How to count distinct values in a list - sql

Question

I am fairly new to writing queries in Snowflake and have run into a hiccup. I am trying to count how many times an item appears in a list all in the same column.
I was able to use the flatten function and then tried to add in the count function with no luck.
Here is a dummy version of my data:
Ticket# Tasks
1 ["cut apple","peel orange","slice cheese"]
2 ["slice cheese","peel orange"]
3 ["cut apple"]
4 ["cut apple","slice cheese"]
5 ["cut apple", "chop kiwi"]
Here is what I want the output to look like:
(hopefully auto populating the distinct list of tasks in desc order)
Tasks Quantity
Cut Apple 4
Slice Cheese 3
Peel Orange 2
Chop Kiwi 1

Answer 1

Step 1: Define a normalized data schema and put the schema into a database.
Your normalized data schema might look something like this:
Step 2: Add your data
Step 3: Then you will be able to use SQL COUNT with DISTINCT to find the unique rows in your data table(s)
SQL COUNT with DISTINCT

Answer 2

Too long for a comment, but guidance for you to look into, then try to write a sample query. While you have the opportunity to do so while learning, I would look into Data Normalization and adjust your "Tasks" column.
You should have a secondary lookup table that has a primary key ID and a description of each unique task (you'll see in the data normalization). So you can follow along from your data context to the document, I will provide the layout examples and see how that helps you.
Starting with your lookup task table...
Tasks Table
TaskID TaskDescription
1 cut apple
2 peel orange
3 slice cheese
4 chop kiwi
Then, you would have another table that has TicketID, and a third table shows multiple records for each TicketID.
Ticket Table
TicketID ExPurchaseDate
1 someDate
2 sameDate
3 etc...
Now, a detail table per ticket.
TicketTasks Table
TicketTaskID TicketID TaskID
1 1 1
2 1 2
3 1 3
4 2 3
5 2 2
6 3 1
7 4 1
7 4 3
8 5 1
9 5 4
Try to digest this some with the normalization and then look into writing a sql query with COUNT(*) and GROUP BY. More than happy to help you more after, but hope this HELPs guide you some.