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若水微香 2025-02-11 12:37:35

我知道它已被标记为解决,但是如果我正确理解,就有一个更容易的解决方案。

来自 poptrox docs ,在您可以通过的锚标签内部

data-poptrox="ignore"

,它将可以作为正常的锚固端工作标签。

所以在OP的情况下

<article class="from-left">
  <a
    href="images/fulls/01.jpg"
    class="image fit"
    data-poptrox="ignore" <!-- ADD THIS LINE -->
  >
    <img src="images/thumbs/01.jpg" title="Superstore project" alt="" />
   </a>
</article>

I know it's been marked as solved, but if I understood correctly there is an easier solution.

From the Poptrox docs, inside the anchor tag you can pass

data-poptrox="ignore"

and it will work as a normal anchor tag.

So in the case of OP

<article class="from-left">
  <a
    href="images/fulls/01.jpg"
    class="image fit"
    data-poptrox="ignore" <!-- ADD THIS LINE -->
  >
    <img src="images/thumbs/01.jpg" title="Superstore project" alt="" />
   </a>
</article>

HTML HREF未打开链接

若水微香 2025-02-11 04:45:33

我不明白sissions的真实类型是什么,

此处错误sigsion> size!== 0,因为您正在尝试将字符串与一个数字进行比较,总是错误的。

如果可以修复,则通过修复类型


type TMap<T> = {
  [index: string]: T
} & {
  size: number
}


type TMap<T> = Map<string, T>

或字符串的映射,您可以


assignments && Object.keys(assignments).length !== 0

根据实际地图类型将映射转换为数组,检查地图大小。
如果MAP然后array.from

如果字符串映射可以使用 in-javascript”>如何将对象{}转换为javaScript中键值对的数组[]的数组[]

I don't understand what is the real type of the assignments

Here error assignments.size !== 0 because you are trying to compare a string with a number, which is always false.

If can be fixed, by fixing type


type TMap<T> = {
  [index: string]: T
} & {
  size: number
}

or


type TMap<T> = Map<string, T>

Or if it map of string, you can check map size with


assignments && Object.keys(assignments).length !== 0

Converting your map to array depending on real map type.
If it Map then Array.from should work

If it map of string you can use How to convert an Object {} to an Array [] of key-value pairs in JavaScript

打电话时打电话给出错误。

若水微香 2025-02-10 21:00:00

EAN列之间并不常见。 DF1更大。

我想到了这样的事情,但这根本不是优雅的...

for x in df1.index:
   for y in df2.index:
      if df1.loc[x, "EAN"] == df2.loc[y, "EAN"]:
         print ("Found!") 
         df1["OLD-DESCRIPTION"] = df2["NEW-DESCRIPTION"]

EAN column is not common between both tables. df1 is way bigger.

I thought about something like this, but this is not elegant at all...

for x in df1.index:
   for y in df2.index:
      if df1.loc[x, "EAN"] == df2.loc[y, "EAN"]:
         print ("Found!") 
         df1["OLD-DESCRIPTION"] = df2["NEW-DESCRIPTION"]

python pandas-在两个数据范围内找到并替换

若水微香 2025-02-10 02:53:46

要删除非空文件夹,您需要一些称为递归的东西。我们通常会读到的是,我们要“ 递归删除一个文件夹”,这意味着要删除一个及其所有内容,包括其他文件夹,以及各自的内容和文件夹,等等。

命令 del 为您提供了。您是否阅读了它的帮助或文档?要看到这一点,您在 cmd 程序中使用的命令是:

del /?

但这并不能直接回答您的问题。您需要放入批处理文件中的内容是:

del /s /q [non empty folder name]
  1. “/s”是递归删除
  2. “/q”是在而无需确认每个文件或文件夹的情况下删除它。如果要选择文件夹中的内容,则可能需要删除此项目。
  3. 您不得编写我在命令示例中写的方括号。只是文件夹名称。

To delete a non empty folder, you need something called recursion. What we usually read is that we want to "delete a folder recursively", which means to delete a folder and all its contents, including other folder, with their respective contents and folders, and so on.

The command del does it for you. Have you read its help or its documentation? To see that, the command you use in the cmd program is:

del /?

But this does not directly answer your question. What you need to put in your batch file is:

del /s /q [non empty folder name]
  1. "/s" is to delete it recursively
  2. "/q" is to delete it without asking for confirmation of each file or folder being deleted. You may want to remove this item, if you want to choose what will be deleted inside the folder.
  3. You must not write the square brackets I wrote in the command example. Just the folder name, as it is.

如何在批处理脚本中删除非空文件夹?

若水微香 2025-02-09 21:24:56

一个选项是使用transform 使用;速度明智,我希望Bert2me的解决方案会更快:

df.assign(UPDATED = df.groupby(grouper)
                      .EVENT
                      .transform(lambda x: set(x) >= {1,3,6}) 
                                           & df.EVENT.eq(1))

     GAME_ID   TIME  PER  EVENT  UPDATED
0   22000394  12:00    1     12    False
1   22000394  12:00    1     10    False
2   22000394  11:36    1      1    False
3   22000394  11:24    1      1    False
4   22000394  11:04    1      1    False
5   22000394  10:41    1      1    False
6   22000394  10:30    1      2    False
7   22000394  10:29    1      4    False
8   22000394  10:17    1      1    False
9   22000394  10:01    1      1    False
10  22000394   9:48    1      2    False
11  22000394   9:46    1      4    False
12  22000394   9:42    1      6    False
13  22000394   9:42    1      3    False
14  22000394   9:42    1      3    False
15  22000394   9:25    1      1    False
16  22000394   9:15    1      1     True
17  22000394   9:15    1      6    False
18  22000394   9:15    1      3    False
19  22000394   8:53    1      1    False
20  22000394   8:33    1      1    False
21  22000394   8:22    1      1    False
22  22000394   8:16    1      2    False
23  22000394   8:16    1      4    False
24  22000394   8:12    1      2    False

One option is using set with transform; speed wise I expect Bert2ME's solution to be faster:

df.assign(UPDATED = df.groupby(grouper)
                      .EVENT
                      .transform(lambda x: set(x) >= {1,3,6}) 
                                           & df.EVENT.eq(1))

     GAME_ID   TIME  PER  EVENT  UPDATED
0   22000394  12:00    1     12    False
1   22000394  12:00    1     10    False
2   22000394  11:36    1      1    False
3   22000394  11:24    1      1    False
4   22000394  11:04    1      1    False
5   22000394  10:41    1      1    False
6   22000394  10:30    1      2    False
7   22000394  10:29    1      4    False
8   22000394  10:17    1      1    False
9   22000394  10:01    1      1    False
10  22000394   9:48    1      2    False
11  22000394   9:46    1      4    False
12  22000394   9:42    1      6    False
13  22000394   9:42    1      3    False
14  22000394   9:42    1      3    False
15  22000394   9:25    1      1    False
16  22000394   9:15    1      1     True
17  22000394   9:15    1      6    False
18  22000394   9:15    1      3    False
19  22000394   8:53    1      1    False
20  22000394   8:33    1      1    False
21  22000394   8:22    1      1    False
22  22000394   8:16    1      2    False
23  22000394   8:16    1      4    False
24  22000394   8:12    1      2    False

通过

若水微香 2025-02-09 20:29:13

不要创建新的历史。
您可以使用使用,而不是react-router-dom,然后获取当前的历史记录。会这样:

  const history = useHistory();

  function doRedirect() {
    history.push("/login");
  }

Do not create new history.
You can use useHistory that exported from react-router-dom instead and get the current history. It would be something like this:

  const history = useHistory();

  function doRedirect() {
    history.push("/login");
  }

无法在React中渲染组件

若水微香 2025-02-09 11:13:01

取决于您问谁。


ecmascript 2021语言标准说:

7.2.2 isarray(gright)

抽象操作isArray采用参数参数。它在调用时执行以下步骤:

  1. 如果类型(参数)不是对象,返回false
  2. 如果参数为array exotic对象,请返回true
  3. 如果参数为代理 exotic对象,则
    一个。如果参数。[[proxyHandler]]null,抛出type> typeerror异常。
    b。令target be 参数。[[ProxyTarget]]
    c。返回? isarray(target)
  4. 返回false

(对此平台进行重新格式化)

可能会帮助知道内置的JavaScript对象具有发动机内部的表示形式。这就是上述伪代码所指的。 JavaScript无法访问伪代码所指的某些内容,但引擎确实可以访问。


另一方面,JavaScript库通常会根据图书馆的年龄,想成为多大的坚固性或目标是什么来做几件事之一。由于它们无法访问引擎内部设备,因此这些猜测很可能在现实世界中正确正确:

  • 现代库将其推迟到array.isarray()
  • polyfills通常使用object 。
  • ​obj.length!= null&amp;&amp; obj.push type ==='函数'
  • 基于构造函数:boolean(obj)&amp;&amp; obj.constructor === array
  • 实例基于:obj实例数组

这不是一个详尽的列表。

Depends on who you ask.


The ECMAScript 2021 Language standard says:

7.2.2 IsArray ( argument )

The abstract operation IsArray takes argument argument. It performs the following steps when called:

  1. If Type(argument) is not Object, return false.
  2. If argument is an Array exotic object, return true.
  3. If argument is a Proxy exotic object, then
    a. If argument.[[ProxyHandler]] is null, throw a TypeError exception.
    b. Let target be argument.[[ProxyTarget]].
    c. Return ? IsArray(target).
  4. Return false.

(Reformatted for this platform)

It may help to know that built-in Javascript objects have a representation that is internal to the engine. This is what the above pseudo-code is referring to. Javascript doesn't have access to some of the things that the pseudo-code refers to but the engine does.


Javascript libraries, on the other hand, typically do one of several things depending on how old the library is, how robust they want to be, or what their goal is. Since they don't have access to the engine internals, these are guesses with a high probability of being correct in the real world:

  • Modern libraries defer to Array.isArray()
  • Polyfills typically use Object.prototype.toString.call(obj) === '[object Array]' (standardized by Mozilla, I believe)
  • Duck typing: Boolean(obj) && obj.length != null && typeof obj.push === 'function'
  • Constructor based: Boolean(obj) && obj.constructor === Array
  • instanceof based: obj instanceof Array

This is not an exhaustive list.

array.isarray()方法如何检查数组?

若水微香 2025-02-09 09:42:36

单线:

filtered = df[df.apply(lambda col: col.isin(pd.Series(dict_filter.get(col.name, [])))).all(axis=1)]

输出:

>>> filtered
       A       B       C
id                      
2   high  medium  bottom

One-liner:

filtered = df[df.apply(lambda col: col.isin(pd.Series(dict_filter.get(col.name, [])))).all(axis=1)]

Output:

>>> filtered
       A       B       C
id                      
2   high  medium  bottom

使用dict中的值过滤熊猫数据框列和行

若水微香 2025-02-09 01:13:58
func imagePickerController(_ picker: UIImagePickerController, didFinishPickingMediaWithInfo info: [UIImagePickerController.InfoKey : Any]) {
       guard let image = info[.originalImage] as? UIImage else {
                  fatalError("Expected a dictionary containing an image, but was provided the following: \(info)")
              }
// make use of the image
   }
func imagePickerController(_ picker: UIImagePickerController, didFinishPickingMediaWithInfo info: [UIImagePickerController.InfoKey : Any]) {
       guard let image = info[.originalImage] as? UIImage else {
                  fatalError("Expected a dictionary containing an image, but was provided the following: \(info)")
              }
// make use of the image
   }

如何从ImagePickerController中返回图像?

若水微香 2025-02-08 18:04:07

您可以下载并将图像存储为对象,不必将其放入您的HTML中。

您可以使用 drawimage function drawimage ,这使您可以按选择的比例绘制图像。您可以使用相同的功能在画布上绘制水印。

一旦您对画布上的图像感到满意,就可以使用 toblob 功能并上传到您的服务器

You can download and store your image as an Image object, it's not necessary to have it in your html.

You can draw it on a canvas using drawImage function, which let you draw the image with the proportion of your choice. You can use the same function to draw a watermark on the canvas.

Once you are happy with the image on the canvas you can get it as a blob using toBlob function and upload to your server

JavaScript画布,调整图像大小

若水微香 2025-02-08 16:27:55

解决!问题是变量fpath = ...在我的〜/.zprofile中声明。感谢@Dan的提示。

Solved! The problem was a variable FPATH=... declared in my ~/.zprofile. Thanks to @dan for the hint.

(HomebrewStalled)ZSH不包括正确的“ /.../ functions”的正确途径。在FPATH变量中

若水微香 2025-02-08 09:40:14

如果您真的想使用lambda,则可以在评论中可以使用您的答案。


private ArrayList<Car> carsInStore ;

public boolean checkIfcarInStore(Car c) {   
    return carsInStore.stream().anyMatch(c::equals);
    }
 }

If you really want to use lambda, this should work else you have your answer within your comment.


private ArrayList<Car> carsInStore ;

public boolean checkIfcarInStore(Car c) {   
    return carsInStore.stream().anyMatch(c::equals);
    }
 }

检查每个对象,在满足第一个条件时返回true,否则返回false

若水微香 2025-02-08 04:50:28

如果您使用jQuery,我建议您使用此插件:

https://github.com/carhartl/jquery-cookie br>
https://github.com/carhartl/jquery-cookie/jquery-cookie/jquie/blob/blob/blob/master/master/jquery/jquery。 cookie.js

<script type="text/javascript"
 src="//cdnjs.cloudflare.com/ajax/libs/jquery-cookie/1.4.1/jquery.cookie.min.js">

这样您就可以阅读这样的cookie:

var value = $.cookie("obligations");

另外,您也可以写cookie:

$.cookie('obligations', 'new_value');
$.cookie('obligations', 'new_value', { expires: 14, path: '/' });

删除cookie:

$.removeCookie('obligations');

If you use jQuery I recommend you to use this plugin:

https://github.com/carhartl/jquery-cookie
https://github.com/carhartl/jquery-cookie/blob/master/jquery.cookie.js

<script type="text/javascript"
 src="//cdnjs.cloudflare.com/ajax/libs/jquery-cookie/1.4.1/jquery.cookie.min.js">

So you can read cookie like this:

var value = $.cookie("obligations");

Also you can write cookie:

$.cookie('obligations', 'new_value');
$.cookie('obligations', 'new_value', { expires: 14, path: '/' });

Delete cookie:

$.removeCookie('obligations');

用名字获取cookie

若水微香 2025-02-08 02:07:01

使用Arrayschema应该为您解决此问题

@ApiResponse(
    content = {@Content(
        array = @ArraySchema(schema = @Schema(implementation = ScanDTO.class))
    )}
)

Using ArraySchema should solve this for you

@ApiResponse(
    content = {@Content(
        array = @ArraySchema(schema = @Schema(implementation = ScanDTO.class))
    )}
)

有什么方法可以使用列表&lt; myclass&gt; .class?

若水微香 2025-02-08 01:24:14

下面的代码应完成工作。但是我不明白为什么您首先没有单独创建这些项目。

foreach ($arr as $i => $data) {
    if (!str_contains($data['id'], ',')) continue;

    $items = explode(',', $data['item']);

    foreach(explode(',', $data['id']) as $i => $id) {
        $new = ['id' => $ids[$i], 'item' => $items[$i]];

        if ($i) $arr[] = $new;
        else $arr[$i] = $new;
    }
}
var_export($arr);

输出:( demo

Warning: Undefined variable $ids

Warning: Trying to access array offset on null

Warning: Undefined variable $ids

Warning: Trying to access array offset on null
array (
  0 => 
  array (
    'id' => NULL,
    'item' => 'Item A',
  ),
  1 => 
  array (
    'id' => 3,
    'item' => 'Item C',
  ),
  2 => 
  array (
    'id' => NULL,
    'item' => ' Item B',
  ),
)

The code down below should do the job. But I didn't understand why you didn't create those items separately in the first place.

foreach ($arr as $i => $data) {
    if (!str_contains($data['id'], ',')) continue;

    $items = explode(',', $data['item']);

    foreach(explode(',', $data['id']) as $i => $id) {
        $new = ['id' => $ids[$i], 'item' => $items[$i]];

        if ($i) $arr[] = $new;
        else $arr[$i] = $new;
    }
}
var_export($arr);

Output: (Demo)

Warning: Undefined variable $ids

Warning: Trying to access array offset on null

Warning: Undefined variable $ids

Warning: Trying to access array offset on null
array (
  0 => 
  array (
    'id' => NULL,
    'item' => 'Item A',
  ),
  1 => 
  array (
    'id' => 3,
    'item' => 'Item C',
  ),
  2 => 
  array (
    'id' => NULL,
    'item' => ' Item B',
  ),
)

当遇到逗号偏置的值对时,在2D数组中的反抛光行值

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