If you want just a rough estimate, you can extrapolate from a sample: SELECT COUNT(*) * 100 FROM sometable SAMPLE (1); For greater speed (but lower accuracy) you can reduce the sample size: SELECT COUNT(*) * 1000 FROM sometable SAMPLE (0.1); For even greater speed (but even worse accuracy) you can use block-wise sampling: SELECT … Read more
SELECT TABLE_NAME, TABLE_ROWS FROM `information_schema`.`tables` WHERE `table_schema` = ‘YOUR_DB_NAME’;
You can use INFORMATION_SCHEMA.TABLES to retrieve information about your database tables. As mentioned in the Microsoft Tables Documentation: INFORMATION_SCHEMA.TABLES returns one row for each table in the current database for which the current user has permissions. The following query, therefore, will return the number of tables in the specified database: USE MyDatabase SELECT COUNT(*) FROM … Read more
Have a look at Determine and count unique values of an array. Or, to count the number of occurrences of 5, simply do sum(your_matrix == 5)
How about something like this: SELECT m.MESSAGEID, sum((case when mp.messageid is not null then 1 else 0 end)) FROM MESSAGE m LEFT JOIN MESSAGEPART mp ON mp.MESSAGEID = m.MESSAGEID GROUP BY m.MESSAGEID; The COUNT() function will count every row, even if it has null. Using SUM() and CASE, you can count only non-null values. EDIT: … Read more
import json json_data = json.dumps({ “result”:[ { “run”:[ { “action”:”stop” }, { “action”:”start” }, { “action”:”start” } ], “find”: “true” } ] }) item_dict = json.loads(json_data) print len(item_dict[‘result’][0][‘run’]) Convert it in dict.
var ListByOwner = list.GroupBy(l => l.Owner) .Select(lg => new { Owner = lg.Key, Boxes = lg.Count(), TotalWeight = lg.Sum(w => w.Weight), TotalVolume = lg.Sum(w => w.Volume) });
Wrap them up and use subqueries: SELECT (SELECT COUNT(*) FROM Toys WHERE little_kid_id = 900)+ (SELECT COUNT(*) from Games WHERE little_kid1 = 900 OR little_kid2 = 900 OR little_kid3 = 900) AS SumCount Voila!
I think you may be after select jobID, JobName, sum(case when Priority = 1 then 1 else 0 end) as priority1, sum(case when Priority = 2 then 1 else 0 end) as priority2, sum(case when Priority = 3 then 1 else 0 end) as priority3, sum(case when Priority = 4 then 1 else 0 end) … Read more
Here’s a solution with the dplyr package – it has n_distinct() as a wrapper for length(unique()). df %>% group_by(color) %>% mutate(unique_types = n_distinct(type))