You could use np.argsort to find the indices of the n largest items for each row: import numpy as np import pandas as pd df = pd.DataFrame({‘id’: [1, 2, 3, 4, 5], ‘p1’: [0, 0, 1, 1, 2], ‘p2’: [9, 2, 3, 5, 3], ‘p3’: [1, 3, 10, 3, 7], ‘p4’: [4, 4, 7, 1, … Read more
head and tail are really useful functions! head(sort(Forbes2000$profits,decreasing=TRUE), n = 50) If you want the first 50 rows of the data.frame, then you can use the arrange function from plyr to sort the data.frame and then use head library(plyr) head(arrange(Forbes2000,desc(profits)), n = 50) Notice that I wrapped profits in a call to desc which means … Read more
The way to do this is with Oracle’s Analytic functions. Your particular scenario is just a variant on the solution I provided in another thread. If you are interested in just selecting the second highest salary then any of DENSE_RANK(), RANK() and ROW_NUMBER() will do the trick: SQL> select * from 2 ( select sal … Read more
I had a similar problem a year ago: Top 3 per group including 0 Using the same approach, this will return the latest three dates for each LoginID – you may get more than three records if there are tied dates for the same LoginID. SELECT PR1.LogInID, PR1.Score, PR1.[Date Taken] FROM Progress AS PR1 WHERE … Read more
The best way to do this is with analytic functions, RANK() or DENSE_RANK() … SQL> select * from ( 2 select empno 3 , sal 4 , rank() over (order by sal desc) as rnk 5 from emp) 6 where rnk <= 5 7 / EMPNO SAL RNK ———- ———- ———- 7839 5000 1 7788 … Read more
Given this data … SQL> select * from qtys 2 / ID TS QTY ———- —————- ———- 1 2010-01-04 11:00 152 2 2010-01-04 11:00 210 1 2010-01-04 10:45 132 2 2010-01-04 10:45 318 4 2010-01-04 10:45 122 1 2010-01-04 10:30 1 3 2010-01-04 10:30 214 2 2010-01-04 10:30 5515 4 2010-01-04 10:30 210 9 rows … Read more
You’ll need to put your current query in subquery as below : SELECT * FROM ( SELECT DISTINCT APP_ID, NAME, STORAGE_GB, HISTORY_CREATED, TO_CHAR(HISTORY_DATE, ‘DD.MM.YYYY’) AS HISTORY_DATE FROM HISTORY WHERE STORAGE_GB IS NOT NULL AND APP_ID NOT IN (SELECT APP_ID FROM HISTORY WHERE TO_CHAR(HISTORY_DATE, ‘DD.MM.YYYY’) =’06.02.2009′) ORDER BY STORAGE_GB DESC ) WHERE ROWNUM <= 10 Oracle … Read more
Did you try df.groupby(‘id’).head(2) Output generated: id value id 1 0 1 1 1 1 2 2 3 2 1 4 2 2 3 7 3 1 4 8 4 1 (Keep in mind that you might need to order/sort before, depending on your data) EDIT: As mentioned by the questioner, use df.groupby(‘id’).head(2).reset_index(drop=True) to remove … Read more