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More effective merging of matched column with duplicates in data.table


More effective merging of matched column with duplicates in data.table

By : tpdev
Date : November 22 2020, 10:38 AM
I hope this helps . I've got two data.tables, both of which share one variable; I'm trying to add a variable that's missing from the second, but which is tied one-for-one to the shared variable. , The only improvements that I could think of is that
code :
setkey(y,let)
y[x[!duplicated(let)], num := i.num, allow.cartesian = TRUE][]
y[unique(x, by = "let"), num := i.num, allow.cartesian = TRUE]
y[x, num := unique(i.num), by = .EACHI, allow.cartesian = TRUE]
y[x[!duplicated(let)], num := i.num]
# or
y[unique(x, by = "let"), num := i.num]
# or (though not recommended in this specific case)
y[x, num := unique(i.num), by = .EACHI]


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Merging 2 data frame with matched column names; each matched column next to each other

Merging 2 data frame with matched column names; each matched column next to each other


By : user57707
Date : March 29 2020, 07:55 AM
hope this fix your issue This seems pretty self-explanatory. Look at the help pages for ?cbind.data.frame, ?make.unique, ?'names<-' and ?"[" for further explanation.
code :
> comb <- cbind(ex1,ex2)
> names(comb) <- make.unique(names(comb))
> comb[sort(names(comb))]
  a a.1 b b.1  c c.1
1 1  12 2  23  9 100
2 2  13 3  24 10 101
3 3  14 4  25 11 102
4 4  15 5  26 12 103
5 5  16 6  27 13 104
MYSQL Move data from one table column to another table column if condition is matched

MYSQL Move data from one table column to another table column if condition is matched


By : apel web
Date : March 29 2020, 07:55 AM
I hope this helps you . i'm not really an mysql guy more like a php guy :) , Try:
code :
UPDATE news n
JOIN news_in_category nic ON n.id = nic.newsId
SET n.categoryID = nic.newsCategoryId
Get the Id of the matched data from other table. No duplicates of ID from both tables

Get the Id of the matched data from other table. No duplicates of ID from both tables


By : Kurt Lobato
Date : March 29 2020, 07:55 AM
it helps some times I think you want a left join. But this is tricky because you have duplicate amounts, but you only want one to match. The solution is to use row_number():
code :
select . . .
from (select a.*, row_number() over (partition by amount order by id) as seqnum
      from a
     ) a left join
     (select b.*, row_number() over (partition by credit order by id) as seqnum
      from b
     )b
     on a.amount = b.credit and a.seqnum = b.seqnum;
SQL Server: getting data from second table when condition is matched wihtout duplicates

SQL Server: getting data from second table when condition is matched wihtout duplicates


By : tumap90tvt
Date : March 29 2020, 07:55 AM
hope this fix your issue I have 2 tables (user, userContact) that are connected by a column userID. , You need a left join of the tables and conditional aggregation:
code :
select u.[User_id], u.[User_Name],
  max(case 
      when c.[UserContact_Type] = 3 then c.[UserContact_Description] 
      else null 
    end 
  ) as UserContact_Description 
from [User] u left join [userContact] c
on c.[User_id] = u.[User_id]
group by u.[User_id], u.[User_Name]
Remove SOME duplicates in data.table column X, but retain duplicates if column Y matches condition

Remove SOME duplicates in data.table column X, but retain duplicates if column Y matches condition


By : user3604579
Date : March 29 2020, 07:55 AM
wish of those help We could create an ordered 'STATUS' based on the levels 'A', 'B', 'C', grouped by 'ID', extract the first present levels (after applying droplevels), do a comparison with 'STATUS' to return logical vector, convert it to a numeric index (.I) also have a condition to check if the number of unique elements in the column is 1, return only the first row
code :
test[test[, .I[if(uniqueN(STATUS) == 1) 1 else 
   STATUS == first(levels(droplevels(ordered(STATUS, 
          levels = c("A", "B", "C")))))], ID]$V1]
#   ID STATUS
#1:  1      B
#2:  2      A
#3:  2      A
#4:  2      A
#5:  2      A
#6:  2      A
#7:  3      C
test <- structure(list(ID = c(1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 
3L, 3L, 3L, 3L), STATUS = c("B", "C", "C", "C", "A", "A", "A", 
"A", "A", "B", "B", "B", "B", "B", "C", "C", "C", "C", "C", "C", 
"C", "C", "C", "C", "C", "C", "C", "C", "C", "C")), class = c("data.table", 
"data.frame"), row.names = c(NA, -30L))
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