You need to cast the text as binary (or use a case-sensitive collation). With temp as ( select ‘Test’ as name UNION ALL select ‘TEST’ UNION ALL select ‘test’ UNION ALL select ‘tester’ UNION ALL select ‘tester’ ) Select Name, COUNT(name) From temp Group By Name, Cast(name As varbinary(100)) Using a collation: Select Name Collate … Read more
As answered by EdChum in the comments. The issue is that apply works column wise by default (see the docs). Therefore, the column names cannot be accessed. To specify that it should be applied to each row instead, axis=1 must be passed: test.apply(lambda x: find_max(x,test,’document_id’,’confidence_level’,’category_id’), axis=1)
Sure, just add the same calculation to the GROUP BY clause: select dateadd(day, -7, Convert(DateTime, mwspp.DateDue) + (7 – datepart(weekday, mwspp.DateDue))), sum(mwspp.QtyRequired) from manufacturingweekshortagepartpurchasing mwspp where mwspp.buildScheduleSimID = 10109 and mwspp.partID = 8366 group by dateadd(day, -7, Convert(DateTime, mwspp.DateDue) + (7 – datepart(weekday, mwspp.DateDue))) order by dateadd(day, -7, Convert(DateTime, mwspp.DateDue) + (7 – datepart(weekday, mwspp.DateDue))) … Read more
Simple, Fast, and Pandaic: ngroups Newer versions of the groupby API (pandas >= 0.23) provide this (undocumented) attribute which stores the number of groups in a GroupBy object. # setup df = pd.DataFrame({‘A’: list(‘aabbcccd’)}) dfg = df.groupby(‘A’) # call `.ngroups` on the GroupBy object dfg.ngroups # 4 Note that this is different from GroupBy.groups which … Read more
You can do both in one query using the OVER clause on another COUNT select count(*) RecordsPerGroup, COUNT(*) OVER () AS TotalRecords from temptable group by column_1, column_2, column_3, column_4
One way is to use the transform function to fill the value column after group by: import pandas as pd a[‘value’] = a.groupby(‘company’)[‘value’].transform(lambda v: v.ffill()) a # company value #level_1 #2010-01-01 a 1.0 #2010-01-01 b 12.0 #2011-01-01 a 2.0 #2011-01-01 b 12.0 #2012-01-01 a 2.0 #2012-01-01 b 14.0 To compare, the original data frame looks … Read more
If I understand what you need, try this: SELECT id, pass, AVG(val) AS val_1 FROM data_r1 GROUP BY id, pass; Or, if you want just one row for every id, this: SELECT d1.id, (SELECT IFNULL(ROUND(AVG(d2.val), 4) ,0) FROM data_r1 d2 WHERE d2.id = d1.id AND pass = 1) as val_1, (SELECT IFNULL(ROUND(AVG(d2.val), 4) ,0) FROM … Read more
A change was made in version 5.7.5 where it will now, by default, reject queries in which you aggregate using a function (sum, avg, max, etc.) in the SELECT clause and fail to put the non-aggregated fields in the GROUP BY clause. This behavior is part and parcel to every other RDBMS and MySQL is … Read more
Use value_counts with normalize=True: df[‘gender’].value_counts(normalize=True) * 100 The result is a fraction in range (0, 1]. We multiply by 100 here in order to get the %.
Your code is almost correct – with two issues: The return type of your function is DataFrame, but the last line is aggregated.show(), which returns Unit. Remove the call to show to return aggregated itself, or just return the result of agg immediately DataFrame.groupBy expects arguments as follows: col1: String, cols: String* – so you … Read more