If by group by you mean adding values with same keys together you could try: let a = [‘A’, ‘B’, ‘C’, ‘D’, ‘E’, ‘F’, ‘G’, ‘A’, ‘F’, ‘C’, ‘A’, ‘E’]; let b = [‘5’, ‘7’, ‘4’, ‘3’, ‘8’, ‘1’, ‘9’, ‘1’, ‘5’, ‘4’, ‘2’, ’10’]; const temp = {}; a.forEach((value, index) => { temp.hasOwnProperty(value) ? … Read more
you should aggregate by column and then sum the results, try this: Notice that, the patient name should should have a numerical counter for keeping track. import pandas as pd import datetime import numpy as np # this a dummy set, you should have already this in your data frame dict_df = {‘Patient’: [1,2,3,4,5], ‘Positive’: … Read more
You want to examine all the objects in your input array, then push it into an array identified by the first letter of the name in your object. var input = [{ name: “John”, id: “123” }, { name: “Josh”, id: “1234” }, { name: “Charlie”, id: “1234253” }, { name: “Charles”, id: “123345” } … Read more
Try this: uniq_accounts = differenc[‘AccountID’].unique() grped = differenc.groupby(‘AccountID’) ## You can get the grouped data for a certain AccountId like this and store in a dictionary: d = {} for account in uniq_accounts: d[account] = grped.get_group(account) ##Now, dictionary has all accounts data in separate dataframes. You can access like this: for key in d.keys(): print(d[key]) … Read more
The ORDER BY clause should be the last, but you have to specify fields to be aggregated. Such as: SELECT surname, count(*) FROM people WHERE user_token=’$user_token’ GROUP BY surname ORDER BY surname 1 solved SELECT * FROM people WHERE user_id=’$user_id’ ORDER BY time GROUP BY surname [closed]
SELECT SUM(how_many) AS per_day, `date` FROM cigaretes GROUP BY DATE(`date`) 0 solved How to sum and group values using MySQL ?
try this, create table #tmp (MR_ID varchar(50),DR_ID int) insert into #tmp VALUES (‘MR_123’, 1),(‘MR_123’, 3),(‘MR_124’, 4),(‘MR_124’, 5) ,(‘MR_124’, 6),(‘MR_125′, 0) declare @DRCol varchar(50) declare @Prefix varchar(20)=’DR_’ ;With CTE as ( select * ,ROW_NUMBER()over(partition by MR_ID order by DR_ID)rn from #tmp ) select top 1 @DRCol=stuff((select ‘,’+'[‘+@Prefix+cast(rn as varchar)+’]’ from cte c1 where c.mr_id=c1.mr_id for xml … Read more
I understand that you want, for each day: the sum of order_items.energy_used for all orders created that day the created_at and order_sum that correspond to the latest order created on that day Your sample data is not very representative of that – it has only one order per day. Also it is unclear why created_at … Read more
In general ORDER BY in a sub-query makes no sense. (It only does when combined with FETCH FIRST/LIMIT/TOP etc.) The solution is to use a correlated sub-query to find the heaviest fish for the “main query”‘s current row’s username, location, species combination. If it’s a tie, both rows will be returned. SELECT * FROM entries … Read more
var results = from kvp in source group kvp by kvp.Key.ToUpper() into g select new { Key= g.Key, Value= g.Average(kvp => kvp.Value) } or var results = source.GroupBy(c=>c.Name) .Select(c => new (c.Key, c.Average(d=>d.Value))); 3 solved Remove duplicate items and calculate average values using LINQ
I suggest you to use the dplyr package. You can try the following code: if df is your dataframe: library(dplyr) summarised_df <- group_by(df, Ticker) %>% summarise( max(tradeDate), sum(SharesBought), max(HighestPxPaid), min(LowestPxPaid) ) 3 solved How to summarise a dataframe in R with different function for different column?
Try this DECLARE @temp TABLE(col1 varchar(20),col2 int, col3 varchar(20)) insert into @temp values (‘data1′, 123 , ’12/03/2009’),(‘data1′, 124 , ’15/09/2009’), (‘data2 ‘,333 ,’02/09/2010’),(‘data2 ‘,323 , ’02/11/2010’), (‘data2 ‘,673 , ’02/09/2014’),(‘data2′,444 , ’05/01/2010’) SELECT (CASE rno WHEN 1 THEN col1 ELSE ” END )AS col1, col2, col3 FROM ( SELECT ROW_NUMBER() OVER(PARTITION BY Col1 ORDER BY … Read more
You can try like following using GROUP BY and UNION ALL. select count(*) CT,Mark from TableName group by Mark union all select Count(*), ‘A+B’ as mark from TableName where mark in(‘A’,’B’) UNION ALL select Count(*), ‘A+C’ as mark from TableName where mark in(‘A’,’C’) union all select Count(*), ‘B+C’ as mark from TableName where mark in(‘B’,’C’) … Read more
Try this using LINQ in C# var result = from tab in dt.AsEnumerable() group tab by tab[“ApplicationNmae”] into groupDt select new { ApplicationNmae = groupDt.Key, Sum = groupDt.Sum((r) => decimal.Parse(r[“Count”].ToString())) }; DataTable dt1 = new DataTable(); dt1 = dt.Clone(); foreach (var item in result) { DataRow newRow = dt1.NewRow(); newRow[“ApplicationNmae”] = item.ApplicationNmae; newRow[“Count”] = item.Sum; … Read more
Do you mean like this? SELECT * FROM ( SELECT a.id, a.name, c.code, a.active_dt, a.inactive_dt, b.category, COUNT(1) OVER(PARTITION BY a.id, a.name, c.code) AS CNT FROM student a, class b, descrip c WHERE a.id=b.id AND a.id=c.id ) WHERE CNT > 1 0 solved how to use ‘GROUP BY’ function with below query