I would like to find public users on Twitter that have 0 followers. I was thinking of using https://developer.twitter.com/en/docs/accounts-and-users/follow-search-get-users/api-reference/get-users-search, but this doesn't have a way to filter by number of followers. Are there any simple alternatives? (Otherwise, I might have to resort to using a graph/search based approach starting from a random point)
Well you didn't specify what library you are using to interact with the Twitter API. But regardless of which technology you're using, the underlying concept is the same. I will use tweepy library in python for my example.
Start by getting the public users using this. The return type is a list of user objects. The user object has several attributes which you can learn about here. For now we are interested in the followers_count attribute. Simply loop through the objects returned and check where the value of this attribute is 0.
Here's how the implementation would look like in python using tweepy library;
search_query = 'your search query here'
get_users = api.search_users(q = search_query)#Returns a list of user objects
for users in get_users:
if users.followers_count ==0:
#Do stuff if the user has 0 followers
Bird SQL by Perplexity AI allows you to do this simply: https://www.perplexity.ai/sql
Query: Users with 0 followers, and 0 following, with at least 5 tweets
SELECT user_url, full_name, followers_count, following_count, tweet_count
FROM users
WHERE (followers_count = 0)
AND (following_count = 0)
AND (tweet_count >= 5)
ORDER BY tweet_count DESC
LIMIT 10
Related
This function suppose to show in which hours adverts are clicked more often.
It works fine however I have problem with sorting it by "HourOfDay". When I add ORDER BY HourOfDay to the end of the query I get en error.
function exportReportToSpreadsheet() {
var spreadsheet = SpreadsheetApp.create('INSERT_REPORT_NAME_HERE');
var report = AdWordsApp.report("SELECT Clicks, Impressions, AverageCpc, HourOfDay FROM ACCOUNT_PERFORMANCE_REPORT DURING LAST_MONTH ORDER BY HourOfDay");
report.exportToSheet(spreadsheet.getActiveSheet());
Logger.log("Report available at " + spreadsheet.getUrl());
}
exportReportToSpreadsheet();
Anyone knows what is wrong with ORDER BY in AdWordsApp.report ?
https://developers.google.com/adwords/scripts/docs/reference/adwordsapp/adwordsapp_report
According to AWQL query language documentation it should work as expected.
https://developers.google.com/adwords/api/docs/guides/awql#using_awql_with_reports
BUG?
You cannot sort reports. From the AWQL documentation:
ORDER BY and LIMIT (sorting and paging) are NOT supported for reports.
Including these clauses in a query will generate an error.
Ordering is only possible when you use the different entities` selectors, e.g. to iterate over campaigns sorted by cost you could do
campaignIterator = AdWordsApp
.campaigns()
.forDateRange("LAST_MONTH")
.orderBy("Clicks DESC");
I'm using a simple follower system in my application and I can get the number of any user's followers by running User.followers.count. However, when I try to sort all users by the number of followers they each have with #orderedUsers = User.all.order("followers.count DESC") it returns the error "ActiveRecord::StatementInvalid: SQLite3::SQLException: no such column: followers.count". Obviously, this is because there is no such column. Is there a way to work around this to do what I wish to achieve?
Thanks.
This code should work (assuming the DB table names are users and followers):
User.joins(:followers).order("count(followers.user_id) desc")
How about something like:
#ordered_users = User.all.sort{|a,b| a.followers.count <=> b.followers.count}
For the reverse order, you can do:
#ordered_users = User.all.sort{|a,b| b.followers.count <=> a.followers.count}
Or, .reverse, as you say in the comments.
EDIT: #Alex Quach left a good alternative in a different post. I've modified it for where it will not include the current user in the list, which may be helpful:
User.all.where('id != ?', current_user.id).sort_by { |u| -u.followers.count }
I would strongly consider using a counter cache on the User model, to hold the count of followers.
This would give a very small performance impact on adding or removing followers, and greatly increase performance when performing sorts:
User.order(followers_count: :desc)
This would be particularly noticeable if you wanted the top-n users by follower count, or finding users with no followers.
I have framed query to submit to solr which is of following format.
id:95154 OR id:68209 OR id:89482 OR id:94233 OR id:112481 OR id:93843
i want to get records according to order from starting. say i need to get document with id 95154 document first then id 68209 next and so on. but its not happening right now its giving last id 93843 first and some times random.i am using solr in grails 2.1 and my solr version is 1.4.0. here is sample way i am getting documents from solr
def server = solrService.getServer('provider')
SolrQuery sponsorSolrQuery = new SolrQuery(solarQuery)
def queryResponse = server.query(sponsorSolrQuery);
documentsList = queryResponse.getResults()
As #injecteer mentions, there is nothing built-in to Lucene to consider the sequence of clauses in a boolean query, but:
You are able to apply boosts to each term, and as long as the field is a basic field (meaning, not a TextField), the boosts will apply cleanly to give you a decent sort by score.
id:95154^6 OR id:68209^5 OR id:89482^4 OR id:94233^3 OR id:112481^2 OR id:93843
there's no such thing in Lucene (I strongly assume, that in Solr as well). In Lucene you can sort the results based on contents of documents' fields, but not on the order of clauses in a query.
that means, that you have to sort the results yourself:
documentsList = queryResponse.getResults()
def sordedByIdOrder = solarQueryAsList.collect{ id -> documentList.find{ it.id == id } }
Does it possible to have an order by "property" with a where clause and now the "index/position" of the result?
I mean, when using order for sorting we need to be able to know the position of the result in the sort.
Imagine a scoreboard with 1 million user node, i do an order by on user node.score with a where "name = user_name" and i wan't to know the current rank of the user. I do not find how to do this using order by ...
start game=node(1)
match game-[:has_child_user]->user
with user
order by user.score
with user
where user.name = "my_user"
return user , "the position in the sort";
the expected result would be :
node_user | rank
(i don't want to fetch one million entries at client side to know the current rank/position of a node in the ORDER BY!)
This functionality does not exist today in Cypher. Do you have an example of what this would look like in SQL? Would the below be something that fits the bill? (just a sketch, not working!)
(your code)
start game=node(1)
match game-[:has_child_user]->user
with user
order by user.score
(+ this code)
with user, index() as rank
return user.name, rank;
If you have more thoughts or want to start hacking on this please open an issue at https://github.com/neo4j/neo4j/issues
For the time being there is a work around that you can do:
start n=node(0),rank_node=node(1)
match n-[r:rank]->rn
where rn.score <= rank_node.score
return rank_node,count(*) as pos;
For live example see: http://console.neo4j.org/?id=bela20
I am not sure if I am going about this the best way, but I will try to explain what I am trying to do.
I have the following domain classes
class User {
static hasMany = [goals: Goal]
}
So each User has a list of Goal objects. I want to be able to take an instance of User and return 5 Users with the highest number of matching Goal objects (with the instance) in their goals list.
Can someone kindly explain how I might go about doing this?
The easiest and most efficient way to achieve this is using plain SQL. Assuming you have these tables
users [id]
goals [id, description]
user_goals [user_id, goal_id]
You can have the following query to do what you need:
set #userId=123;
select user_id, count(*) as matched from user_goals
where user_id!=#userId
and goal_id in (select ug.goal_id from user_goals ug where ug.user_id=#userId)
group by user_id order by matched desc limit 5;
This takes a user id and returns a list of other users with matching goals, sorted by the number of matches. Wrap it up in a GoalService and you're done!
class GoalService {
def findUsersWithSimilarGoals(user) {
// ...
}
}
It may also be possible to do this with criteria or HQL, but with queries like this it's usually easier to use SQL.
If you're looking for a simple match, perhaps the easiest way would be to do a findAll for each Goal and then count the number of results that each other User appears in:
Map user2Count = [:]
for (goal in myUser.goals){
for (u in User.findAllByGoal(goal)){
def count = user2Count.containsKey(u) ? user2Count.get(u) : 0
count++
user2Count.put(u, count)
}
}
// get the top 5 users
def topUsers = user2Count.entrySet().sort({ it.value }).reverse()[0..5]
This may be too slow, depending on your needs, but it is simple. If many users share the same goals then you could cache the results of findAllByGoal.