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# How can i count the numbers in every subset?

By : user2951357
Date : November 18 2020, 01:01 AM
hop of those help? I have several subset like this , Make a list then use lapply
code :
``````MyList <- list("s1"=s1, "s2"=s2, "s3"=s3)
lapply(MyList,function(x) length(x[x == -4])
``````

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## How can I get both the count of a subset as well as the count of the total set in one query?

By : osmaneralp
Date : March 29 2020, 07:55 AM
this will help In the simple case, assume I have a table that looks like this: , If enabled is always 1 or 0, you can do:
code :
``````SELECT
COUNT(*) as total_count,
SUM(enabled) as enabled_count
FROM widget
``````
``````SELECT
COUNT(*) as total_count,
SUM( CASE WHEN enabled in ('enabled value 1', 'enabled value 2')
THEN 1
ELSE 0
END
) as enabled_count
FROM widget
``````

## Find the largest subset of a set of numbers such that the sum of any 2 numbers is not divisible by a given number

By : Rahul Singh
Date : March 29 2020, 07:55 AM
around this issue Instead of generating all pairs, this could be done by counting modulus.
Time Complexity: O(n + k)

## total count and subset count inside a group by

By : Ilia Shibaev
Date : March 29 2020, 07:55 AM
it should still fix some issue One way to do it is using conditional aggregation:
Create and populate sample table (Please save us this step in your future questions)
code :
``````CREATE TABLE trip
(
id int,
start_date datetime,
end_date datetime
)

INSERT INTO trip VALUES
(7454, '2013-09-01 01:01:00.000', '2013-09-01 01:05:00.000'),
(7457, '2013-09-01 01:09:00.000', '2013-09-01 01:12:00.000'),
(7458, '2013-09-01 02:01:00.000', '2013-09-01 02:08:00.000'),
(7459, '2013-09-01 02:04:00.000', '2013-09-01 02:23:00.000'),
(7460, '2013-09-01 02:04:00.000', '2013-09-01 02:25:00.000'),
(7461, '2013-09-01 02:09:00.000', '2013-09-01 02:12:00.000'),
(7463, '2013-09-01 02:19:00.000', '2013-09-01 02:29:00.000'),
(7465, '2013-09-01 02:27:00.000', '2013-09-01 02:29:00.000'),
(7466, '2013-09-01 04:06:00.000', '2013-09-02 15:08:00.000'),
(7467, '2013-09-01 05:24:00.000', '2013-09-02 05:37:00.000')
``````
``````SELECT  DATEADD(MONTH,DATEDIFF(MONTH,0,[start_date]),0) As TripMonth,
COUNT(id) As 'NumTrips',
SUM
(
CASE WHEN DATEDIFF(DAY, Start_Date, End_Date) > 0 THEN 1 ELSE 0 END
) As 'NumTrips>Day'
FROM Trip
``````
``````TripMonth               NumTrips    NumTrips>Day
01.09.2013 00:00:00     10          2
``````

## Get contiguous subset from given N numbers using less numbers of loop

By : user3583504
Date : March 29 2020, 07:55 AM
fixed the issue. Will look into that further You can't reduce the time complexity as you want to generate all subarrays. However, you can make use of range function in PHP to reduce the nested loops to 2. Note that this doesn't technically reduce the nested length to 2, see range function code but the code looks more idiomatic.
code :
``````<?php

function getContigousSubSet(\$n){
\$result = array();
for(\$i=1; \$i<=\$n;\$i++){
for(\$j=\$i;\$j<=\$n;++\$j){
\$result[] = range(\$i,\$j);
}
}
return \$result;
}

print_r(getContigousSubSet(10));
``````

## Django - Count a subset of related models - Need to annotate count of active Coupons for each Item

By : Luke Gaddie
Date : March 29 2020, 07:55 AM