people.csv")){
  var bits = line.Split(',');
  dt.Rows.Add(new object[]{ 
    bits[0], //Name
    int.Parse(bits[1]), //Age
    DateTime.Parse(bits[2]), //HireDate
    bool.Parse(bits[3]) //IsManager
  });
}

当然，如果您是从CSV读取的话，最好使用CSVhelper（例如CSVHelper） - 它可以直接读取到DataTables中，并且比这更复杂得多。例子。这只是为了显示“制作数据台，添加列，通过提供列的值添加行”的过程，

最好为此创建一个强大的型号数据词：

• 将新的数据集类型添加到您的项目中并进行goive IT 好名称
• 表面
• 打开它，右键单击“
• 一个 使用此列在代码中完成的列

与上述大多相同，除了您制作了表格。这是一个内部类，因此您也使用数据集的名称来创建它：

var dt = new YourDataSetNameHere.YourDataTableNameHere_DataTable();

//the columns are already added, you don't need to add them

foreach(var line in ...){
  ...

  dt.Add_YourDataTableNameHere_Row(
    bits[0], //Name
    int.Parse(bits[1]), //Age
    DateTime.Parse(bits[2]), //HireDate
    bool.Parse(bits[3]) //IsManager
  );
}

它们比常规弱键入数据表更好地使用

Answer 11

您需要节点V16+和任何Linux OS

1. git clone jitsi-meet
2. npm install
3. 学习react，因为您需要知识来更改功能，但是您可以从简单的内容开始，例如更改文本因此转到jitsi-meet/lang/main.json更改您要更改的文本。
4. 然后在终端中键入make命令
5. ，然后使用服务器上的原始代码更改整个源代码，您将看到结果

Answer 12

我会尽力而为。

我正在尝试代码，并使用第一个功能做点事
它看起来像这样：

def func(x):
    return lambda y: (x + y + 1)
print(func(3)(1))

有趣的部分是您实际上可以将lambda值放在函数值旁边，因此lambda值将为1，函数值将为3，然后结果将是：

3 + 1 + 1 = 5

现在很难我花了很长时间才非常了解它
因为它很棘手。

def func(x):
    return lambda y: (x + y + 2)
print(func(3)(1))



def func1(x):
    return lambda y : (func(x)(x)+y+1)
print(func1(4)(1))

我在此代码中看到一些棘手的东西，例如：

4 + 4 + 1 + 1 = 10，

就像将数字成为12
这就是从上述函数中获得值2的值的地方，结果将是：

4 + 4 + 4 + 1 + 1 + 2 = 12 = 12

这就是我所知道的。

现在我的大脑很热。

Answer 13

我看过的大多数论坛似乎都表明必须重新订购证书

它们是正确的。该证书由CA数字签名，任何编辑证书的尝试都将使签名无效。您必须从CA（以您的情况为例）重新订购CA（第三方）的证书。

Answer 14

使用@luis MiguelMejíaSuárez解决方案：

import org.joda.time.LocalDate

val numberOfMonthsBack = 5
val date = new LocalDate("2022-05-24")
val dateSeq = Seq(date)
val allDates = dateSeq.toStream ++ Stream.continually(dateSeq)
  .zip(Stream.from(1))
  .flatMap(ticks => 
    ticks._1.map(x => x.minusMonths(ticks._2))
  )
allDates
  .take(numberOfMonthsBack).toList

Answer 15

您可以通过完全合格的登录字符串来摆脱提示。

.logon my_server/user，password

如果您仍然只想使用.logon中的服务器详细信息。与.set logonprompt Off一起，将环境变量设置为guilogon为no。

在CMD中：setx guilogon no

smippet来自 td文档：

请注意，设置logonprompt to Off有时不会
使用Windows时，足以抑制所有不必要的提示
BTEQ。您可能还需要指示CLI抑制其产生
所谓的Guilogon对话框。

这可以通过设置环境变量来完成
至no。