new code added and changes

Author: ankurbhambri

consequitive.sql

@@ -12,26 +12,24 @@ WITH cte AS (
         LAST_VALUE(model) OVER (PARTITION BY category ORDER BY price DESC RANGE BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING) AS least_expensive_product_name
     FROM products
 )
-SELECT 
-    category, 
-    model, 
-    price, 
-    expensive_product_name, 
-    least_expensive_product_name 
-FROM cte;
+SELECT * FROM cte;
+
 
 -- Query to find users who have logged in consecutively 3 or more times
 
 SELECT DISTINCT repeated_names
 FROM (
-    SELECT *,
-           -- Check if the current user_name is the same as the next two user_name values
-           CASE 
-               WHEN user_name = LEAD(user_name) OVER (ORDER BY login_id)
-                    AND user_name = LEAD(user_name, 2) OVER (ORDER BY login_id)
-               THEN user_name 
-               ELSE NULL 
-           END AS repeated_names
+    SELECT
+        *,
+        -- Check if the current user_name is the same as the next two user_name values
+        CASE
+            WHEN
+                user_name = LEAD(user_name) OVER (ORDER BY login_id) AND
+                user_name = LEAD(user_name, 2) OVER (ORDER BY login_id)
+            THEN
+                user_name
+            END AS repeated_names
+
     FROM login_details
 ) AS x
 -- Filter out rows where repeated_names is not null

median.sql

@@ -1,20 +1,84 @@
--- median (odd set of numbers) = ((n+1)/2)th term
--- median (even set of numbers) = ((n/2)th term + ((n/2)+1)th term)/2
-
-with cte as (
-    select student_id, sat_writing, rank() over(order by sat_writing) rn
-    from sat_scores
-),
-cte2 as (
-    select 
-    case 
-        when
-            (count(*) % 2) = 0 then (count(*) / 2) + (count(*) / 2) + 1 -- even median
-        else
-            (count(*) + 1 ) / 2 end as cn -- odd median
-    from cte
+/*
+    Median (odd set of numbers) = ((n+1)/2)th term
+
+    Median (even set of numbers) = ((n/2)th term + ((n/2)+1)th term)/2
+
+    The median is also known as the 50th percentile
+
+
+*/
+
+-- Using Window Function
+
+
+-- This is more accurate one in fraction
+
+
+SELECT 
+  PERCENTILE_CONT(0.5) WITHIN GROUP (ORDER BY column_name) AS median
+FROM
+  table_name;
+
+
+-- One more way to calculate median is by using the following query:
+
+SELECT
+	percentile_disc(0.5) WITHIN GROUP (
+	ORDER BY temperature)
+FROM city_data;
+
+-- Because the query used percentile_disc(), the result is a value that exists in the dataset.
+
+
+-- Calculating Multiple Percentiles
+
+
+SELECT
+	device_id,
+	percentile_cont(0.25) WITHIN GROUP(
+ORDER BY
+	humidity) AS percentile_25,
+	percentile_cont(0.50) WITHIN GROUP(
+ORDER BY
+	humidity) AS percentile_50,
+	percentile_cont(0.75) WITHIN GROUP(
+ORDER BY
+	humidity) AS percentile_75,
+	percentile_cont(0.95) WITHIN GROUP(
+ORDER BY
+	humidity) AS percentile_95
+FROM
+	conditions
+GROUP BY
+	device_id; 
+
+
+-- Calculating a Series of Percentiles
+
+SELECT
+	city,
+	percentile,
+	percentile_cont(p) WITHIN GROUP (
+ORDER BY
+	temperature)
+FROM
+	city_data,
+	generate_series(0.01, 1, 0.01) AS percentile
+GROUP BY
+	city, percentile;
+
+
+-- Without Window Function
+
+WITH get_median AS (
+  SELECT
+    y, row_number() OVER(ORDER BY y ASC) AS rn_asc, 
+    COUNT(*) OVER() AS ct
+  FROM dataset
 )
-select cte.student_id, cte2.cn
-from cte, cte2
-where cte.rn=cte2.cn
-order by cte.student_id
+SELECT
+  AVG(y) AS median
+FROM
+  get_median
+WHERE
+  rn_asc BETWEEN ct/2.0 AND ct / 2.0 + 1;