Haskell List Comprehension

[MGibson019]

Hi

Basically I've got the following code.
- What it does is it takes two input the itemName and the sectionName
- checks if the input is equal to any item in the list (test database) and increment it
- prints the modified list

It does what it needs to do but when I try to increment the amount of sales for apple the output doubles the record of the apple. someone told me that I should try list comprehension but got no idea of how to use it, im a total beginner and would appreciate a step by step explanation :)

Input Code:

items :: [Grocery]
items = [("Water", "Drink Section", 1),
         ("Squash", "Drink Section", 1),
         ("Apple", "Fruit Section", 1),
         ("Plates", "Disposable Section", 1),
         ("Plates", "Ceramic Section", 1)]

sales:: [database] -> String -> String-> [database]
sales db itemName sectionName = []
sales ((item, section, qty): xs) itemName sectionName 
     | item == itemName && section== sectionName = [(item, section, qty + 1)] sales xs itemName sectionName
     | otherwise = sales xs itemName sectionName

Current Output:

[("Apple", "Fruit Section", 2),
 ("Water", "Drink Section", 1),
 ("Squash", "Drink Section", 1), 
 ("Apple", "Fruit Section", 1),
 ("Plates", "Disposable Section", 1),
 ("Plates", "Ceramic Section", 1)]

Thank You!

[sepp2k]

I can't reproduce the behavior you're describing.

Your code doesn't produce the output you say it does - your code doesn't even compile. If I correct the code's syntax error, the result will always be the empty list because you have overlapping cases with the less specific case coming first. If I rearrenge the order of your cases, I get the correct output and no entries are duplicated.

Regarding list comprehensions: In its simplest form a list comprehension has the form [expressionInvolving item | item <- items]. This will evaluate the expression expressionInvolving item for every item in the list items. For example you can use a list comprehension to replace each occurence of the string "foo" with "bar" in a list of strings like this:

strings = ["foo", "lala", "lulu", "foo", "baz"]
newstrings = [if string == "foo" then "bar" else "string" | string <- strings]

Now newstrings will be ["bar", "lala", "lulu", "bar", "baz"]. The same logic can be applied to solve your problem.

Note that in simple cases like this a single call to map can do the same thing as the list comprehension. Also in real code I would move the if-then bit in its own named function for readability.

[MGibson019]

hi again
I was able to create it and get a record just wondering how to get the rest of the list?
thank you!

[sepp2k]

I'm not sure what you mean by that. Both using a list comprehension or fixing the code in your originial post will give you a whole list as the result, not a single record.

[MGibson019]

sales testDb item sec = [ (i,s,q+1) | (i,s,q) <- testDb, i == item && sec == s ]
hi sorry, i'm using list comprehension now and have the code above and it only returns a single record

[sepp2k]

It contains a list with a single record in it. That's because you added a condition to the generator. That's not what I did in my example.

In my example I had a simple generator without a condtion attached to it. I then used if-then-else in the body of the list comprehension to replace matching items and keep all other items unchanged. If you do the same thing, you'll get the result you want.

[MGibson019]

done it sir! :) thanks for your help appreciate it!