's questions

谷歌表格图片

如何在 A1、B1、C1 和 D1（分别）处使用数组公式来提供所需的输出列，如图所示。

Google 表格链接

注意： “主标题”（亲戚、朋友、商业朋友）和“姓名”之间的区别在于“主标题”在开始和结束处有 1 个空格字母。

A、B、C 列三者相同，唯一的区别在于“主标题”A 列提供字母，B 列提供数字，C 列提供罗马数字。

序列编号条件：
（1）序列编号完全基于列：NAMES 和 PAX。
（2）如果 PAX 为 0，则输出为空白。

对于 A、B、C 列：
当 PAX 为空白时：
-> 如果它是“主标题”，则输出为字母 OR
-> 如果它是“名称”，则输出为空白。

对于 D 列：
当 PAX 为空白时：
-> 如果它是“主标题”，则输出为空白 OR
-> 如果它是“名称”，则输出为序列号，直到下一个“主标题”，然后再次从 1 重新开始，直到下一个“主标题”等等。

编辑：
列名称和 PAX 是输入，其余的是所需的输出。

数据：

我有一个复杂的嵌套字典结构，需要根据嵌套字典中的值删除元素。我的字典如下所示：

my_dict = {
    'item1': {'name': 'Apple', 'price': 1.0, 'category': {'id': 1, 'name': 'Fruit'}},
    'item2': {'name': 'Banana', 'price': 0.5, 'category': {'id': 1, 'name': 'Fruit'}},
    'item3': {'name': 'Carrot', 'price': 0.75, 'category': {'id': 2, 'name': 'Vegetable'}},
    'item4': {'name': 'Broccoli', 'price': 1.5, 'category': {'id': 2, 'name': 'Vegetable'}}
}

我想过滤这个词典，使其仅包含属于“水果”类别的项目。我尝试了以下代码：

new_dict = {}
for key, value in my_dict.items():
    if value['category']['name'] == 'Fruit':
        new_dict[key] = value
print(new_dict)

这是可行的，但我想知道是否有更简洁或更 Pythonic 的方法来实现这一点，也许使用字典理解或类似的过滤函数filter()。

鉴于

class Foo {
  x = 1;
  y = 2;
}
class Bar extends Foo {
  override x = 11;
  z = 3;
}

是否可以自动推导出Foo给定值Bar？比如Superclass<Bar> === Foo？

我目前正在创建一个应用程序，其中我编写了一个函数以便能够从具有特定扩展名的 Web 获取图像，但这里有一个错误，我不知道如何解决帮帮我

文字错误

/C:/Users/sina%20saffar/AppData/Local/Pub/Cache/hosted/pub.dev/flutter_dropzone_web-3.0.13/lib/flutter_dropzone_plugin.dart:64:25: Error: The return type of the method 'FlutterDropzonePlugin.pickFiles' is 'Future<List<dynamic>>', which does not match the return type, 'Future<List<DropzoneFileInterface>>', of the overridden method, 'FlutterDropzonePlatform.pickFiles'.
 - 'Future' is from 'dart:async'.
 - 'List' is from 'dart:core'.
 - 'DropzoneFileInterface' is from 'package:flutter_dropzone_platform_interface/dropzone_file_interface.dart' ('/C:/Users/sina%20saffar/AppData/Local/Pub/Cache/hosted/pub.dev/flutter_dropzone_platform_interface-2.2.0/lib/dropzone_file_interface.dart').
Change to a subtype of 'Future<List<DropzoneFileInterface>>'.
  Future<List<dynamic>> pickFiles(bool multiple,
                        ^
/C:/Users/sina%20saffar/AppData/Local/Pub/Cache/hosted/pub.dev/flutter_dropzone_platform_interface-2.2.0/lib/flutter_dropzone_platform_interface.dart:97:39: Context: This is the overridden method ('pickFiles').
  Future<List<DropzoneFileInterface>> pickFiles(bool multiple,

选择本地图像功能：

 Future<void> selectLocalImages() async {
    final files = await dropzoneViewController.pickFiles(
      multiple: true,
      mime: ['image/jpeg', 'image/png'],
    );

    if (files.isNotEmpty) {
      for (var file in files) {
        if (file is html.File) {
          final bytes = await dropzoneViewController.getFileData(file);
          final image = ImageModel(
            url: '',
            file: file,
            folder: '',
            filename: file.name,
            localImageToDisplay: Uint8List.fromList(bytes),
          );
          selectedImagesToUpload.add(image);
        }
      }
    }
  }

我在 Flutter iOS 应用中尝试使用 Firebase Messaging 获取令牌时遇到错误。错误消息如下：

FLTFirebaseMessaging: An error occurred while calling method Messaging#getToken, errorOrNil => {    NSLocalizedFailureReason = "Invalid fetch response, expected 'token' or 'Error' key";}

[ERROR:flutter/runtime/dart_vm_initializer.cc(41)] Unhandled Exception: [firebase_messaging/unknown] An unknown error has occurred.
#0      StandardMethodCodec.decodeEnvelope (package:flutter/src/services/message_codecs.dart:651:7)
#1      MethodChannel._invokeMethod (package:flutter/src/services/platform_channel.dart:334:18)
<asynchronous suspension>
#2      MethodChannel.invokeMapMethod (package:flutter/src/services/platform_channel.dart:534:43)
<asynchronous suspension>
#3      MethodChannelFirebaseMessaging.getToken (package:firebase_messaging_platform_interface/src/method_channel/method_channel_messaging.dart:248:11)
<asynchronous suspension>
#4      FCMToken.getFcm (package:instahire/firebase/fcm_token.dart:5:21)
<asynchronous suspension>
#5      SplashController.getToken (package:instahire/controllers/splash_controller.dart:42:23)
<asynchronous suspension>

直到昨天它还运行正常，突然出现此错误。如何解决？

我正在我的项目中设置 Stripe 用于付款，但在管理客户方面遇到了一些问题。

我试图做的是创建一个新客户，并在用户创建结帐会话时将条纹客户 ID 存储在我的数据库中。问题是，在用户完成付款后，正在创建另一个客户。

@app.post('/create-checkout-session')
async def create_checkout_session(request: Request):
  

    data = await request.json()

    price_id = data['price_id']
    user_id = data['user_id']
    base_url = data['base_url']
    stripe_customer_id = data.get('stripe_customer_id')

    try:
        if stripe_customer_id is not None:
            stripe_customer = stripe.Customer.retrieve(stripe_customer_id)

            if stripe_customer:
                print('stripe_customer', stripe_customer)
                return  
        else:
            print('request', request)
            stripe_customer = stripe.Customer.create(metadata={"user_id": user_id})
            customer_id = stripe_customer.id
            await update_user_with_stripe_customer_id(user_id, customer_id)  
        
        checkout_session = stripe.checkout.Session.create(
            customer=stripe_customer_id,
            line_items=[
                {
                    'price': price_id,
                    'quantity': 1,
                },
            ],
            mode='subscription',
            success_url=base_url + '/payment-success',
            cancel_url=base_url + '?canceled=true',
            client_reference_id=user_id,
        )
        
        logger.info(f"Checkout session created successfully: {checkout_session.id}")
        return JSONResponse(content={"url": checkout_session.url}, status_code=200)
    except stripe.error.StripeError as e:
        logger.error(f"Stripe error: {str(e)}")
        raise HTTPException(status_code=400, detail=str(e))
    except Exception as e:
        logger.error(f"Error creating checkout session: {str(e)}")
        raise HTTPException(status_code=500, detail=str(e))

async def update_user_with_stripe_customer_id(user_id: str, stripe_customer_id: str):
    db = firestore.client()
    user_ref = db.collection('customers').where(filter=FieldFilter("user_id", "==", user_id))
    user_docs = user_ref.get()

    if user_docs:
        user_doc = user_docs[0]
        user_doc.reference.update({
            'stripe_customer_id': stripe_customer_id
        })
        print(f"Updated user {user_id} with Stripe customer ID {stripe_customer_id}")
    else:
        logger.error(f"User {user_id} not found in database")
        raise HTTPException(status_code=404, detail="User not found") 

结账会话完成且付款成功后，我想更新数据库以向用户提供分配的信用额度。但是我收到 404 未找到用户错误。

@app.post('/webhook')
async def webhook_received(request: Request):
    print("Webhook received!")
    payload = await request.body()
    sig_header = request.headers.get('stripe-signature')
    logger.info(f"Stripe signature: {sig_header}")

    try:
        event = stripe.Webhook.construct_event(
            payload, sig_header, STRIPE_WEBHOOK_SECRET
        )
        print(f"Event constructed: {event['type']}")
    except ValueError as e:
        logger.info(f"Invalid payload: {str(e)}")
        raise HTTPException(status_code=400, detail='Invalid payload')
    except stripe.error.SignatureVerificationError as e:
        logger.info(f"Invalid signature: {str(e)}")
        raise HTTPException(status_code=400, detail='Invalid signature')

    if event['type'] == 'checkout.session.completed':
        logger.info("Checkout session completed event received")
        return
        session = event['data']['object']
        try:
            result = await handle_checkout_session_completed(session)
            return JSONResponse(content={"status": "success", "result": result}, status_code=200)
        except Exception as e:
            logger.error(f"Error handling checkout session: {str(e)}")
            return JSONResponse(content={"status": "error", "message": str(e)}, status_code=500)
    elif event['type'] =='invoice.paid':
        logger.info("Invoice paid event received")
        session = event['data']['object']
        try:
            result = await handle_invoice_paid(session)
            return JSONResponse(content={"status": "success", "result": result}, status_code=200)
        except Exception as e:
            logger.error(f"Error handling invoice paid: {str(e)}")
            return JSONResponse(content={"status": "error", "message": str(e)}, status_code=500)
    else:
        logger.info(f"Unhandled event type: {event['type']}")

async def handle_invoice_paid(session):
    print('session', session)
    credits = session.get('lines').data[0].plan.metadata.credits
    tier = session.get('lines').data[0].plan.metadata.tier
    stripe_customer_id = session.get('customer')
    print('credits', credits, tier, stripe_customer_id)

    if credits:
        db = firestore.client()
        user_ref = db.collection('customers').where(filter=FieldFilter("stripe_customer_id", "==", stripe_customer_id))
        user_docs = user_ref.get()

        if user_docs:
            user_doc = user_docs[0]
            user_doc.reference.update({
                'credits': credits,
                'current_subscription_tier': tier,
                'stripe_customer_id': stripe_customer_id
            })
            updated_user_doc = user_doc.reference.get()
            updated_user_data = updated_user_doc.to_dict()
            return JSONResponse(content={"updated_user": updated_user_data}, status_code=200)
        else:
            print(f"User document not found for user_id: {stripe_customer_id}")
            raise HTTPException(status_code=404, detail="User not found")
    else:
        print(f"credits: {credits} not found")  

我尝试过不同的方法来创建客户。最终我需要用客户 ID 来更新数据库，但我的问题是，由于要创建多个用户，因此有多个客户 ID。

为什么相同的查询在 Windows MySQL 实例上有效，但在 Ubuntu MySQL 实例上无效？

实例版本非常接近：

• Windows：8.0.16
• Ubuntu：8.0.39-0ubuntu0.24.04.2

Windows：@@sql_mode：STRICT_TRANS_TABLES，NO_ENGINE_SUBSTITUTION

Ubuntu：@@sql_mode：ONLY_FULL_GROUP_BY、STRICT_TRANS_TABLES、NO_ZERO_IN_DATE、NO_ZERO_DATE、ERROR_FOR_DIVISION_BY_ZERO、NO_ENGINE_SUBSTITUTION

Ubuntu 上的错误：

错误 3065 (HY000)：ORDER BY 子句的表达式 #1 不在 SELECT 列表中，引用的列“p.priority”不在 SELECT 列表中；这与 DISTINCT 不兼容

查询：

SELECT DISTINCT 
    `id`, `date_created`, `date_last_update`, `text_id_map`, `address_search_aid`, `lat`, `lng`, 
    `street_line1`, `street_line2`, `zip`, `company_status`, `compassion_level`, `emails`, 
    `phones`, `social_media`, `urls`, `description`, `importance`, `name`, `object_status`, 
    `search_aid`, `short_name`, `created_by_id`, `last_update_by_id`, `alternate_country_id`,
    `city_id`, `parent_id`, `city_region_id`, `autocomplete`, `sitemap_xml`
FROM (
    
    SELECT p.*, 0 AS `priority`, CASE WHEN `parent_id` IS NULL THEN 1 ELSE 0 END AS `is_parent`
    FROM `provider` p 
    CROSS JOIN JSON_TABLE( JSON_KEYS(`name`), '$[*]' COLUMNS (locale VARCHAR(10) PATH '$') ) AS l 
    CROSS JOIN JSON_TABLE( JSON_EXTRACT(`name`, CONCAT('$.', l.locale)), '$' COLUMNS (`jsonval` VARCHAR(2048) PATH '$') ) AS j
    WHERE
        INSTR( CONCAT(' ', REPLACE(j.jsonval, '-', ' '), ' '), CONCAT(' ', 'web', ' ') ) > 0
        AND
        INSTR( CONCAT(' ', REPLACE(j.jsonval, '-', ' '), ' '), CONCAT(' ', 'kon', ' ') ) > 0 
    
    UNION
    
    SELECT *, 
        CASE 
            WHEN 
                INSTR( CONCAT(' ', `autocomplete`, ' '), CONCAT(' ', 'web', ' ') ) > 0 
                AND 
                INSTR( CONCAT(' ', `autocomplete`, ' '), CONCAT(' ', 'kon', ' ') ) > 0 
            THEN 1 
            WHEN INSTR( `autocomplete`, 'web' ) > 0 AND INSTR( `autocomplete`, 'kon' ) > 0 THEN 2 
            ELSE 3 
        END `priority`, 
        CASE 
            WHEN parent_id IS NULL THEN 1 
            ELSE 0 
        END AS `is_parent`
    FROM `provider`
    
) p
WHERE `priority` < 3
ORDER BY `priority`, `is_parent` DESC
LIMIT 10;

表是这样创建的：

CREATE TABLE `provider` (
  `id` int NOT NULL,
  `date_created` varchar(80) CHARACTER SET utf8mb4 COLLATE utf8mb4_unicode_ci DEFAULT NULL,
  `date_last_update` varchar(80) CHARACTER SET utf8mb4 COLLATE utf8mb4_unicode_ci DEFAULT NULL,
  `text_id_map` text COLLATE utf8mb4_unicode_ci,
  `address_search_aid` varchar(255) CHARACTER SET utf8mb4 COLLATE utf8mb4_unicode_ci DEFAULT NULL,
  `lat` varchar(255) CHARACTER SET utf8mb4 COLLATE utf8mb4_unicode_ci DEFAULT NULL,
  `lng` varchar(255) CHARACTER SET utf8mb4 COLLATE utf8mb4_unicode_ci DEFAULT NULL,
  `street_line1` varchar(255) CHARACTER SET utf8mb4 COLLATE utf8mb4_unicode_ci DEFAULT NULL,
  `street_line2` varchar(255) CHARACTER SET utf8mb4 COLLATE utf8mb4_unicode_ci DEFAULT NULL,
  `zip` varchar(255) CHARACTER SET utf8mb4 COLLATE utf8mb4_unicode_ci DEFAULT NULL,
  `company_status` varchar(255) CHARACTER SET utf8mb4 COLLATE utf8mb4_unicode_ci DEFAULT NULL,
  `compassion_level` varchar(255) CHARACTER SET utf8mb4 COLLATE utf8mb4_unicode_ci DEFAULT NULL,
  `emails` varchar(255) COLLATE utf8mb4_unicode_ci DEFAULT NULL,
  `phones` varchar(255) COLLATE utf8mb4_unicode_ci DEFAULT NULL,
  `social_media` text COLLATE utf8mb4_unicode_ci,
  `urls` varchar(255) COLLATE utf8mb4_unicode_ci DEFAULT NULL,
  `description` text COLLATE utf8mb4_unicode_ci,
  `importance` varchar(255) CHARACTER SET utf8mb4 COLLATE utf8mb4_unicode_ci DEFAULT NULL,
  `name` text COLLATE utf8mb4_unicode_ci,
  `object_status` varchar(16) CHARACTER SET utf8mb4 COLLATE utf8mb4_unicode_ci DEFAULT NULL,
  `search_aid` text COLLATE utf8mb4_unicode_ci,
  `short_name` text COLLATE utf8mb4_unicode_ci,
  `created_by_id` int DEFAULT NULL,
  `last_update_by_id` int DEFAULT NULL,
  `alternate_country_id` int DEFAULT NULL,
  `city_id` int DEFAULT NULL,
  `parent_id` int DEFAULT NULL,
  `city_region_id` int DEFAULT NULL,
  `autocomplete` text COLLATE utf8mb4_unicode_ci,
  `sitemap_xml` text CHARACTER SET utf8mb4 COLLATE utf8mb4_unicode_ci,
  PRIMARY KEY (`id`),
  KEY `FKotfd80u3xayhxxjrvtpclqwe6` (`created_by_id`),
  KEY `FK7gqnbcylk4ghwciki0779g32u` (`last_update_by_id`),
  KEY `FKk6jlj4kwu3hi5kl4a4qqso7j3` (`alternate_country_id`),
  KEY `FKrlh8uhcluf8si4vfbgdw3w6p1` (`city_id`),
  KEY `FK6psluxn6a0b64bcxggfuum2l0` (`parent_id`),
  KEY `FKegsn16mm766i6j5du01dlvn6` (`city_region_id`),
  CONSTRAINT `FK6psluxn6a0b64bcxggfuum2l0` FOREIGN KEY (`parent_id`) REFERENCES `provider` (`id`),
  CONSTRAINT `FK7gqnbcylk4ghwciki0779g32u` FOREIGN KEY (`last_update_by_id`) REFERENCES `veg_user` (`id`),
  CONSTRAINT `FKegsn16mm766i6j5du01dlvn6` FOREIGN KEY (`city_region_id`) REFERENCES `city_region` (`id`),
  CONSTRAINT `FKk6jlj4kwu3hi5kl4a4qqso7j3` FOREIGN KEY (`alternate_country_id`) REFERENCES `country` (`id`),
  CONSTRAINT `FKotfd80u3xayhxxjrvtpclqwe6` FOREIGN KEY (`created_by_id`) REFERENCES `veg_user` (`id`),
  CONSTRAINT `FKrlh8uhcluf8si4vfbgdw3w6p1` FOREIGN KEY (`city_id`) REFERENCES `city` (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8mb4 COLLATE=utf8mb4_unicode_ci

来自 Windows 上的 Heidi 的屏幕截图；查询从这个 SO 问题复制而来：

这个想法是创建一个查询，它将用作自动完成，但它会将具有精确关键字的结果移动到顶部。

查询的第一部分 p1 搜索精确匹配，而最后一部分 p2 搜索任何子字符串的出现。这意味着在第一部分中将选择名称中包含“ One ”的提供商，而在第二部分中将选择名称中包含“ Done ”的提供商。

问题是第二部分也从第一部分中选择了项目。
因此，我尝试为 union 的每个部分添加别名，并尝试为 union 的最后一部分添加附加条件：AND p2.id != p1.id

这对于 MySQL 不起作用：

SQL 错误（1054）：‘where 子句’中未知列‘p1.id’。

也尝试过HAVING，但根本没有运气。

问题是：有没有办法一次性获得这样的结果？

SELECT * FROM (

    (SELECT DISTINCT 
        p1.*, 
        1 AS priority, 
        CASE WHEN parent_id IS NULL THEN 1 ELSE 2 END AS is_parent
    FROM provider p1
    WHERE CONCAT(' ', name, ' ') LIKE '% one %'
    LIMIT 100)
    
    UNION  
    
    (SELECT DISTINCT 
        p2.*, 
        2 AS priority, 
        CASE WHEN parent_id IS NULL THEN 1 ELSE 2 END AS is_parent
    FROM provider p2
    WHERE name LIKE '%one%' AND p2.id != p1.id
    LIMIT 100)

    ORDER BY priority, is_parent
    
) AS t LIMIT 100;

样本数据：

ID | Parent ID | Name          
---------------------------
 1 | <NULL> | One dove
 2 |      1 | One play
 3 | <NULL> | Monitor
 4 |      1 | Day one
 5 | <NULL> | Drone
 6 | <NULL> | Screen
 7 | <NULL> | Done with you
 8 | <NULL> | Not done
 9 | <NULL> | All as one

预期结果：

ID | Parent ID | Name           | Priority | Is parent         
------------------------------------------------------
 1 |    <NULL> | One dove       |        1 |         1
 9 |    <NULL> | All as one     |        1 |         1
 2 |         1 | One play       |        1 |         2
 4 |         1 | Day one        |        1 |         2
 5 |    <NULL> | Drone          |        2 |         1
 7 |    <NULL> | Done with you  |        2 |         1
 8 |    <NULL> | Not done       |        2 |         1

当TooltipBox打开时，单击屏幕上的任何 Composable 都不会产生任何效果，除了关闭 Popup。它是如何做到的，以及在打开时无论当前 Composable 在 Composition 树中的什么位置，实现它的有效方法是什么，Popup这样其他 Composable 都不会Popup接收任何手势。

@Preview
@Composable
fun TooltipBoxSample() {
    Column(
        modifier = Modifier.fillMaxSize().border(2.dp, Color.Blue)
    ) {
        Button(
            onClick = {}
        ){
            Text("Some button")
        }

        val state = rememberTooltipState(
            isPersistent = true
        )
        val coroutineScope = rememberCoroutineScope()

        Spacer(modifier = Modifier.height(120.dp))

        Row(
            modifier = Modifier.weight(1f)
        ) {
            Spacer(modifier = Modifier.width(100.dp))

            TooltipBox(
                positionProvider = rememberPlainTooltipPositionProvider(
                    spacingBetweenTooltipAndAnchor = 16.dp
                ),
                state = state,
                tooltip = {

                    PlainTooltip(
                        caretProperties = CaretProperties(
                            caretWidth = 16.dp,
                            caretHeight = 16.dp
                        ),
                        shape = RoundedCornerShape(16.dp),
                        containerColor = Color.Red
                    ) {
                        Text(
                            text = "Tooltip Content for testing...",
                            modifier = Modifier.padding(16.dp)
                        )
                    }

                },
                content = {
                    Icon(
                        modifier = Modifier
                            .size(60.dp)
                            .clickable {
                                coroutineScope.launch {
                                    state.show()
                                }
                            },
                        imageVector = Icons.Default.Info,
                        contentDescription = null
                    )
                }
            )
        }

    }
}

如果你使用PopupComposables 触摸屏幕上的任意位置，则会收到点击事件

@Preview
@Composable
fun PopupOpenCloseTest() {

    var showPopup by remember {
        mutableStateOf(false)
    }


    Column(
        modifier = Modifier.fillMaxSize()
    ) {

        Button(
            onClick = {}
        ) {
            Text("Some button")
        }

        Box {
            Icon(
                modifier = Modifier
                    .border(2.dp, Color.Blue)
                    .size(60.dp)
                    .clickable {
                        if (showPopup.not()) {
                            showPopup = true
                        }
                        println("CLICKED showPopup: $showPopup")
                    },
                imageVector = Icons.Default.Info,
                contentDescription = null
            )

            if (showPopup) {
                Popup(
                    offset = IntOffset(300, 300),
                    onDismissRequest = {
                        if (showPopup) {
                            showPopup = false
                        }
                    }
                ) {
                    Box(
                        modifier = Modifier.background(Color.White).padding(16.dp)
                    ) {
                        Text("Popup Content")
                    }
                }
            }
        }

    }
}

我检查了Tooltip和BasicTooltip的源代码，但没有地方创建取消手势的窗口或视图，或者使用任何手势。

我有一台 mac m2，我正在尝试使用一些钛模块运行苹果模拟器，但出现以下错误：

[ERROR] Error: The app is using native modules that do not support arm64 simulators and you are on an arm64 device:
- de.marcelpociot.imagefromgif
- ti.safaridialog
- com.movento.splitwindow
- net.iamyellow.tiws
    at iOSBuilder.invokeXcodeBuild (/Users/programador1/Library/Application Support/Titanium/mobilesdk/osx/12.2.1.GA/iphone/cli/commands/_build.js:7096:17)
Process exited with 1

请帮我。

之前的模块没有 m2 的更新，或者至少我没有找到最近的更新。

目前，要重新初始化 的模型AutoModelForSequenceClassification，我们可以这样做：

from transformers import AutoModel, AutoConfig, AutoModelForSequenceClassification

m = "moussaKam/frugalscore_tiny_bert-base_bert-score"
config = AutoConfig.from_pretrained(m)
model_from_scratch = AutoModel(config)

model_from_scratch.save_pretrained("frugalscore_tiny_bert-from_scratch")

model = AutoModelForSequenceClassification(
  "frugalscore_tiny_bert-from_scratch", local_files_only=True
)

有没有某种方法可以重新初始化模型权重而不保存用 初始化的新预训练模型AutoConfig？

model = AutoModelForSequenceClassification(
  "moussaKam/frugalscore_tiny_bert-base_bert-score", 
  local_files_only=True
  reinitialize_weights=True
)

或类似的东西：

model = AutoModelForSequenceClassification(
  "moussaKam/frugalscore_tiny_bert-base_bert-score", 
  local_files_only=True
)

model.reinitialize_parameters()

在比较 2 个表之间的数据的查询中，我经常使用 和 的组合COALESCE来FULL OUTER JOIN显示仅在其中 1 个表中可用的记录。

这可以用更少的语法糖来完成吗？（我的意思并不是替换COALESCE为之NVL类的。）

WITH Dataset1 AS (
    SELECT
        id,
        SUM(amount) AS amount
    FROM
        table1
    GROUP BY
        id
),

Dataset2 AS (
    SELECT
        id,
        SUM(amount) AS amount
    FROM
        table2
    GROUP BY
        id
)

SELECT
    COALESCE(d1.id, d2.id) AS ID,
    COALESCE(d1.amount, 0) AS D1_AMOUNT,
    COALESCE(d2.amount, 0) AS D2_AMOUNT,
    COALESCE(d1.amount, 0) - COALESCE(d2.amount, 0) AS DELTA
FROM
    Dataset1 d1
FULL OUTER JOIN
    Dataset2 d2 c ON
        d2.id = d1.id
WHERE
    ABS(COALESCE(d1.amount, 0) - COALESCE(d2.amount, 0)) >= 5
ORDER BY
    ID

我目前有一个包含多个对象的数组。for 循环使用模型的 for 循环基于数组对象执行查询集。

search_post = ["1990", "2023"]
for query in search_post:
    vehicle_result = Vehicle.objects.filter(
        Q(title__icontains=query) |
        Q(details__icontains=query) |
        Q(year__icontains=query) |
        Q(notes__icontains=query)
    )
    print('vehicle_result: %s' % vehicle_result)

我遇到的问题是，当数组有多个项目时，它会为每个对象执行 for 循环，并输出每个对象的结果。如果我在代码中的其他地方调用此变量，则仅保留最后一个对象结果。

示例输出：

vehicle_result: <QuerySet [<Vehicle: BMW>]>
vehicle_result: <QuerySet [<Vehicle: Toyota>, <Vehicle: Mercedes>, <Vehicle: Honda>]>

我想将它们全部存储/合并到一个变量中 - 防止丢失所有结果（我猜现在只保留最后一个结果，因为变量被覆盖了）

我怎样才能实现？