在您的特定示例中，.drop_duplicates(）應(yīng)該很好吧？ In [71]: dfOut[71]: ACTIVITY_DATE OWNNER_ID OWNER_NAME0 1/1/2020 23344 JAMES NELSON1 2/1/2020 33445 NIGEL THOMAS2 1/1/2020 23344 JAMES NELSON3 2/1/2020 33445 NIGEL THOMASIn [72]: df.drop_duplicates()Out[72]: ACTIVITY_DATE OWNNER_ID OWNER_NAME0 1/1/2020 23344 JAMES NELSON1 2/1/2020 33445 NIGEL THOMAS 或者，如果您只想提供要比較的列的子集，可以這樣做： In [77]: df.drop_duplicates(subset=['ACTIVITY_DATE','OWNNER_ID'])Out[77]: ACTIVITY_DATE OWNNER_ID OWNER_NAME0 1/1/2020 23344 JAMES NELSON1 2/1/2020 33445 NIGEL THOMAS

在插入新項(xiàng)之前，請(qǐng)檢查是否已經(jīng)存在具有相同titleName的項(xiàng)。如果沒(méi)有，請(qǐng)插入新項(xiàng)目。 像這樣的 @IBAction func buttonClicked(_ sender: UIBarButtonItem) { let fetchRequest: NSFetchRequest<Bookmarks> = Bookmarks.fetchRequest() fetchRequest.predicate = NSPredicate(format: "titleName == %@", titleName!) if (try? context.fetch(fetchRequest))?.first != nil { bookmarkButton.image = UIImage(systemName: "bookmark.fill") } else { let bookmark = Bookmarks(context: context) bookmarkButton.image = UIImage(systemName: "bookmark") bookmark.urlToImage = urlImage bookmark.source = sourceName bookmark.titleName = titleName bookmark.urlLink = url boo

如果您使用的是mysql 8.0，那么row_number(）和公共表表達(dá)式是一種方便的方法： Schema: create table user(UserID int, UserName varchar(50), UserEmail varchar(50), FlagUser varchar(50)); insert into user values(1, 'Penny' , 'penny@gmail.com' ,'A1'); insert into user values(2, 'Maika' , 'maika@gmail.com' ,'A1'); insert into user values(3, 'Laila' , 'laila@gmail.com' ,'A1'); insert into user values(4, 'Laura' , 'laura@gmail.com' ,'A2'); insert into user values(5, 'Penny' , 'penny@gmail.com' ,'A2'); insert into user values(6, 'Maika' , 'maika@gmail.com' ,'A2'); Query with cte as ( select *,row_number()over(partition by username

Try this: DELETE FROM #A WHERE #A.EmployeeID > #a.CoworkerIDAND EXISTS(SELECT 1 FROM #A A2 where a2.CoworkerID = #A.EmployeeID and a2.employeeID = #a.coworkerID) See SqlFiddle

因?yàn)槟偸窃趂etchSavedStationItems（）中附加數(shù)組 itemsList += try persistenceService.fetchStationItems() 嘗試在附加數(shù)組之前清除它 itemsList.removeAll()itemsList += try persistenceService.fetchStationItems()

您需要連接表值數(shù)據(jù)以獲得所需的行數(shù)，例如： select sum(stats.power)from ( select 489191 as template union all select 489191 union all select 489100) as vlistjoin stats on vlist.template = stats.template 如果您使用的是MySQL 8或更高版本，則可以使用values表值構(gòu)造函數(shù)，而不是union all： select sum(stats.power)from (values row(489191), row(489191) row(489100)) as vlist(template)join stats on vlist.template = stats.template

$map = [];foreach ($arr as $v) { if(array_key_exists($v['attr_id'],$map)){ $map[$v['attr_id']]++; }else{ $map[$v['attr_id']]=1; }}$rs = [];for ($i=0; $i < count($arr); $i++) { if($map[$arr[$i]['attr_id']] == 1){ array_push($rs,$arr[$i]); }}print_r($rs);

使用data.table，按地址分組，循環(huán)列，刪除NAs，獲取第一個(gè)值： data[, lapply(.SD, function(i) (na.omit(i)[ 1 ])), by = address]# address revenue castord versaze# 1: AA 22 21 124# 2: BB 121 22 22# 3: CC 33 3 33# 4: DD 44 33 43# 5: EE 33 223 44

定義包含登錄和密碼數(shù)據(jù)的字典 Login = {'Login1': 'Password1', 'Login2': 'Password2', 'Login3': 'Password3', 'Login4': 'Password4', 'Login5': 'Password5', 'Login6': 'Password6', 'Login7': 'Password7', 'Login8': 'Password8', 'Login9': 'Password9', 'Login10': 'Password10', } 像這樣使用for-loop： for login, password in Login.items(): # Use login and password instead of login1 and password1

我想，這就是你要找的問(wèn)題 select sum(variety), sum(number_grafts), date from <table_name> group by date

當(dāng)CipherText持有對(duì)它的引用時(shí)，不能對(duì)string_table進(jìn)行變異。這只是Rust的核心不變量之一。所以你不能使用Cow。圍繞它的典型方式通常是： 使用索引/鍵來(lái)引用另一個(gè)結(jié)構(gòu)，不幸的是，在這個(gè)場(chǎng)景中，我看不到一個(gè)好的方法來(lái)做到這一點(diǎn)。 使用Rc使密文和表共享字節(jié)的所有權(quán)。這確實(shí)意味著密文可以獨(dú)立存在，但在明智地使用get_or_insert之前，您可以始終將表用作代理來(lái)刪除重復(fù)元素。#![feature(hash_set_entry)]use std::rc::Rc;use ahash::AHashSet;#[derive(Debug)]struct CipherText { ciphertext: Rc<Vec<u8>>, ciphers: Vec<Cipher>,}#[derive(Debug)]struct Cipher { cipher_id: Rc<Vec<u8>>, keys: Vec<Rc<Vec<u8>>>,}fn main() { let mut string_table: AHashSet<Rc<Vec<u8>>> = vec![ Rc::new("Hello World!".as_bytes().to_vec())

在提到coalesce時(shí)，我曾假設(shè)目標(biāo)是將4列壓縮為一個(gè)值列表，而預(yù)期結(jié)果示例現(xiàn)在顯示的情況并非如此。 有幾種方法可以解決這個(gè)問(wèn)題，一種是使用window functions為每個(gè)值分配一個(gè)序列號(hào)；為此，必須將這些值組合到一個(gè)列表中，您可以使用union，然后將組合值partitioned作為該列表的結(jié)果，并且每個(gè)重復(fù)項(xiàng)都獲得一個(gè)連續(xù)的行號(hào)。那就是只保留那些值為1的。 with Alltel as ( select id, [group], tel1 from MT_TEMP_TEL2 union all select id, [group], tel2 from MT_TEMP_TEL2 union all select id, [group], tel3 from MT_TEMP_TEL2 union all select id, [group], tel4 from MT_TEMP_TEL2 ), rn as ( select id, tel1, Row_Number() over (partition by tel1 order by [group]) n from alltel where tel1 !='')select * from MT_TEMP_TEL2 twhere not exists (select * from rn where rn.id=t.id an

這是否有效： df %>% group_by(category, name, date, checkup_complete) %>% filter(if(all(checkup_complete == 'Y')) row_number() == n() else TRUE)# A tibble: 9 x 5# Groups: category, name, date, checkup_complete [8] index name category date checkup_complete <dbl> <chr> <chr> <chr> <chr> 1 2 name1 cat1 date1 N 2 3 name1 cat2 date1 Y 3 4 name1 cat1 date1 Y 4 5 name1 cat1 date2 N 5 6 name2 cat1 date1 Y 6 7 name3 cat1 date2 Y 7 9 name3 cat1 date2 N 8 10 name2 cat2 date1 Y 9 11 name1 cat1 date1 N

因?yàn)槟诘總€(gè)項(xiàng)目鏈接，所以無(wú)需再次重申lop的含義，您必須將各個(gè)頁(yè)面與各個(gè)url一起帶到各個(gè)頁(yè)面，并從該頁(yè)面中選擇所需的數(shù)據(jù)項(xiàng)目。 Code: import scrapyclass WatchbotSpider(scrapy.Spider): name = 'watchbot' start_urls = ['https://www.watch.de/germany/rolex.html'] def parse(self, response, **kwargs): for link in response.css('div.product-item-link a::attr(href)'): url = link.get() yield scrapy.Request(url, callback=self.parse_categories) def parse_categories(self, response): yield { 'id': response.css('div.product-ref-item.product-sku.col-auto>span[itemprop="sku"]::text').get(), 'product_name': response.css('h1.product-name::text').get().strip(), 'price'

我認(rèn)為這是一個(gè)gap-and-islands問(wèn)題。為此，您可以使用lag()和累積和： select dd.*, sum(case when prev_currentpos = currentpos then 0 else 1 end) over (partition by userid order by occurredon ) as locationChangeGroupfrom (select dd.*, lag(currentpos) over (partition by userid order by occurredon) as prev_currentpos from dummydata dd ) dd