笙痞

文章 评论 浏览 30

笙痞 2025-02-21 00:45:58

不知道您会使用什么日期。尽管您可以使用array_zip,序列和日间隔生成。

df=spark.createDataFrame([(1,'0001-01-01','9999-12-31')], ('id','start','end'))
df.withColumn('start',to_date('start')).withColumn('end',to_date('end')).selectExpr('id',"inline(arrays_zip(sequence(start,end,interval 1 day)))")

Not sure what you would use the dates for. You can though generate using array_zip, sequence and day interval.

df=spark.createDataFrame([(1,'0001-01-01','9999-12-31')], ('id','start','end'))
df.withColumn('start',to_date('start')).withColumn('end',to_date('end')).selectExpr('id',"inline(arrays_zip(sequence(start,end,interval 1 day)))")

如何在Apache Spark或Spark SQL中的两个范围之间生成日期?

笙痞 2025-02-20 15:54:37
string(REGEX MATCHALL "[a-zA-Z]+\ |[a-zA-Z]+$" SEXY_LIST "${SEXY_STRING}")
string(REGEX MATCHALL "[a-zA-Z]+\ |[a-zA-Z]+$" SEXY_LIST "${SEXY_STRING}")

用CMAKE将字符串分为列表的常见方法是什么?

笙痞 2025-02-20 12:09:49

我如何使它更有效,尽可能便携。

  1. 更多错误检查。示例:失败的fopen()很容易发生。

  2. 意识到在fopen()中会发生大量时间,本地代码可能会改善时间很少。

  3. 避免重写前缀。

  4. 使用辅助功能。

示例:

// Return count of successfully written files.
int create_many_files(const char *prefix, int count, const char *suffix) {
  int n;
  int len = snprintf(NULL, 0, "%s%n%d%s", prefix, &n, count, suffix);
  if (len < 0) {
    return 0;
  }

  // Use len to determine longest name and use a VLA or allocation.
  // Consider using a fixed array when len is not too big.
  char *filename = malloc((size_t)len + 1u);
  if (filename == NULL) {
    return 0;
  }
  strcpy(filename, prefix);
  char *offset = filename + n;

  for (int i = 0; i < count; i++) {
    sprintf(offset, "%d%s", i + 1, suffix);
    FILE *fp = fopen(filename, "w");
    if (fp == NULL) {
      free(filename);
      return i;
    }
    fprintf(fp, "This is file no %d", i + 1);
    fclose(fp);
  }

  free(filename);
  return count;
}

其他潜在错误检查:

  • prefix == null
  • cout&lt; 0
  • suffix == null
  • fprintf()&lt; 0
  • fclose()!= 0

how can I make this more efficient and as portable as possible.

  1. More error checking. Example: a failed fopen() can readily occur.

  2. Realize that a huge amount of time will occur in fopen() and local code likely will have scant time improvements.

  3. Avoid re-writing the prefix.

  4. Use a helper function.

Example:

// Return count of successfully written files.
int create_many_files(const char *prefix, int count, const char *suffix) {
  int n;
  int len = snprintf(NULL, 0, "%s%n%d%s", prefix, &n, count, suffix);
  if (len < 0) {
    return 0;
  }

  // Use len to determine longest name and use a VLA or allocation.
  // Consider using a fixed array when len is not too big.
  char *filename = malloc((size_t)len + 1u);
  if (filename == NULL) {
    return 0;
  }
  strcpy(filename, prefix);
  char *offset = filename + n;

  for (int i = 0; i < count; i++) {
    sprintf(offset, "%d%s", i + 1, suffix);
    FILE *fp = fopen(filename, "w");
    if (fp == NULL) {
      free(filename);
      return i;
    }
    fprintf(fp, "This is file no %d", i + 1);
    fclose(fp);
  }

  free(filename);
  return count;
}

Other potential error checks:

  • prefix == NULL
  • cout < 0
  • suffix == NULL
  • fprintf() < 0
  • fclose() != 0

在C中创建1000个文本文件

笙痞 2025-02-20 07:08:20

我找到了一个解决方案

connect()
  async function connect() {
  try {
    const con = await mysql.createConnection({
      "host": ,
      "port": ,
      "user": ,
      "password" : ,
      "database" : 
    })

    const insertsql = await con.query(
      "INSERT INTO `b_info`(`title`, `subtitle`, `item_price`, `ros_info`, `img`, `p`, `info_items`, `link`, `theopeninfo2`) VALUES (?, ?, ?, ?, ?, ?, ?, ?, ?)",
      [title , subtitle , item_price , ros_info , img , p , info_items , link , theopeninfo2 ]
    )
    console.table(insertsql[0])
  }
  catch(ex)
  {
    console.error(ex)
  }

I found a solution

connect()
  async function connect() {
  try {
    const con = await mysql.createConnection({
      "host": ,
      "port": ,
      "user": ,
      "password" : ,
      "database" : 
    })

    const insertsql = await con.query(
      "INSERT INTO `b_info`(`title`, `subtitle`, `item_price`, `ros_info`, `img`, `p`, `info_items`, `link`, `theopeninfo2`) VALUES (?, ?, ?, ?, ?, ?, ?, ?, ?)",
      [title , subtitle , item_price , ros_info , img , p , info_items , link , theopeninfo2 ]
    )
    console.table(insertsql[0])
  }
  catch(ex)
  {
    console.error(ex)
  }

我试图从puppeteer(async)插入数据,但我获得了“连接”。

笙痞 2025-02-19 18:01:19

我认为弄清楚此方法是否存在任何理由的关键是确认在调用该方法之前是否启动任务。

  1. 如果未启动任务,则该方法可能应称为〜当alleSequention时。
  2. 如果启动了任务,则此方法似乎不会添加任何好东西。

无论如何,我都会反对将此方法作为扩展方法:

  1. 在自动完成提示中显示此方法迟早会使某人通过已经启动任务,并认为任务将被顺序执行,但这是此方法的呼叫者任务启动时负责谁。
  2. 甚至官方时,不是扩展方法。

I think the key to figuring out if this method has any reason to exist is to confirm if the tasks are started before the method is called.

  1. If the tasks are not started then the method should probably be called ~WhenAllSequential.
  2. If the tasks are started then this method seems not to add anything good.

In any case, I would argue against having this method as an extension method:

  1. Having this method shown in auto-completion hints will sooner or later make someone pass already started tasks and think the tasks will be executed sequentially but it's the caller of this method who is in charge when the tasks are started.
  2. Even the official WhenAll is not an extension method.

我的同事是否复制了任务的确切行为。

笙痞 2025-02-19 16:12:38

您的构建应该看起来像这样。有一些填充。不要使用Appbar,因为它太高了。

  @override
  Widget build(BuildContext context) {
    return SafeArea(
      child: Scaffold(
        backgroundColor: Color(0xffF24004),
        body: Column(
          children: [
            SizedBox(height: 20,),
            Text(
              "P O M O D O N E",
              style: TextStyle(
                fontWeight: FontWeight.bold,
                fontSize: 30,
                color: Color(0xffF3F5F4),
              ),
            ),
            Container(
              alignment: Alignment.center,
              decoration: BoxDecoration(
                  image: DecorationImage(
                      image: AssetImage("lib/img/pomodoro.png"),
                      fit: BoxFit.cover)),
            ),
            Spacer(),
            FloatingActionButton(onPressed: () {}),

          ],
        ),
      ),
    );
  }

Your build should looks like this. With some padding. Don't use the appBar because it's too high.

  @override
  Widget build(BuildContext context) {
    return SafeArea(
      child: Scaffold(
        backgroundColor: Color(0xffF24004),
        body: Column(
          children: [
            SizedBox(height: 20,),
            Text(
              "P O M O D O N E",
              style: TextStyle(
                fontWeight: FontWeight.bold,
                fontSize: 30,
                color: Color(0xffF3F5F4),
              ),
            ),
            Container(
              alignment: Alignment.center,
              decoration: BoxDecoration(
                  image: DecorationImage(
                      image: AssetImage("lib/img/pomodoro.png"),
                      fit: BoxFit.cover)),
            ),
            Spacer(),
            FloatingActionButton(onPressed: () {}),

          ],
        ),
      ),
    );
  }

如何将PNG图像分层在颤动中的背景颜色上?

笙痞 2025-02-19 06:35:58

添加kotlinversion =“ 1.6.0” in android/build.gradle

    buildscript {
    ext {
        buildToolsVersion = "31.0.0"
        minSdkVersion = 21
        compileSdkVersion = 31
        targetSdkVersion = 31
        kotlinVersion = "1.6.0"

        if (System.properties['os.arch'] == "aarch64") {
            // For M1 Users we need to use the NDK 24 which added support for aarch64
            ndkVersion = "24.0.8215888"
        } else {
            // Otherwise we default to the side-by-side NDK version from AGP.
            ndkVersion = "21.4.7075529"
        }
    }
    repositories {
        google()
        mavenCentral()
    }
    dependencies {
        classpath("com.android.tools.build:gradle:7.0.4")
        classpath("com.facebook.react:react-native-gradle-plugin")
        classpath("de.undercouch:gradle-download-task:4.1.2")
        // NOTE: Do not place your application dependencies here; they belong
        // in the individual module build.gradle files
    }
}

Add kotlinVersion = "1.6.0" in android/build.gradle

    buildscript {
    ext {
        buildToolsVersion = "31.0.0"
        minSdkVersion = 21
        compileSdkVersion = 31
        targetSdkVersion = 31
        kotlinVersion = "1.6.0"

        if (System.properties['os.arch'] == "aarch64") {
            // For M1 Users we need to use the NDK 24 which added support for aarch64
            ndkVersion = "24.0.8215888"
        } else {
            // Otherwise we default to the side-by-side NDK version from AGP.
            ndkVersion = "21.4.7075529"
        }
    }
    repositories {
        google()
        mavenCentral()
    }
    dependencies {
        classpath("com.android.tools.build:gradle:7.0.4")
        classpath("com.facebook.react:react-native-gradle-plugin")
        classpath("de.undercouch:gradle-download-task:4.1.2")
        // NOTE: Do not place your application dependencies here; they belong
        // in the individual module build.gradle files
    }
}

class&#x27; kotlin.unit&#x27;用不兼容的Kotlin编译。其元数据的二进制版本为1.6.0,预期版本为1.1.15

笙痞 2025-02-19 02:42:37

如本评论所建议: https://stackoverflow.com/a/68394617/2371266 设置脚本壳PowerShell 似乎有效。之后,我尝试删除脚本壳配置并重试脚本,然后再次遇到原始错误,因此我假设它默认为其他错误,这使事情变得有些怪异。我想我现在就把它指向Powershell。

As suggested in this comment: https://stackoverflow.com/a/68394617/2371266, running npm config set script-shell powershell seems to have worked. I tried deleting the script-shell config afterwards and retrying and got the original error again so I'm assuming it is defaulting to something else and that's making things freak out a bit. I guess I'll just keep it pointed at powershell for now.

NPM开始使用Create-React-App创建新的React项目后失败

笙痞 2025-02-18 22:23:31

问题似乎是由于对模板功能的误解而出现的。它主要用于显示目的而不是过滤目的,并且您面临着一个过滤问题:

我想过滤满足特定条件的列表的3个第一个元素

是您的主要问题,并且模板框架在解决您的问题时会很差,而这正是Python和Django的Orm擅长的。

为什么不首先过滤Django,然后在模板中显示?例如,如下:

...
products = queryset.filter(abc=condition)[:3]
context = Context({ 'products': products }) 
return HttpResponse(template.render(context))

The problem seems to arise due to a misconception of the templating functionality. It is to be used mainly for display purposes and not filtering purposes, and you are facing a filtering problem:

I want to filter the 3 first elements of a list that fulfill a specific condition

is your main issue and templating frameworks will be poor at solving your issue, whereas this is exactly what python and Django's ORM are good at.

Why don't you first filter in Django, then display in template? For example as follows:

...
products = queryset.filter(abc=condition)[:3]
context = Context({ 'products': products }) 
return HttpResponse(template.render(context))

如何在Django模板中进行柜台上的循环中断?

笙痞 2025-02-18 18:45:09

也许这会有所帮助:

with calcflag as (
select customer_id, IFF( sum(flag_1) over (PARTITION by customer_id order by order_date rows between 3 preceding and 1 preceding) = 3, 1, 0 ) as new_flag 
from tmp_Test)
select customer_id, max(new_flag) flag_2
from calcflag
group by 1
order by 1;

+-------------+--------+
| CUSTOMER_ID | FLAG_2 |
+-------------+--------+
|           1 |      1 |
|           2 |      0 |
+-------------+--------+

Maybe this can be help:

with calcflag as (
select customer_id, IFF( sum(flag_1) over (PARTITION by customer_id order by order_date rows between 3 preceding and 1 preceding) = 3, 1, 0 ) as new_flag 
from tmp_Test)
select customer_id, max(new_flag) flag_2
from calcflag
group by 1
order by 1;

+-------------+--------+
| CUSTOMER_ID | FLAG_2 |
+-------------+--------+
|           1 |      1 |
|           2 |      0 |
+-------------+--------+

如何通过连续日期变量检查值

笙痞 2025-02-18 06:17:53

这是一种可能的方法的草图(如果您可以将输入调整为8个字节的倍数,或者您知道前方的向量大小,则可以对此进行很多改进。但是它使您有了一个想法,如何使用爆米花。 (对于32bits体系结构,模式是相同的,但性能更少)

// #include <intrin.h> // for popcnt which counts number of set bits in a variable
#include <array>
#include <iostream>
#include <vector>
#include <cstdint>
#include <memory>


// work on copy on purpose so we can pad memory of vector to 64bit alignment (kind of a quick hack for now, the extra memory (re)allocations might slow you down too much)
auto vertical_popcount(std::vector<std::uint8_t> values)
{
    // use 64bit architecture to do 8 values per cycle
    static constexpr std::array<std::uint64_t, 8> masks
    {
        0x0101010101010101,
        0x0202020202020202,
        0x0404040404040404,
        0x0808080808080808,
        0x1010101010101010,
        0x2020202020202020,
        0x4040404040404040,
        0x8080808080808080
    };

    //using an array instead of vector safes at least one dynamic allocation
    std::array<std::size_t, 8> counts{};

    // align data to multiple of 8 bytes
    // add a few extra 0 bytes, they wont impact the counting
    for (std::size_t n = 0; n < values.size() % 8; ++n)
        values.push_back(0);

    // make a uint64_t pointer into 8 bit data
    // to pickup 8 bytes at a time for masking
    auto ptr = reinterpret_cast<std::uint64_t*>(values.data());
    for (std::size_t n = 0; n < values.size() / 8; ++n)
    {
        for (std::size_t m = 0; m < 8; ++m)
        {
            // mask 8 bytes at a time
            auto masked_value = (*ptr & masks[m]);
            
            // count bits of 8 uint8_t's in one loop
            auto bitcount = std::popcount(masked_value);
            counts[m] += bitcount;
        }
        ++ptr;
    }

    
    return counts;
}

int main() 
{
    std::vector<uint8_t> data({ 0b00001001, 0b00000100, 0b00000101 });
    auto result = vertical_popcount(data);
    std::cout << "occurrence of bits: " << result[0] << ", " << result[1] << ", " << result[2] << ", " << result[3] << ", " << result[4] << ", " << result[5] << ", " << result[6] << ", " << result[7] << "\n";
    return 0;
}

Here is a sketch of a possible approach (you can improve on this a lot if you can tweak your input to multiples of 8 bytes, or if you know the size of your vectors up front). But it gives you an idea how to use popcount. (For 32bits architecture the pattern is the same but you get less performance)

// #include <intrin.h> // for popcnt which counts number of set bits in a variable
#include <array>
#include <iostream>
#include <vector>
#include <cstdint>
#include <memory>


// work on copy on purpose so we can pad memory of vector to 64bit alignment (kind of a quick hack for now, the extra memory (re)allocations might slow you down too much)
auto vertical_popcount(std::vector<std::uint8_t> values)
{
    // use 64bit architecture to do 8 values per cycle
    static constexpr std::array<std::uint64_t, 8> masks
    {
        0x0101010101010101,
        0x0202020202020202,
        0x0404040404040404,
        0x0808080808080808,
        0x1010101010101010,
        0x2020202020202020,
        0x4040404040404040,
        0x8080808080808080
    };

    //using an array instead of vector safes at least one dynamic allocation
    std::array<std::size_t, 8> counts{};

    // align data to multiple of 8 bytes
    // add a few extra 0 bytes, they wont impact the counting
    for (std::size_t n = 0; n < values.size() % 8; ++n)
        values.push_back(0);

    // make a uint64_t pointer into 8 bit data
    // to pickup 8 bytes at a time for masking
    auto ptr = reinterpret_cast<std::uint64_t*>(values.data());
    for (std::size_t n = 0; n < values.size() / 8; ++n)
    {
        for (std::size_t m = 0; m < 8; ++m)
        {
            // mask 8 bytes at a time
            auto masked_value = (*ptr & masks[m]);
            
            // count bits of 8 uint8_t's in one loop
            auto bitcount = std::popcount(masked_value);
            counts[m] += bitcount;
        }
        ++ptr;
    }

    
    return counts;
}

int main() 
{
    std::vector<uint8_t> data({ 0b00001001, 0b00000100, 0b00000101 });
    auto result = vertical_popcount(data);
    std::cout << "occurrence of bits: " << result[0] << ", " << result[1] << ", " << result[2] << ", " << result[3] << ", " << result[4] << ", " << result[5] << ", " << result[6] << ", " << result[7] << "\n";
    return 0;
}

是否有任何算法或位骇客来计算每个n位的发生在整数上的数组?

笙痞 2025-02-18 06:13:31

您需要添加到路径中的剧作家。

但是,做到这一点的更好方法(如果将Python添加到您的路径中)而不将其添加到路径中的方法是运行:

python -m playwright install

这将剧作家模块作为脚本运行。使用Python -H有关这些标志的更多信息,以及Python -M playwright有关Playwright专门支持的标志的更多信息

You need playwright added to your PATH.

However, a better way to do this (if python is added to your path) without adding it to your PATH is by running:

python -m playwright install

This runs the playwright module as a script. Use python -h for more information on these flags, and python -m playwright for more information on the flags supported specifically by playwright

安装Microsoft剧作家时出错

笙痞 2025-02-17 22:46:59

如果要将DOB与当前日期进行比较,则需要首先将其变成日期对象:

function invalidDOB(dateofbirth){
 var pattern = /^\d{4}-(0[1-9]|1[0-2])-(0[1-9]|[12][0-9]|3[01])$/, dob;
 var date= new Date();
 if (dateofbirth == "" || dateofbirth == null||!pattern.test(dateofbirth)) {
  return "Invalid date format, should be: yyyy-mm-dd";
 }
 else if((dob=new Date(dateofbirth))>date){
  return "The date is in the future: "+dob.toLocaleString();
 }
 else{
  return undefined;
 }
}

const test=["01.02.1974","1974-02-01","02/01/1974","1 Feb 1974","1974-02-31","2023-02-31"];

test.forEach(d=>console.log(`${d}: ${invalidDOB(d)??"OK"}`));

正如最后两个测试用例所示,这也不是“完美”:JavaScript接受类似“ 1974-02-31”之类的输入,并将其解释为“ 1974年2月28日之后的3天”,即当年3月3日。

If you want to compare the DOB with the current date you need to turn it into a date object first:

function invalidDOB(dateofbirth){
 var pattern = /^\d{4}-(0[1-9]|1[0-2])-(0[1-9]|[12][0-9]|3[01])$/, dob;
 var date= new Date();
 if (dateofbirth == "" || dateofbirth == null||!pattern.test(dateofbirth)) {
  return "Invalid date format, should be: yyyy-mm-dd";
 }
 else if((dob=new Date(dateofbirth))>date){
  return "The date is in the future: "+dob.toLocaleString();
 }
 else{
  return undefined;
 }
}

const test=["01.02.1974","1974-02-01","02/01/1974","1 Feb 1974","1974-02-31","2023-02-31"];

test.forEach(d=>console.log(`${d}: ${invalidDOB(d)??"OK"}`));

As the last two test cases show, this is not yet "perfect" either: JavaScript accepts input like "1974-02-31" and will interpret it as "3 days after 28 February 1974", which is 3 March of that year.

验证JavaScript中出生日期

笙痞 2025-02-17 21:19:48

也许更好的方法是将令牌存储在因此,您无需访问Redux商店即可获得它。只需获取令牌并以这种方式存储的方式:

localStorage.setItem('token', JSON.stringify(token))

以后在您的服务中,只需取回这样的方法:

const token = JSON.parse(localStorage.getItem('token'));

如果您不感到安全地将令牌存储在本地存储中,还有 secure-ls package,该软件包允许对保存到本地存储的数据进行加密。

Maybe a better approach would be to store the token on the localStorage so that you don't need to access the Redux store to obtain it. Just fetch the token and store it this way:

localStorage.setItem('token', JSON.stringify(token))

Later in your service, just retrieve like this:

const token = JSON.parse(localStorage.getItem('token'));

If you don’t feel safe storing the token in the local storage, there's also the secure-ls package, that allows encryption for the data saved to the local storage.

React-Boilerplate:实现JWT身份验证。尝试从服务派遣行动

笙痞 2025-02-17 13:16:58

确保检查安全转移是否启用。转到Azure存储帐户 - &GT;设置转到您会找到安全传输的配置。如果没有,请启用它。安全传输仅通过安全连接允许向存储帐户的请求提供存储帐户的安全性。

是否有其他可用选项?

有不同的替代选项可用:直接从Azure Databricks笔记本中读取Azure Blob存储文件。

  • 选项1: 从Gauri Mahajan

    访问Azure Blob存储

  • Gauri -sources/azure/azure-storage#access-azure-blob-storage-inder-the-dataframe-api“ rel =“ nofollow noreferrer”>使用Microsoft提供的SAS代币访问Azure Blob Storage >

Make sure to check whether Secure transfer Enable or not. Go to azure storage account -> Settings go to the configuration you will find Secure transfer. If not, enable it. The secure transfer provides the security of your storage account by only allowing requests to the storage account by a secure connection.

Is there any alternative options available ?

There are different alternative options available: Reading Azure Blob Storage file directly from Azure Databricks Notebook.

我如何从Azure Databricks笔记本中读取Azure Blob存储文件Dieclty

更多

推荐作者

櫻之舞

文章 0 评论 0

弥枳

文章 0 评论 0

m2429

文章 0 评论 0

野却迷人

文章 0 评论 0

我怀念的。

文章 0 评论 0

    我们使用 Cookies 和其他技术来定制您的体验包括您的登录状态等。通过阅读我们的 隐私政策 了解更多相关信息。 单击 接受 或继续使用网站,即表示您同意使用 Cookies 和您的相关数据。
    原文