I have an sql select query that has a group by. I want to count all the records after the group by statement. Is there a way for this directly from sql? For example, having a table with users I want to select the different towns and the total number of users select town, count(*) … Read more
I am playing around with MongoDB trying to figure out how to do a simple SELECT province, COUNT(*) FROM contest GROUP BY province But I can’t seem to figure it out using the aggregate function. I can do it using some really weird group syntax db.user.group({ “key”: { “province”: true }, “initial”: { “count”: 0 … Read more
This is pretty simple but I’m at a loss: Given this type of data set: UserInfo(name, metric, day, other_metric) and this sample data set: joe 1 01/01/2011 5 jane 0 01/02/2011 9 john 2 01/03/2011 0 jim 3 01/04/2011 1 jean 1 01/05/2011 3 jill 2 01/06/2011 5 jeb 0 01/07/2011 3 jenn 0 01/08/2011 … Read more
There are plenty of similar questions to be found on here but I don’t think that any answer the question adequately. I’ll continue from the current most popular question and use their example if that’s alright. The task in this instance is to get the latest post for each author in the database. The example … Read more
AM using MySQL 5.7.13 on my windows PC with WAMP Server Here my Problem is While executing this query SELECT * FROM `tbl_customer_pod_uploads` WHERE `load_id` = ’78’ AND `status` = ‘Active’ GROUP BY `proof_type` Am getting always error like this Expression #1 of SELECT list is not in GROUP BY clause and contains nonaggregated column … Read more
If I have a table CREATE TABLE users ( id int(10) unsigned NOT NULL auto_increment, name varchar(255) NOT NULL, profession varchar(255) NOT NULL, employer varchar(255) NOT NULL, PRIMARY KEY (id) ) and I want to get all unique values of profession field, what would be faster (or recommended): SELECT DISTINCT u.profession FROM users u or … Read more
Someone sent me a SQL query where the GROUP BY clause consisted of the statement: GROUP BY 1. This must be a typo right? No column is given the alias 1. What could this mean? Am I right to assume that this must be a typo? 6 Answers 6
This question already has answers here: Pandas ‘count(distinct)’ equivalent (10 answers) Closed 3 years ago. I need to count unique ID values in every domain. I have data: ID, domain 123, ‘vk.com’ 123, ‘vk.com’ 123, ‘twitter.com’ 456, ‘vk.com’ 456, ‘facebook.com’ 456, ‘vk.com’ 456, ‘google.com’ 789, ‘twitter.com’ 789, ‘vk.com’ I try df.groupby([‘domain’, ‘ID’]).count() But I want … Read more
This question already has answers here: GROUP BY / aggregate function confusion in SQL (5 answers) Closed 2 years ago. I got an error – Column ‘Employee.EmpID’ is invalid in the select list because it is not contained in either an aggregate function or the GROUP BY clause. select loc.LocationID, emp.EmpID from Employee as emp … Read more
I am using this data frame: Fruit Date Name Number Apples 10/6/2016 Bob 7 Apples 10/6/2016 Bob 8 Apples 10/6/2016 Mike 9 Apples 10/7/2016 Steve 10 Apples 10/7/2016 Bob 1 Oranges 10/7/2016 Bob 2 Oranges 10/6/2016 Tom 15 Oranges 10/6/2016 Mike 57 Oranges 10/6/2016 Bob 65 Oranges 10/7/2016 Tony 1 Grapes 10/7/2016 Bob 1 Grapes … Read more
